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9.E: Hypothesis Testing with One Sample (Exercises)

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These are homework exercises to accompany the Textmap created for "Introductory Statistics" by OpenStax.

9.1: Introduction

9.2: null and alternative hypotheses.

Some of the following statements refer to the null hypothesis, some to the alternate hypothesis.

State the null hypothesis, $H_{0}$, and the alternative hypothesis. $H_{a}$, in terms of the appropriate parameter $(\mu \text{or} p)$.

The mean number of years Americans work before retiring is 34.
At most 60% of Americans vote in presidential elections.
The mean starting salary for San Jose State University graduates is at least $100,000 per year.
Twenty-nine percent of high school seniors get drunk each month.
Fewer than 5% of adults ride the bus to work in Los Angeles.
The mean number of cars a person owns in her lifetime is not more than ten.
About half of Americans prefer to live away from cities, given the choice.
Europeans have a mean paid vacation each year of six weeks.
The chance of developing breast cancer is under 11% for women.
Private universities' mean tuition cost is more than $20,000 per year.
$H_{0}: \mu = 34; H_{a}: \mu \neq 34$
$H_{0}: p \leq 0.60; H_{a}: p > 0.60$
$H_{0}: \mu \geq 100,000; H_{a}: \mu < 100,000$
$H_{0}: p = 0.29; H_{a}: p \neq 0.29$
$H_{0}: p = 0.05; H_{a}: p < 0.05$
$H_{0}: \mu \leq 10; H_{a}: \mu > 10$
$H_{0}: p = 0.50; H_{a}: p \neq 0.50$
$H_{0}: \mu = 6; H_{a}: \mu \neq 6$
$H_{0}: p ≥ 0.11; H_{a}: p < 0.11$
$H_{0}: \mu \leq 20,000; H_{a}: \mu > 20,000$

Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than thirty percent of the teen girls smoke to stay thin? The alternative hypothesis is:

$p < 0.30$
$p \leq 0.30$
$p \geq 0.30$
$p > 0.30$

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is:

$p = 0.20$
$p > 0.20$
$p < 0.20$
$p \leq 0.20$

Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are:

$H_{0}: \bar{x} = 4.5, H_{a}: \bar{x} > 4.5$
$H_{0}: \mu \geq 4.5, H_{a}: \mu < 4.5$
$H_{0}: \mu = 4.75, H_{a}: \mu > 4.75$
$H_{0}: \mu = 4.5, H_{a}: \mu > 4.5$

9.3: Outcomes and the Type I and Type II Errors

State the Type I and Type II errors in complete sentences given the following statements.

The mean number of cars a person owns in his or her lifetime is not more than ten.
Private universities mean tuition cost is more than $20,000 per year.
Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years.
Type I error: We conclude that more than 60% of Americans vote in presidential elections, when the actual percentage is at most 60%.Type II error: We conclude that at most 60% of Americans vote in presidential elections when, in fact, more than 60% do.
Type I error: We conclude that the mean starting salary is less than $100,000, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000.
Type I error: We conclude that the proportion of high school seniors who get drunk each month is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who get drunk each month is 29% when, in fact, it is not 29%.
Type I error: We conclude that fewer than 5% of adults ride the bus to work in Los Angeles, when the percentage that do is really 5% or more. Type II error: We conclude that 5% or more adults ride the bus to work in Los Angeles when, in fact, fewer that 5% do.
Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10.
Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half.
Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not.
Type I error: We conclude that the proportion is less than 11%, when it is really at least 11%. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11%, when in fact it is less than 11%.
Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000.

For statements a-j in Exercise 9.109 , answer the following in complete sentences.

State a consequence of committing a Type I error.
State a consequence of committing a Type II error.

When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is “the drug is unsafe.” What is the Type II Error?

To conclude the drug is safe when in, fact, it is unsafe.
Not to conclude the drug is safe when, in fact, it is safe.
To conclude the drug is safe when, in fact, it is safe.
Not to conclude the drug is unsafe when, in fact, it is unsafe.

A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________.

at least 20%, when in fact, it is less than 20%.
20%, when in fact, it is 20%.
less than 20%, when in fact, it is at least 20%.
less than 20%, when in fact, it is less than 20%.

It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5%, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average?

The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours

is more than seven hours.
is at most seven hours.
is at least seven hours.
is less than seven hours.

to conclude that the current mean hours per week is higher than 4.5, when in fact, it is higher
to conclude that the current mean hours per week is higher than 4.5, when in fact, it is the same
to conclude that the mean hours per week currently is 4.5, when in fact, it is higher
to conclude that the mean hours per week currently is no higher than 4.5, when in fact, it is not higher

9.4: Distribution Needed for Hypothesis Testing

$N\left(7.24, \frac{1.93}{\sqrt{22}}\right)$
$N\left(7.24, 1.93\right)$

9.5: Rare Events, the Sample, Decision and Conclusion

The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.

Is this a test of one mean or proportion?
State the null and alternative hypotheses. $H_{0}$ : ____________________ $H_{a}$ : ____________________
Is this a right-tailed, left-tailed, or two-tailed test?
What symbol represents the random variable for this test?
In words, define the random variable for this test.
$x =$ ________________
$n =$ ________________
$p′ =$ _____________
Calculate $\sigma_{x} =$ __________. Show the formula set-up.
State the distribution to use for the hypothesis test.
Find the $p\text{-value}$.
Reason for the decision:
Conclusion (write out in a complete sentence):

9.6: Additional Information and Full Hypothesis Test Examples

For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in [link] . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files.

If you are using a Student's $t$ - distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.)

A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using $\alpha = 0.05$, is the data highly inconsistent with the claim?

$H_{0}: \mu \geq 50,000$
$H_{a}: \mu < 50,000$
Let $\bar{X} =$ the average lifespan of a brand of tires.
normal distribution
$z = -2.315$
$p\text{-value} = 0.0103$
Check student’s solution.
alpha: 0.05
Decision: Reject the null hypothesis.
Reason for decision: The $p\text{-value}$ is less than 0.05.
Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles.
$(43,537, 49,463)$

From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?

The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1% level?

$H_{0}: \mu = $1.00$
$H_{a}: \mu \neq $1.00$
Let $\bar{X} =$ the average cost of a daily newspaper.
$z = –0.866$
$p\text{-value} = 0.3865$
$\alpha: 0.01$
Decision: Do not reject the null hypothesis.
Reason for decision: The $p\text{-value}$ is greater than 0.01.
Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1.
$($0.84, $1.06)$

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

The mean number of sick days an employee takes per year is believed to be about ten. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let $x =$ the number of sick days they took for the past year. Should the personnel team believe that the mean number is ten?

$H_{0}: \mu = 10$
$H_{a}: \mu \neq 10$
Let $\bar{X}$ the mean number of sick days an employee takes per year.
Student’s t -distribution
$t = –1.12$
$p\text{-value} = 0.300$
$\alpha: 0.05$
Reason for decision: The $p\text{-value}$ is greater than 0.05.
Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten.
$(4.9443, 11.806)$

In 1955, Life Magazine reported that the 25 year-old mother of three worked, on average, an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level?

Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think?

$H_{0}: p \geq 0.6$
$H_{a}: p < 0.6$
Let $P′ =$ the proportion of students who feel more enriched as a result of taking Elementary Statistics.
normal for a single proportion
$p\text{-value} = 0.1308$
Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched.

The “plus-4s” confidence interval is $(0.411, 0.648)$

A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5; 4; 7; 3; 6; 4; 5; 3; 6; 3; 8; 5. Conduct a hypothesis test of your belief.

Refer to Exercise 9.119 . Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four.

$H_{0}: \mu = 4$
$H_{a}: \mu \neq 4$
Let $\bar{X}$ the average I.Q. of a set of brown trout.
two-tailed Student's t-test
$t = 1.95$
$p\text{-value} = 0.076$
Reason for decision: The $p\text{-value}$ is greater than 0.05
Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four.
$(3.8865,5.9468)$

According to an article in Newsweek , the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7% girls). Suppose you don’t believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7?

A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results.

$H_{a}: p < 0.13$
Let $P′ =$ the proportion of Americans who have seen or sensed angels
–2.688
$p\text{-value} = 0.0036$
Reason for decision: The $p\text{-value}$e is less than 0.05.
Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%.

The“plus-4s” confidence interval is (0.0022, 0.0978)

The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours?

Data (length of mean work week): 70; 45; 55; 60; 65; 55; 55; 60; 50; 55.

Use the “Lap time” data for Lap 4 (see [link] ) to test the claim that Terri finishes Lap 4, on average, in less than 129 seconds. Use all twenty races given.

$H_{0}: \mu \geq 129$
$H_{a}: \mu < 129$
Let $\bar{X} =$ the average time in seconds that Terri finishes Lap 4.
Student's t -distribution
$t = 1.209$
Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds.
$(128.63, 130.37)$

Use the “Initial Public Offering” data (see [link] ) to test the claim that the mean offer price was $18 per share. Do not use all the data. Use your random number generator to randomly survey 15 prices.

The following questions were written by past students. They are excellent problems!

"Asian Family Reunion," by Chau Nguyen

Every two years it comes around.

We all get together from different towns.

In my honest opinion,

It's not a typical family reunion.

Not forty, or fifty, or sixty,

But how about seventy companions!

The kids would play, scream, and shout

One minute they're happy, another they'll pout.

The teenagers would look, stare, and compare

From how they look to what they wear.

The men would chat about their business

That they make more, but never less.

Money is always their subject

And there's always talk of more new projects.

The women get tired from all of the chats

They head to the kitchen to set out the mats.

Some would sit and some would stand

Eating and talking with plates in their hands.

Then come the games and the songs

And suddenly, everyone gets along!

With all that laughter, it's sad to say

That it always ends in the same old way.

They hug and kiss and say "good-bye"

And then they all begin to cry!

I say that 60 percent shed their tears

But my mom counted 35 people this year.

She said that boys and men will always have their pride,

So we won't ever see them cry.

I myself don't think she's correct,

So could you please try this problem to see if you object?

$H_{0}: p = 0.60$
$H_{a}: p < 0.60$
Let $P′ =$ the proportion of family members who shed tears at a reunion.
–1.71
Reason for decision: $p\text{-value} < \alpha$
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the $p\text{-value}$ and alpha are quite close, so other tests should be done.
We are 95% confident that between 38.29% and 61.71% of family members will shed tears at a family reunion. $(0.3829, 0.6171)$. The“plus-4s” confidence interval (see chapter 8) is $(0.3861, 0.6139)$

Note that here the “large-sample” $1 - \text{PropZTest}$ provides the approximate $p\text{-value}$ of 0.0438. Whenever a $p\text{-value}$ based on a normal approximation is close to the level of significance, the exact $p\text{-value}$ based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course.

"The Problem with Angels," by Cyndy Dowling

Although this problem is wholly mine,

The catalyst came from the magazine, Time.

On the magazine cover I did find

The realm of angels tickling my mind.

Inside, 69% I found to be

In angels, Americans do believe.

Then, it was time to rise to the task,

Ninety-five high school and college students I did ask.

Viewing all as one group,

Random sampling to get the scoop.

So, I asked each to be true,

"Do you believe in angels?" Tell me, do!

Hypothesizing at the start,

Totally believing in my heart

That the proportion who said yes

Would be equal on this test.

Lo and behold, seventy-three did arrive,

Out of the sample of ninety-five.

Now your job has just begun,

Solve this problem and have some fun.

"Blowing Bubbles," by Sondra Prull

Studying stats just made me tense,

I had to find some sane defense.

Some light and lifting simple play

To float my math anxiety away.

Blowing bubbles lifts me high

Takes my troubles to the sky.

POIK! They're gone, with all my stress

Bubble therapy is the best.

The label said each time I blew

The average number of bubbles would be at least 22.

I blew and blew and this I found

From 64 blows, they all are round!

But the number of bubbles in 64 blows

Varied widely, this I know.

20 per blow became the mean

They deviated by 6, and not 16.

From counting bubbles, I sure did relax

But now I give to you your task.

Was 22 a reasonable guess?

Find the answer and pass this test!

$H_{0}: \mu \geq 22$
$H_{a}: \mu < 22$
Let $\bar{X} =$ the mean number of bubbles per blow.
–2.667
$p\text{-value} = 0.00486$
Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22.
$(18.501, 21.499)$

"Dalmatian Darnation," by Kathy Sparling

A greedy dog breeder named Spreckles

Bred puppies with numerous freckles

The Dalmatians he sought

Possessed spot upon spot

The more spots, he thought, the more shekels.

His competitors did not agree

That freckles would increase the fee.

They said, “Spots are quite nice

But they don't affect price;

One should breed for improved pedigree.”

The breeders decided to prove

This strategy was a wrong move.

Breeding only for spots

Would wreak havoc, they thought.

His theory they want to disprove.

They proposed a contest to Spreckles

Comparing dog prices to freckles.

In records they looked up

One hundred one pups:

Dalmatians that fetched the most shekels.

They asked Mr. Spreckles to name

An average spot count he'd claim

To bring in big bucks.

Said Spreckles, “Well, shucks,

It's for one hundred one that I aim.”

Said an amateur statistician

Who wanted to help with this mission.

“Twenty-one for the sample

Standard deviation's ample:

They examined one hundred and one

Dalmatians that fetched a good sum.

They counted each spot,

Mark, freckle and dot

And tallied up every one.

Instead of one hundred one spots

They averaged ninety six dots

Can they muzzle Spreckles’

Obsession with freckles

Based on all the dog data they've got?

"Macaroni and Cheese, please!!" by Nedda Misherghi and Rachelle Hall

As a poor starving student I don't have much money to spend for even the bare necessities. So my favorite and main staple food is macaroni and cheese. It's high in taste and low in cost and nutritional value.

One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02 !!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook:

Price per box of Mac and Cheese:

5 stores @ $2.02
15 stores @ $0.25
3 stores @ $1.29
6 stores @ $0.35
4 stores @ $2.27
7 stores @ $1.50
5 stores @ $1.89
8 stores @ 0.75.

I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. (After all, as a poor starving student I can't be expected to feed our class of animals!)

$H_{0}: \mu \leq 1$
$H_{a}: \mu > 1$
Let $\bar{X} =$ the mean cost in dollars of macaroni and cheese in a certain town.
Student's $t$-distribution
$t = 0.340$
$p\text{-value} = 0.36756$
Conclusion: The mean cost could be $1, or less. At the 5% significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1.
$(0.8291, 1.241)$

"William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi

THE CHARACTERS (in order of appearance):

HAMLET, Prince of Denmark and student of Statistics
POLONIUS, Hamlet’s tutor
HOROTIO, friend to Hamlet and fellow student

Scene: The great library of the castle, in which Hamlet does his lessons

(The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage.)

POLONIUS: My Lord, how cans’t thou admit that thou hast seen a ghost! It is but a figment of your imagination!

HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable.

POLONIUS If thou doest insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What piece of work is man, how noble in reason, how infinite in faculties? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true.

HORATIO (to Hamlet): What should we do, my Lord?

HAMLET: Go to thy purpose, Horatio.

HORATIO: To what end, my Lord?

HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no.

(Horatio exits, followed by Polonius, leaving Hamlet to ponder alone.)

(The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters.)

POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations?

HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes.

POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is not our test. (Horatio and Polonius leave together.)

HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.)

(Curtain falls)

"Untitled," by Stephen Chen

I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1%.

So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right?

$H_{0}: p = 0.01$
$H_{a}: p > 0.01$
Let $P′ =$ the proportion of errors generated
Normal for a single proportion
Decision: Reject the null hypothesis
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01.

The“plus-4s” confidence interval is $(0.004, 0.144)$.

"Japanese Girls’ Names"

by Kumi Furuichi

It used to be very typical for Japanese girls’ names to end with “ko.” (The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation.) “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters which have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on.

However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names which end with “ko.” More and more, parents seem to have become creative, modernized, and, sometimes, westernized in naming their children.

I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, co-workers, and acquaintances’ names that I could remember. Following are the names. (Some are repeats.) Test to see if the proportion has dropped for this generation.

Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko.

"Phillip’s Wish," by Suzanne Osorio

My nephew likes to play

Chasing the girls makes his day.

He asked his mother

If it is okay

To get his ear pierced.

She said, “No way!”

To poke a hole through your ear,

Is not what I want for you, dear.

He argued his point quite well,

Says even my macho pal, Mel,

Has gotten this done.

It’s all just for fun.

C’mon please, mom, please, what the hell.

Again Phillip complained to his mother,

Saying half his friends (including their brothers)

Are piercing their ears

And they have no fears

He wants to be like the others.

She said, “I think it’s much less.

We must do a hypothesis test.

And if you are right,

I won’t put up a fight.

But, if not, then my case will rest.”

We proceeded to call fifty guys

To see whose prediction would fly.

Nineteen of the fifty

Said piercing was nifty

And earrings they’d occasionally buy.

Then there’s the other thirty-one,

Who said they’d never have this done.

So now this poem’s finished.

Will his hopes be diminished,

Or will my nephew have his fun?

$H_{0}: p = 0.50$
$H_{a}: p < 0.50$
Let $P′ =$ the proportion of friends that has a pierced ear.
–1.70
$p\text{-value} = 0.0448$
Reason for decision: The $p\text{-value}$ is less than 0.05. (However, they are very close.)
Conclusion: There is sufficient evidence to support the claim that less than 50% of his friends have pierced ears.
Confidence Interval: $(0.245, 0.515)$: The “plus-4s” confidence interval is $(0.259, 0.519)$.

"The Craven," by Mark Salangsang

Once upon a morning dreary

In stats class I was weak and weary.

Pondering over last night’s homework

Whose answers were now on the board

This I did and nothing more.

While I nodded nearly napping

Suddenly, there came a tapping.

As someone gently rapping,

Rapping my head as I snore.

Quoth the teacher, “Sleep no more.”

“In every class you fall asleep,”

The teacher said, his voice was deep.

“So a tally I’ve begun to keep

Of every class you nap and snore.

The percentage being forty-four.”

“My dear teacher I must confess,

While sleeping is what I do best.

The percentage, I think, must be less,

A percentage less than forty-four.”

This I said and nothing more.

“We’ll see,” he said and walked away,

And fifty classes from that day

He counted till the month of May

The classes in which I napped and snored.

The number he found was twenty-four.

At a significance level of 0.05,

Please tell me am I still alive?

Or did my grade just take a dive

Plunging down beneath the floor?

Upon thee I hereby implore.

Toastmasters International cites a report by Gallop Poll that 40% of Americans fear public speaking. A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test to determine if the percent at her school is less than 40%.

$H_{0}: p = 0.40$
$H_{a}: p < 0.40$
Let $P′ =$ the proportion of schoolmates who fear public speaking.
–1.01
$p\text{-value} = 0.1563$
Conclusion: There is insufficient evidence to support the claim that less than 40% of students at the school fear public speaking.
Confidence Interval: $(0.3241, 0.4240)$: The “plus-4s” confidence interval is $(0.3257, 0.4250)$.

Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California, was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data.

According to an article in Bloomberg Businessweek , New York City's most recent adult smoking rate is 14%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 14% or if it has decreased.

$H_{0}: p = 0.14$
$H_{a}: p < 0.14$
Let $P′ =$ the proportion of NYC residents that smoke.
–0.2756
$p\text{-value} = 0.3914$
At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14.
Confidence Interval: $(0.0502, 0.2070)$: The “plus-4s” confidence interval (see chapter 8) is $(0.0676, 0.2297)$.

The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test.

Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test.

$H_{0}: \mu = 69,110$
$H_{0}: \mu > 69,110$
Let $\bar{X} =$ the mean salary in dollars for California registered nurses.
$t = 1.719$
$p\text{-value}: 0.0466$
Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110.
$($68,757, $73,485)$

La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the U.S. is less than four years old.

After conducting the test, your decision and conclusion are

Reject $H_{0}$: There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
Do not reject $H_{0}$: There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.
Do not reject $H_{0}$: There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin.
Reject $H_{0}$: There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin.

At a 1% level of significance, an appropriate conclusion is:

There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20%.
There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20%.
There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20%.

At a significance level of $a = 0.05$, what is the correct conclusion?

There is enough evidence to conclude that the mean number of hours is more than 4.75
There is enough evidence to conclude that the mean number of hours is more than 4.5
There is not enough evidence to conclude that the mean number of hours is more than 4.5
There is not enough evidence to conclude that the mean number of hours is more than 4.75

Instructions: For the following ten exercises,

Hypothesis testing: For the following ten exercises, answer each question.

State the null and alternate hypothesis.

State the $p\text{-value}$.

State $\alpha$.

What is your decision?

Write a conclusion.

Answer any other questions asked in the problem.

According to the Center for Disease Control website, in 2011 at least 18% of high school students have smoked a cigarette. An Introduction to Statistics class in Davies County, KY conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions.

A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?

$H_{0}: p = 0.488$ $H_{a}: p \neq 0.488$
$p\text{-value} = 0.0114$
$\alpha = 0.05$
Reject the null hypothesis.
At the 5% level of significance, there is enough evidence to conclude that 48.8% of families own stocks.
The survey does not appear to be accurate.

Driver error can be listed as the cause of approximately 54% of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using $\alpha = 0.05$, is the AAA proportion accurate?

The US Department of Energy reported that 51.7% of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the $\alpha = 0.05$ level in Kentucky? Are the results applicable across the country? Why?

$H_{0}: p = 0.517$ $H_{0}: p \neq 0.517$
$p\text{-value} = 0.9203$.
$\alpha = 0.05$.
Do not reject the null hypothesis.
At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517.
However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation.

For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use $\alpha = 0.01$ level of significance. What is the possible proportion of patrons that do borrow books from the Owensboro Library?

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the $\alpha = 0.05 level$, can it be concluded that the mean rainfall was below the reported average? What if $\alpha = 0.01$? Assume the amount of summer rainfall follows a normal distribution.

$H_{0}: \mu \geq 11.52$ $H_{a}: \mu < 11.52$
$p\text{-value} = 0.000002$ which is almost 0.
At the 5% significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average.
We would make the same conclusion if alpha was 1% because the $p\text{-value}$ is almost 0.

A survey in the N.Y. Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the $\alpha = 0.10$ level, is the Austin, TX commute significantly less than the mean commute time for the 15 largest US cities?

A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals

3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1

At the $\alpha = 0.05$ level can it be concluded that the sample mean is higher than 5.8 visits per year?

$H_{0}: \mu \leq 5.8$ $H_{a}: \mu > 5.8$
$p\text{-value} = 0.9987$
At the 5% level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year.

According to the N.Y. Times Almanac the mean family size in the U.S. is 3.18. A sample of a college math class resulted in the following family sizes:

5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2

At $\alpha = 0.05$ level, is the class’ mean family size greater than the national average? Does the Almanac result remain valid? Why?

The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct?

$H_{0}: \mu \geq 150$ $H_{0}: \mu < 150$
$p\text{-value} = 0.0622$
$\alpha = 0.01$
At the 1% significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average.
The student academic group’s claim appears to be correct.

9.7: Hypothesis Testing of a Single Mean and Single Proportion

9.4 Full Hypothesis Test Examples

Tests on means, example 9.8.

Jeffrey, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds . His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims . For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal.

Set up the Hypothesis Test:

Since the problem is about a mean, this is a test of a single population mean .

H 0 : μ = 16.43 H a : μ < 16.43

For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed.

Determine the distribution needed:

Random variable: X ¯ X ¯ = the mean time to swim the 25-yard freestyle.

Distribution for the test: X ¯ X ¯ is normal (population standard deviation is known: σ = 0.8)

X ¯ ~ N ( μ , σ X n ) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) X ¯ ~ N ( 16.43 , 0.8 15 )

μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15.

Calculate the p -value using the normal distribution for a mean:

p -value = P ( x ¯ x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16.

p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16.

μ = 16.43 comes from H 0 . Our assumption is μ = 16.43.

Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.

Compare α and the p -value:

α = 0.05 p -value = 0.0187 α > p -value

Make a decision: Since α > α > p -value, reject H 0 .

This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds.

Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles.

The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in at least 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.)

The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.)

The mean throwing distance of a football for Marco, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal.

First, determine what type of test this is, set up the hypothesis test, find the p -value, sketch the graph, and state your conclusion.

Example 9.9

Jasmine has just begun her new job on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of 108 dollars with a standard deviation of 12 dollars. Test at 5% significance that the population mean is at least 100 dollars against the alternative that it is less than 100 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jasmine has met this requirement at the significance level of 95%?

H 0 : µ ≤ 100 H a : µ > 100 The null and alternative hypothesis are for the parameter µ because the number of dollars of the contracts is a continuous random variable. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not "too high" a number. This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis.
Test statistic: t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67 t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67
Critical value: t a = 1.753 t a = 1.753 with n-1 degrees of freedom= 15

The test statistic is a Student's t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t t ( t a ) ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jasmine's performance meets company standards.

It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, state your conclusion, and identify the Type I errors.

Example 9.10

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, s 2 s 2 . Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of significance in the analysis should be 99%.

Again we will follow the steps in our analysis of this problem.

STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus:

STEP 2 : Decide the level of significance and draw the graph showing the critical value.

This problem has already set the level of significance at 99%. The decision seems an appropriate one and shows the thought process when setting the significance level. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table or the t-table at 0.005 column and infinite degrees of freedom. We draw the graph and mark these points.

STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph.

STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis.

STEP 5 : Reach a Conclusion

Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of significance we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of significance we conclude that the machine is under filling the bottles and is in need of repair”.

Try It 9.10

A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2).

Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 .

Hypothesis Test for Proportions

Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p .

The population parameter for the binomial is p . The estimated value (point estimate) for p is p′ where p′ = x/n , x is the number of successes in the sample and n is the sample size.

When you perform a hypothesis test of a population proportion p , you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five ( np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = np μ = np and σ = npq σ = npq . Remember that q = 1 – p q = 1 – p . There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p .

Again, we begin with the standardizing formula modified because this is the distribution of a binomial.

Substituting p 0 p 0 , the hypothesized value of p , we have:

This is the test statistic for testing hypothesized values of p , where the null and alternative hypotheses take one of the following forms:

The decision rule stated above applies here also: if the calculated value of Z c shows that the sample proportion is "too many" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is "too many" is pre-determined by the analyst depending on the level of significance required in the test.

Example 9.11

The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50% . They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance.

STEP 1 : Set the null and alternative hypothesis.

H 0 : p = 0.50 H a : p ≠ 0.50

The words "is the same or different from" tell you this is a two-tailed test. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of borrowers is different from 50% when, in fact, the proportion is actually 50%. (Reject the null hypothesis when the null hypothesis is true). The Type II error is there is not enough evidence to conclude that the proportion of first time borrowers differs from 50% when, in fact, the proportion does differ from 50%. (You fail to reject the null hypothesis when the null hypothesis is false.)

STEP 2 : Decide the level of significance and draw the graph showing the critical value

The level of significance has been set by the problem at the 5% level. Because this is two-tailed test one-half of the alpha value will be in the upper tail and one-half in the lower tail as shown on the graph. The critical value for the normal distribution at the 95% level of confidence is 1.96. This can easily be found on the student’s t-table at the very bottom at infinite degrees of freedom remembering that at infinity the t-distribution is the normal distribution. Of course the value can also be found on the normal table but you have go looking for one-half of 95 (0.475) inside the body of the table and then read out to the sides and top for the number of standard deviations.

STEP 3 : Calculate the sample parameters and critical value of the test statistic.

The test statistic is a normal distribution, Z, for testing proportions and is:

For this case, the sample of 100 found 53 of these loans were smaller than those of other borrowers. The sample proportion, p′ = 53/100= 0.53 The test question, therefore, is : “Is 0.53 significantly different from .50?” Putting these values into the formula for the test statistic we find that 0.53 is only 0.60 standard deviations away from .50. This is barely off of the mean of the standard normal distribution of zero. There is virtually no difference from the sample proportion and the hypothesized proportion in terms of standard deviations.

STEP 4 : Compare the test statistic and the critical value.

The calculated value is well within the critical values of ± 1.96 standard deviations and thus we cannot reject the null hypothesis. To reject the null hypothesis we need significant evident of difference between the hypothesized value and the sample value. In this case the sample value is very nearly the same as the hypothesized value measured in terms of standard deviations.

STEP 5 : Reach a conclusion

The formal conclusion would be “At a 5% level of significance we cannot reject the null hypothesis that 50% of first-time borrowers take out smaller loans than other borrowers.” Notice the length to which the conclusion goes to include all of the conditions that are attached to the conclusion. Statisticians, for all the criticism they receive, are careful to be very specific even when this seems trivial. Statisticians cannot say more than they know, and the data constrain the conclusion to be within the metes and bounds of the data.

Try It 9.11

A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.

Example 9.12

Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones.

Here is an abbreviate version of the system to solve hypothesis tests applied to a test on a proportions.

Try It 9.12

Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors.

Example 9.13

The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.

1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98; 1.02; .95; .95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05.

Let’s follow a four-step process to answer this statistical question.

H 0 : μ ≤ 1
H a : μ > 1
Plan : We are testing a sample mean without a known population standard deviation with less than 30 observations. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal.
Do the calculations and draw the graph .
State the Conclusions : We cannot accept the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one.

Try It 9.13

The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows:

205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205

Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal.

Example 9.14

In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users (the rate of brain cancer for non-cell phone users is 0.0340%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

H 0 : p ≤ 0.00034
H a : p > 0.00034

If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.

We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np' = 420,019(0.00034) = 142.8, nq' = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p' = 0.00034. Thus we will be able to generalize our results to the population.

Try It 9.14

In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.

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S.3 hypothesis testing.

In reviewing hypothesis tests, we start first with the general idea. Then, we keep returning to the basic procedures of hypothesis testing, each time adding a little more detail.

The general idea of hypothesis testing involves:

Making an initial assumption.
Collecting evidence (data).
Based on the available evidence (data), deciding whether to reject or not reject the initial assumption.

Every hypothesis test — regardless of the population parameter involved — requires the above three steps.

Example S.3.1

Is normal body temperature really 98.6 degrees f section .

Consider the population of many, many adults. A researcher hypothesized that the average adult body temperature is lower than the often-advertised 98.6 degrees F. That is, the researcher wants an answer to the question: "Is the average adult body temperature 98.6 degrees? Or is it lower?" To answer his research question, the researcher starts by assuming that the average adult body temperature was 98.6 degrees F.

Then, the researcher went out and tried to find evidence that refutes his initial assumption. In doing so, he selects a random sample of 130 adults. The average body temperature of the 130 sampled adults is 98.25 degrees.

Then, the researcher uses the data he collected to make a decision about his initial assumption. It is either likely or unlikely that the researcher would collect the evidence he did given his initial assumption that the average adult body temperature is 98.6 degrees:

If it is likely , then the researcher does not reject his initial assumption that the average adult body temperature is 98.6 degrees. There is not enough evidence to do otherwise.
either the researcher's initial assumption is correct and he experienced a very unusual event;
or the researcher's initial assumption is incorrect.

In statistics, we generally don't make claims that require us to believe that a very unusual event happened. That is, in the practice of statistics, if the evidence (data) we collected is unlikely in light of the initial assumption, then we reject our initial assumption.

Example S.3.2

Criminal trial analogy section .

One place where you can consistently see the general idea of hypothesis testing in action is in criminal trials held in the United States. Our criminal justice system assumes "the defendant is innocent until proven guilty." That is, our initial assumption is that the defendant is innocent.

In the practice of statistics, we make our initial assumption when we state our two competing hypotheses -- the null hypothesis ( H 0 ) and the alternative hypothesis ( H A ). Here, our hypotheses are:

H 0 : Defendant is not guilty (innocent)
H A : Defendant is guilty

In statistics, we always assume the null hypothesis is true . That is, the null hypothesis is always our initial assumption.

The prosecution team then collects evidence — such as finger prints, blood spots, hair samples, carpet fibers, shoe prints, ransom notes, and handwriting samples — with the hopes of finding "sufficient evidence" to make the assumption of innocence refutable.

In statistics, the data are the evidence.

The jury then makes a decision based on the available evidence:

If the jury finds sufficient evidence — beyond a reasonable doubt — to make the assumption of innocence refutable, the jury rejects the null hypothesis and deems the defendant guilty. We behave as if the defendant is guilty.
If there is insufficient evidence, then the jury does not reject the null hypothesis . We behave as if the defendant is innocent.

In statistics, we always make one of two decisions. We either "reject the null hypothesis" or we "fail to reject the null hypothesis."

Errors in Hypothesis Testing Section

Did you notice the use of the phrase "behave as if" in the previous discussion? We "behave as if" the defendant is guilty; we do not "prove" that the defendant is guilty. And, we "behave as if" the defendant is innocent; we do not "prove" that the defendant is innocent.

This is a very important distinction! We make our decision based on evidence not on 100% guaranteed proof. Again:

If we reject the null hypothesis, we do not prove that the alternative hypothesis is true.
If we do not reject the null hypothesis, we do not prove that the null hypothesis is true.

We merely state that there is enough evidence to behave one way or the other. This is always true in statistics! Because of this, whatever the decision, there is always a chance that we made an error .

Let's review the two types of errors that can be made in criminal trials:

Table S.3.2 shows how this corresponds to the two types of errors in hypothesis testing.

Note that, in statistics, we call the two types of errors by two different names -- one is called a "Type I error," and the other is called a "Type II error." Here are the formal definitions of the two types of errors:

There is always a chance of making one of these errors. But, a good scientific study will minimize the chance of doing so!

Making the Decision Section

Recall that it is either likely or unlikely that we would observe the evidence we did given our initial assumption. If it is likely , we do not reject the null hypothesis. If it is unlikely , then we reject the null hypothesis in favor of the alternative hypothesis. Effectively, then, making the decision reduces to determining "likely" or "unlikely."

In statistics, there are two ways to determine whether the evidence is likely or unlikely given the initial assumption:

We could take the " critical value approach " (favored in many of the older textbooks).
Or, we could take the " P -value approach " (what is used most often in research, journal articles, and statistical software).

In the next two sections, we review the procedures behind each of these two approaches. To make our review concrete, let's imagine that μ is the average grade point average of all American students who major in mathematics. We first review the critical value approach for conducting each of the following three hypothesis tests about the population mean $\mu$:

In Practice

We would want to conduct the first hypothesis test if we were interested in concluding that the average grade point average of the group is more than 3.
We would want to conduct the second hypothesis test if we were interested in concluding that the average grade point average of the group is less than 3.
And, we would want to conduct the third hypothesis test if we were only interested in concluding that the average grade point average of the group differs from 3 (without caring whether it is more or less than 3).

Upon completing the review of the critical value approach, we review the P -value approach for conducting each of the above three hypothesis tests about the population mean $\mu$. The procedures that we review here for both approaches easily extend to hypothesis tests about any other population parameter.

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Statistics and probability

Course: statistics and probability > unit 12, simple hypothesis testing.

Idea behind hypothesis testing
Examples of null and alternative hypotheses
Writing null and alternative hypotheses
P-values and significance tests
Comparing P-values to different significance levels
Estimating a P-value from a simulation
Estimating P-values from simulations
Using P-values to make conclusions
Your answer should be
an integer, like 6 ‍
an exact decimal, like 0.75 ‍
a simplified proper fraction, like 3 / 5 ‍
a simplified improper fraction, like 7 / 4 ‍
a mixed number, like 1 3 / 4 ‍
a percent, like 12.34 % ‍
(Choice A) We cannot reject the hypothesis. A We cannot reject the hypothesis.
(Choice B) We should reject the hypothesis. B We should reject the hypothesis.

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Hypothesis Testing Solved Examples(Questions and Solutions)

Here is a list hypothesis testing exercises and solutions. Try to solve a question by yourself first before you look at the solution.

Question 1 In the population, the average IQ is 100 with a standard deviation of 15. A team of scientists want to test a new medication to see if it has either a positive or negative effect on intelligence, or not effect at all. A sample of 30 participants who have taken the medication has a mean of 140. Did the medication affect intelligence? View Solution to Question 1

A professor wants to know if her introductory statistics class has a good grasp of basic math. Six students are chosen at random from the class and given a math proficiency test. The professor wants the class to be able to score above 70 on the test. The six students get the following scores:62, 92, 75, 68, 83, 95. Can the professor have 90% confidence that the mean score for the class on the test would be above 70. Solution to Question 2

Question 3 In a packaging plant, a machine packs cartons with jars. It is supposed that a new machine would pack faster on the average than the machine currently used. To test the hypothesis, the time it takes each machine to pack ten cartons are recorded. The result in seconds is as follows.

Do the data provide sufficient evidence to conclude that, on the average, the new machine packs faster? Perform the required hypothesis test at the 5% level of significance. Solution to Question 3

Question 4 We want to compare the heights in inches of two groups of individuals. Here are the measurements: X: 175, 168, 168, 190, 156, 181, 182, 175, 174, 179 Y: 120, 180, 125, 188, 130, 190, 110, 185, 112, 188 Solution to Question 4

Question 5 A clinic provides a program to help their clients lose weight and asks a consumer agency to investigate the effectiveness of the program. The agency takes a sample of 15 people, weighing each person in the sample before the program begins and 3 months later. The results a tabulated below

Determine is the program is effective. Solution to Question 5

Question 6 A sample of 20 students were selected and given a diagnostic module prior to studying for a test. And then they were given the test again after completing the module. . The result of the students scores in the test before and after the test is tabulated below.

We want to see if there is significant improvement in the student’s performance due to this teaching method Solution to Question 6

Question 7 A study was performed to test wether cars get better mileage on premium gas than on regular gas. Each of 10 cars was first filled with regular or premium gas, decided by a coin toss, and the mileage for the tank was recorded. The mileage was recorded again for the same cars using other kind of gasoline. Determine wether cars get significantly better mileage with premium gas.

Mileage with regular gas: 16,20,21,22,23,22,27,25,27,28 Mileage with premium gas: 19, 22,24,24,25,25,26,26,28,32 Solution to Question 7

Question 8 An automatic cutter machine must cut steel strips of 1200 mm length. From a preliminary data, we checked that the lengths of the pieces produced by the machine can be considered as normal random variables with a 3mm standard deviation. We want to make sure that the machine is set correctly. Therefore 16 pieces of the products are randomly selected and weight. The figures were in mm: 1193,1196,1198,1195,1198,1199,1204,1193,1203,1201,1196,1200,1191,1196,1198,1191 Examine wether there is any significant deviation from the required size Solution to Question 8

Question 9 Blood pressure reading of ten patients before and after medication for reducing the blood pressure are as follows

Patient: 1,2,3,4,5,6,7,8,9,10 Before treatment: 86,84,78,90,92,77,89,90,90,86 After treatment: 80,80,92,79,92,82,88,89,92,83

Test the null hypothesis of no effect agains the alternate hypothesis that medication is effective. Execute it with Wilcoxon test Solution to Question 9

Question on ANOVA Sussan Sound predicts that students will learn most effectively with a constant background sound, as opposed to an unpredictable sound or no sound at all. She randomly divides 24 students into three groups of 8 each. All students study a passage of text for 30 minutes. Those in group 1 study with background sound at a constant volume in the background. Those in group 2 study with nose that changes volume periodically. Those in group 3 study with no sound at all. After studying, all students take a 10 point multiple choice test over the material. Their scores are tabulated below.

Group1: Constant sound: 7,4,6,8,6,6,2,9 Group 2: Random sound: 5,5,3,4,4,7,2,2 Group 3: No sound at all: 2,4,7,1,2,1,5,5 Solution to Question 10

Question 11 Using the following three groups of data, perform a one-way analysis of variance using α = 0.05.

Solution to Question 11

Question 12 In a packaging plant, a machine packs cartons with jars. It is supposed that a new machine would pack faster on the average than the machine currently used. To test the hypothesis, the time it takes each machine to pack ten cartons are recorded. The result in seconds is as follows.

New Machine: 42,41,41.3,41.8,42.4,42.8,43.2,42.3,41.8,42.7 Old Machine: 42.7,43.6,43.8,43.3,42.5,43.5,43.1,41.7,44,44.1

Perform an F-test to determine if the null hypothesis should be accepted. Solution to Question 12

Question 13 A random sample 500 U.S adults are questioned about their political affiliation and opinion on a tax reform bill. We need to test if the political affiliation and their opinon on a tax reform bill are dependent, at 5% level of significance. The observed contingency table is given below.

Solution to Question 13

Question 14 Can a dice be considered regular which is showing the following frequency distribution during 1000 throws?

Solution to Question 14

Solution to Question 15

Question 16 A newly developed muesli contains five types of seeds (A, B, C, D and E). The percentage of which is 35%, 25%, 20%, 10% and 10% according to the product information. In a randomly selected muesli, the following volume distribution was found.

Lets us decide about the null hypothesis whether the composition of the sample corresponds to the distribution indicated on the packaging at alpha = 0.1 significance level. Solution to Question 16

Question 17 A research team investigated whether there was any significant correlation between the severity of a certain disease runoff and the age of the patients. During the study, data for n = 200 patients were collected and grouped according to the severity of the disease and the age of the patient. The table below shows the result

Let us decided about the correlation between the age of the patients and the severity of disease progression. Solution to Question 17

Question 18 A publisher is interested in determine which of three book cover is most attractive. He interviews 400 people in each of the three states (California, Illinois and New York), and asks each person which of the cover he or she prefers. The number of preference for each cover is as follows:

Do these data indicate that there are regional differences in people’s preferences concerning these covers? Use the 0.05 level of significance. Solution to Question 18

Question 19 Trees planted along the road were checked for which ones are healthy(H) or diseased (D) and the following arrangement of the trees were obtained:

H H H H D D D H H H H H H H D D H H D D D

Test at the = 0.05 significance wether this arrangement may be regarded as random

Solution to Question 19

Question 20 Suppose we flip a coin n = 15 times and come up with the following arrangements

H T T T H H T T T T H H T H H

(H = head, T = tail)

Test at the alpha = 0.05 significance level whether this arrangement may be regarded as random.

Solution to Question 20

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Knowledge Base
Choosing the Right Statistical Test | Types & Examples

Choosing the Right Statistical Test | Types & Examples

Published on January 28, 2020 by Rebecca Bevans . Revised on June 22, 2023.

Statistical tests are used in hypothesis testing . They can be used to:

determine whether a predictor variable has a statistically significant relationship with an outcome variable.
estimate the difference between two or more groups.

Statistical tests assume a null hypothesis of no relationship or no difference between groups. Then they determine whether the observed data fall outside of the range of values predicted by the null hypothesis.

If you already know what types of variables you’re dealing with, you can use the flowchart to choose the right statistical test for your data.

Statistical tests flowchart

What does a statistical test do, when to perform a statistical test, choosing a parametric test: regression, comparison, or correlation, choosing a nonparametric test, flowchart: choosing a statistical test, other interesting articles, frequently asked questions about statistical tests.

Statistical tests work by calculating a test statistic – a number that describes how much the relationship between variables in your test differs from the null hypothesis of no relationship.

It then calculates a p value (probability value). The p -value estimates how likely it is that you would see the difference described by the test statistic if the null hypothesis of no relationship were true.

If the value of the test statistic is more extreme than the statistic calculated from the null hypothesis, then you can infer a statistically significant relationship between the predictor and outcome variables.

If the value of the test statistic is less extreme than the one calculated from the null hypothesis, then you can infer no statistically significant relationship between the predictor and outcome variables.

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You can perform statistical tests on data that have been collected in a statistically valid manner – either through an experiment , or through observations made using probability sampling methods .

For a statistical test to be valid , your sample size needs to be large enough to approximate the true distribution of the population being studied.

To determine which statistical test to use, you need to know:

whether your data meets certain assumptions.
the types of variables that you’re dealing with.

Statistical assumptions

Statistical tests make some common assumptions about the data they are testing:

Independence of observations (a.k.a. no autocorrelation): The observations/variables you include in your test are not related (for example, multiple measurements of a single test subject are not independent, while measurements of multiple different test subjects are independent).
Homogeneity of variance : the variance within each group being compared is similar among all groups. If one group has much more variation than others, it will limit the test’s effectiveness.
Normality of data : the data follows a normal distribution (a.k.a. a bell curve). This assumption applies only to quantitative data .

If your data do not meet the assumptions of normality or homogeneity of variance, you may be able to perform a nonparametric statistical test , which allows you to make comparisons without any assumptions about the data distribution.

If your data do not meet the assumption of independence of observations, you may be able to use a test that accounts for structure in your data (repeated-measures tests or tests that include blocking variables).

Types of variables

The types of variables you have usually determine what type of statistical test you can use.

Quantitative variables represent amounts of things (e.g. the number of trees in a forest). Types of quantitative variables include:

Continuous (aka ratio variables): represent measures and can usually be divided into units smaller than one (e.g. 0.75 grams).
Discrete (aka integer variables): represent counts and usually can’t be divided into units smaller than one (e.g. 1 tree).

Categorical variables represent groupings of things (e.g. the different tree species in a forest). Types of categorical variables include:

Ordinal : represent data with an order (e.g. rankings).
Nominal : represent group names (e.g. brands or species names).
Binary : represent data with a yes/no or 1/0 outcome (e.g. win or lose).

Choose the test that fits the types of predictor and outcome variables you have collected (if you are doing an experiment , these are the independent and dependent variables ). Consult the tables below to see which test best matches your variables.

Parametric tests usually have stricter requirements than nonparametric tests, and are able to make stronger inferences from the data. They can only be conducted with data that adheres to the common assumptions of statistical tests.

The most common types of parametric test include regression tests, comparison tests, and correlation tests.

Regression tests

Regression tests look for cause-and-effect relationships . They can be used to estimate the effect of one or more continuous variables on another variable.

Comparison tests

Comparison tests look for differences among group means . They can be used to test the effect of a categorical variable on the mean value of some other characteristic.

T-tests are used when comparing the means of precisely two groups (e.g., the average heights of men and women). ANOVA and MANOVA tests are used when comparing the means of more than two groups (e.g., the average heights of children, teenagers, and adults).

Correlation tests

Correlation tests check whether variables are related without hypothesizing a cause-and-effect relationship.

These can be used to test whether two variables you want to use in (for example) a multiple regression test are autocorrelated.

Non-parametric tests don’t make as many assumptions about the data, and are useful when one or more of the common statistical assumptions are violated. However, the inferences they make aren’t as strong as with parametric tests.

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This flowchart helps you choose among parametric tests. For nonparametric alternatives, check the table above.

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

Normal distribution
Descriptive statistics
Measures of central tendency
Correlation coefficient
Null hypothesis

Methodology

Cluster sampling
Stratified sampling
Types of interviews
Cohort study
Thematic analysis

Research bias

Implicit bias
Cognitive bias
Survivorship bias
Availability heuristic
Nonresponse bias
Regression to the mean

Statistical tests commonly assume that:

the data are normally distributed
the groups that are being compared have similar variance
the data are independent

If your data does not meet these assumptions you might still be able to use a nonparametric statistical test , which have fewer requirements but also make weaker inferences.

A test statistic is a number calculated by a statistical test . It describes how far your observed data is from the null hypothesis of no relationship between variables or no difference among sample groups.

The test statistic tells you how different two or more groups are from the overall population mean , or how different a linear slope is from the slope predicted by a null hypothesis . Different test statistics are used in different statistical tests.

Statistical significance is a term used by researchers to state that it is unlikely their observations could have occurred under the null hypothesis of a statistical test . Significance is usually denoted by a p -value , or probability value.

Statistical significance is arbitrary – it depends on the threshold, or alpha value, chosen by the researcher. The most common threshold is p < 0.05, which means that the data is likely to occur less than 5% of the time under the null hypothesis .

When the p -value falls below the chosen alpha value, then we say the result of the test is statistically significant.

Quantitative variables are any variables where the data represent amounts (e.g. height, weight, or age).

Categorical variables are any variables where the data represent groups. This includes rankings (e.g. finishing places in a race), classifications (e.g. brands of cereal), and binary outcomes (e.g. coin flips).

You need to know what type of variables you are working with to choose the right statistical test for your data and interpret your results .

Discrete and continuous variables are two types of quantitative variables :

Discrete variables represent counts (e.g. the number of objects in a collection).
Continuous variables represent measurable amounts (e.g. water volume or weight).

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Lesson 10 of 24 By Avijeet Biswal

A Complete Guide on Hypothesis Testing in Statistics

In today’s data-driven world , decisions are based on data all the time. Hypothesis plays a crucial role in that process, whether it may be making business decisions, in the health sector, academia, or in quality improvement. Without hypothesis & hypothesis tests, you risk drawing the wrong conclusions and making bad decisions. In this tutorial, you will look at Hypothesis Testing in Statistics.

What Is Hypothesis Testing in Statistics?

Hypothesis Testing is a type of statistical analysis in which you put your assumptions about a population parameter to the test. It is used to estimate the relationship between 2 statistical variables.

Let's discuss few examples of statistical hypothesis from real-life -

A teacher assumes that 60% of his college's students come from lower-middle-class families.
A doctor believes that 3D (Diet, Dose, and Discipline) is 90% effective for diabetic patients.

Now that you know about hypothesis testing, look at the two types of hypothesis testing in statistics.

Hypothesis Testing Formula

Z = ( x̅ – μ0 ) / (σ /√n)

Here, x̅ is the sample mean,
μ0 is the population mean,
σ is the standard deviation,
n is the sample size.

How Hypothesis Testing Works?

An analyst performs hypothesis testing on a statistical sample to present evidence of the plausibility of the null hypothesis. Measurements and analyses are conducted on a random sample of the population to test a theory. Analysts use a random population sample to test two hypotheses: the null and alternative hypotheses.

The null hypothesis is typically an equality hypothesis between population parameters; for example, a null hypothesis may claim that the population means return equals zero. The alternate hypothesis is essentially the inverse of the null hypothesis (e.g., the population means the return is not equal to zero). As a result, they are mutually exclusive, and only one can be correct. One of the two possibilities, however, will always be correct.

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Null Hypothesis and Alternate Hypothesis

The Null Hypothesis is the assumption that the event will not occur. A null hypothesis has no bearing on the study's outcome unless it is rejected.

H0 is the symbol for it, and it is pronounced H-naught.

The Alternate Hypothesis is the logical opposite of the null hypothesis. The acceptance of the alternative hypothesis follows the rejection of the null hypothesis. H1 is the symbol for it.

Let's understand this with an example.

A sanitizer manufacturer claims that its product kills 95 percent of germs on average.

To put this company's claim to the test, create a null and alternate hypothesis.

H0 (Null Hypothesis): Average = 95%.

Alternative Hypothesis (H1): The average is less than 95%.

Another straightforward example to understand this concept is determining whether or not a coin is fair and balanced. The null hypothesis states that the probability of a show of heads is equal to the likelihood of a show of tails. In contrast, the alternate theory states that the probability of a show of heads and tails would be very different.

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Hypothesis Testing Calculation With Examples

Let's consider a hypothesis test for the average height of women in the United States. Suppose our null hypothesis is that the average height is 5'4". We gather a sample of 100 women and determine that their average height is 5'5". The standard deviation of population is 2.

To calculate the z-score, we would use the following formula:

z = ( x̅ – μ0 ) / (σ /√n)

z = (5'5" - 5'4") / (2" / √100)

z = 0.5 / (0.045)

We will reject the null hypothesis as the z-score of 11.11 is very large and conclude that there is evidence to suggest that the average height of women in the US is greater than 5'4".

Steps of Hypothesis Testing

Step 1: specify your null and alternate hypotheses.

It is critical to rephrase your original research hypothesis (the prediction that you wish to study) as a null (Ho) and alternative (Ha) hypothesis so that you can test it quantitatively. Your first hypothesis, which predicts a link between variables, is generally your alternate hypothesis. The null hypothesis predicts no link between the variables of interest.

Step 2: Gather Data

For a statistical test to be legitimate, sampling and data collection must be done in a way that is meant to test your hypothesis. You cannot draw statistical conclusions about the population you are interested in if your data is not representative.

Step 3: Conduct a Statistical Test

Other statistical tests are available, but they all compare within-group variance (how to spread out the data inside a category) against between-group variance (how different the categories are from one another). If the between-group variation is big enough that there is little or no overlap between groups, your statistical test will display a low p-value to represent this. This suggests that the disparities between these groups are unlikely to have occurred by accident. Alternatively, if there is a large within-group variance and a low between-group variance, your statistical test will show a high p-value. Any difference you find across groups is most likely attributable to chance. The variety of variables and the level of measurement of your obtained data will influence your statistical test selection.

Step 4: Determine Rejection Of Your Null Hypothesis

Your statistical test results must determine whether your null hypothesis should be rejected or not. In most circumstances, you will base your judgment on the p-value provided by the statistical test. In most circumstances, your preset level of significance for rejecting the null hypothesis will be 0.05 - that is, when there is less than a 5% likelihood that these data would be seen if the null hypothesis were true. In other circumstances, researchers use a lower level of significance, such as 0.01 (1%). This reduces the possibility of wrongly rejecting the null hypothesis.

Step 5: Present Your Results

The findings of hypothesis testing will be discussed in the results and discussion portions of your research paper, dissertation, or thesis. You should include a concise overview of the data and a summary of the findings of your statistical test in the results section. You can talk about whether your results confirmed your initial hypothesis or not in the conversation. Rejecting or failing to reject the null hypothesis is a formal term used in hypothesis testing. This is likely a must for your statistics assignments.

Types of Hypothesis Testing

To determine whether a discovery or relationship is statistically significant, hypothesis testing uses a z-test. It usually checks to see if two means are the same (the null hypothesis). Only when the population standard deviation is known and the sample size is 30 data points or more, can a z-test be applied.

A statistical test called a t-test is employed to compare the means of two groups. To determine whether two groups differ or if a procedure or treatment affects the population of interest, it is frequently used in hypothesis testing.

Chi-Square

You utilize a Chi-square test for hypothesis testing concerning whether your data is as predicted. To determine if the expected and observed results are well-fitted, the Chi-square test analyzes the differences between categorical variables from a random sample. The test's fundamental premise is that the observed values in your data should be compared to the predicted values that would be present if the null hypothesis were true.

Hypothesis Testing and Confidence Intervals

Both confidence intervals and hypothesis tests are inferential techniques that depend on approximating the sample distribution. Data from a sample is used to estimate a population parameter using confidence intervals. Data from a sample is used in hypothesis testing to examine a given hypothesis. We must have a postulated parameter to conduct hypothesis testing.

Bootstrap distributions and randomization distributions are created using comparable simulation techniques. The observed sample statistic is the focal point of a bootstrap distribution, whereas the null hypothesis value is the focal point of a randomization distribution.

A variety of feasible population parameter estimates are included in confidence ranges. In this lesson, we created just two-tailed confidence intervals. There is a direct connection between these two-tail confidence intervals and these two-tail hypothesis tests. The results of a two-tailed hypothesis test and two-tailed confidence intervals typically provide the same results. In other words, a hypothesis test at the 0.05 level will virtually always fail to reject the null hypothesis if the 95% confidence interval contains the predicted value. A hypothesis test at the 0.05 level will nearly certainly reject the null hypothesis if the 95% confidence interval does not include the hypothesized parameter.

Simple and Composite Hypothesis Testing

Depending on the population distribution, you can classify the statistical hypothesis into two types.

Simple Hypothesis: A simple hypothesis specifies an exact value for the parameter.

Composite Hypothesis: A composite hypothesis specifies a range of values.

A company is claiming that their average sales for this quarter are 1000 units. This is an example of a simple hypothesis.

Suppose the company claims that the sales are in the range of 900 to 1000 units. Then this is a case of a composite hypothesis.

One-Tailed and Two-Tailed Hypothesis Testing

The One-Tailed test, also called a directional test, considers a critical region of data that would result in the null hypothesis being rejected if the test sample falls into it, inevitably meaning the acceptance of the alternate hypothesis.

In a one-tailed test, the critical distribution area is one-sided, meaning the test sample is either greater or lesser than a specific value.

In two tails, the test sample is checked to be greater or less than a range of values in a Two-Tailed test, implying that the critical distribution area is two-sided.

If the sample falls within this range, the alternate hypothesis will be accepted, and the null hypothesis will be rejected.

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Right Tailed Hypothesis Testing

If the larger than (>) sign appears in your hypothesis statement, you are using a right-tailed test, also known as an upper test. Or, to put it another way, the disparity is to the right. For instance, you can contrast the battery life before and after a change in production. Your hypothesis statements can be the following if you want to know if the battery life is longer than the original (let's say 90 hours):

The null hypothesis is (H0 <= 90) or less change.
A possibility is that battery life has risen (H1) > 90.

The crucial point in this situation is that the alternate hypothesis (H1), not the null hypothesis, decides whether you get a right-tailed test.

Left Tailed Hypothesis Testing

Alternative hypotheses that assert the true value of a parameter is lower than the null hypothesis are tested with a left-tailed test; they are indicated by the asterisk "<".

Suppose H0: mean = 50 and H1: mean not equal to 50

According to the H1, the mean can be greater than or less than 50. This is an example of a Two-tailed test.

In a similar manner, if H0: mean >=50, then H1: mean <50

Here the mean is less than 50. It is called a One-tailed test.

Type 1 and Type 2 Error

A hypothesis test can result in two types of errors.

Type 1 Error: A Type-I error occurs when sample results reject the null hypothesis despite being true.

Type 2 Error: A Type-II error occurs when the null hypothesis is not rejected when it is false, unlike a Type-I error.

Suppose a teacher evaluates the examination paper to decide whether a student passes or fails.

H0: Student has passed

H1: Student has failed

Type I error will be the teacher failing the student [rejects H0] although the student scored the passing marks [H0 was true].

Type II error will be the case where the teacher passes the student [do not reject H0] although the student did not score the passing marks [H1 is true].

Level of Significance

The alpha value is a criterion for determining whether a test statistic is statistically significant. In a statistical test, Alpha represents an acceptable probability of a Type I error. Because alpha is a probability, it can be anywhere between 0 and 1. In practice, the most commonly used alpha values are 0.01, 0.05, and 0.1, which represent a 1%, 5%, and 10% chance of a Type I error, respectively (i.e. rejecting the null hypothesis when it is in fact correct).

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A p-value is a metric that expresses the likelihood that an observed difference could have occurred by chance. As the p-value decreases the statistical significance of the observed difference increases. If the p-value is too low, you reject the null hypothesis.

Here you have taken an example in which you are trying to test whether the new advertising campaign has increased the product's sales. The p-value is the likelihood that the null hypothesis, which states that there is no change in the sales due to the new advertising campaign, is true. If the p-value is .30, then there is a 30% chance that there is no increase or decrease in the product's sales. If the p-value is 0.03, then there is a 3% probability that there is no increase or decrease in the sales value due to the new advertising campaign. As you can see, the lower the p-value, the chances of the alternate hypothesis being true increases, which means that the new advertising campaign causes an increase or decrease in sales.

Why is Hypothesis Testing Important in Research Methodology?

Hypothesis testing is crucial in research methodology for several reasons:

Provides evidence-based conclusions: It allows researchers to make objective conclusions based on empirical data, providing evidence to support or refute their research hypotheses.
Supports decision-making: It helps make informed decisions, such as accepting or rejecting a new treatment, implementing policy changes, or adopting new practices.
Adds rigor and validity: It adds scientific rigor to research using statistical methods to analyze data, ensuring that conclusions are based on sound statistical evidence.
Contributes to the advancement of knowledge: By testing hypotheses, researchers contribute to the growth of knowledge in their respective fields by confirming existing theories or discovering new patterns and relationships.

Limitations of Hypothesis Testing

Hypothesis testing has some limitations that researchers should be aware of:

It cannot prove or establish the truth: Hypothesis testing provides evidence to support or reject a hypothesis, but it cannot confirm the absolute truth of the research question.
Results are sample-specific: Hypothesis testing is based on analyzing a sample from a population, and the conclusions drawn are specific to that particular sample.
Possible errors: During hypothesis testing, there is a chance of committing type I error (rejecting a true null hypothesis) or type II error (failing to reject a false null hypothesis).
Assumptions and requirements: Different tests have specific assumptions and requirements that must be met to accurately interpret results.

After reading this tutorial, you would have a much better understanding of hypothesis testing, one of the most important concepts in the field of Data Science . The majority of hypotheses are based on speculation about observed behavior, natural phenomena, or established theories.

If you are interested in statistics of data science and skills needed for such a career, you ought to explore Simplilearn’s Post Graduate Program in Data Science.

If you have any questions regarding this ‘Hypothesis Testing In Statistics’ tutorial, do share them in the comment section. Our subject matter expert will respond to your queries. Happy learning!

1. What is hypothesis testing in statistics with example?

Hypothesis testing is a statistical method used to determine if there is enough evidence in a sample data to draw conclusions about a population. It involves formulating two competing hypotheses, the null hypothesis (H0) and the alternative hypothesis (Ha), and then collecting data to assess the evidence. An example: testing if a new drug improves patient recovery (Ha) compared to the standard treatment (H0) based on collected patient data.

2. What is hypothesis testing and its types?

Hypothesis testing is a statistical method used to make inferences about a population based on sample data. It involves formulating two hypotheses: the null hypothesis (H0), which represents the default assumption, and the alternative hypothesis (Ha), which contradicts H0. The goal is to assess the evidence and determine whether there is enough statistical significance to reject the null hypothesis in favor of the alternative hypothesis.

Types of hypothesis testing:

One-sample test: Used to compare a sample to a known value or a hypothesized value.
Two-sample test: Compares two independent samples to assess if there is a significant difference between their means or distributions.
Paired-sample test: Compares two related samples, such as pre-test and post-test data, to evaluate changes within the same subjects over time or under different conditions.
Chi-square test: Used to analyze categorical data and determine if there is a significant association between variables.
ANOVA (Analysis of Variance): Compares means across multiple groups to check if there is a significant difference between them.

3. What are the steps of hypothesis testing?

The steps of hypothesis testing are as follows:

Formulate the hypotheses: State the null hypothesis (H0) and the alternative hypothesis (Ha) based on the research question.
Set the significance level: Determine the acceptable level of error (alpha) for making a decision.
Collect and analyze data: Gather and process the sample data.
Compute test statistic: Calculate the appropriate statistical test to assess the evidence.
Make a decision: Compare the test statistic with critical values or p-values and determine whether to reject H0 in favor of Ha or not.
Draw conclusions: Interpret the results and communicate the findings in the context of the research question.

4. What are the 2 types of hypothesis testing?

One-tailed (or one-sided) test: Tests for the significance of an effect in only one direction, either positive or negative.
Two-tailed (or two-sided) test: Tests for the significance of an effect in both directions, allowing for the possibility of a positive or negative effect.

The choice between one-tailed and two-tailed tests depends on the specific research question and the directionality of the expected effect.

5. What are the 3 major types of hypothesis?

The three major types of hypotheses are:

Null Hypothesis (H0): Represents the default assumption, stating that there is no significant effect or relationship in the data.
Alternative Hypothesis (Ha): Contradicts the null hypothesis and proposes a specific effect or relationship that researchers want to investigate.
Nondirectional Hypothesis: An alternative hypothesis that doesn't specify the direction of the effect, leaving it open for both positive and negative possibilities.

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About the author.

Avijeet is a Senior Research Analyst at Simplilearn. Passionate about Data Analytics, Machine Learning, and Deep Learning, Avijeet is also interested in politics, cricket, and football.

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Statistics Interview Questions for Intermediate Level

19. How to check if a Distribution is Normal?

20. What is Measure of Position? How is it helpful in Descriptive Statistics?

21. What are the different types of Probability Sampling methods?

22. What is Confidence level? How is Confidence level related to Width of Confidence interval?

23. What are the different types of Probability Distributions?

24. What is Hypothesis testing? Where do we use it?

25. What is Interquartile range?

26. What is Conditional Probability? How is it related to Bayes Theorem?

27. Explain the Joint and Marginal Probability.

28. What is the difference Between Probability Mass function and Probability Distribution Function?

29. What is Z-score? How do you calculate it?

30. What is meant by standardization? Why do we sometimes standardize Normal Distribution?

31. What are Axioms of Probability?

32. What is Empirical Rule?

33. What is the difference Between Null Hypothesis and Alternate Hypothesis?

34. What is Standard Error? How is it related to Variance of a Data Distribution?

Statistics Interview Questions for Expert Level

35. What is Sampling Bias? How would you avoid bias in your dataset?

36. What is meant by Type I and Type II error? How does it affect your decision making?

37. What is Statistical Significance ? How does Hypothesis Testing help prove statistical Significance? Illustrate the steps in doing so.

38. What is p-value? How is its value related to Confidence Interval?

39. What is the difference between t-score and z-score?

40. What is One sample test? How is it different from Two Sample test? Give a scenario for each of these types of hypothesis testing?

41. What are the assumptions made in One sample Z-test?

42. What is the difference between One tailed and Two tailed test?

43. Why is t-test not used for Two sampling test of Proportions?

44. What are the assumptions made in Chi-square test?

45. What are the assumptions made in t-test hypothesis testing?

46. What are different types of t-test?

47. How can you determine if two samples have equal variance?

48. What is F-test? What are the steps involved in an F-test?

49. What is the usage of Box plot in Statistical Analysis?

50. You are given a task to measure the average height of all the trees in the world. How would you approach this problem with the help of statistics?

Statistics is a branch of mathematics that deals with large amounts of data and the analysis of that data across various industries. Now, if you are looking for career opportunities as a data analyst or data scientist , then knowledge of statistics is very important. Because in most of these interviews, you will encounter statistical questions.

Hence, this blog post aims to delve into some of the most frequently asked interview questions for statistics. Moreover, by the end of this write-up, you will gain comprehensive insights at all levels, ranging from beginners to advanced statistical interview inquiries.

Table of Content

As we aware, statistics is a branch of math that deals with the collection of data, data analysis, interpretation of data, and organization of data. Beciaclly, it is used in various fields, including business, economics, government, medicine, science, and the social sciences.

Now, if we talk about statistics types , then it has two types: descriptive statistics and inferential statistics . Descriptive statistics is used to summarize and describe data, while inferential statistics is used to draw conclusions about populations based on samples.

Random Sampling is a process of selecting a subset from a population such that it ensures every member of a group in that population has equal chance of getting selected. Random Sampling is used to:

it helps in making generalizations about the population
it helps in reducing bias
helps in extracting meaningful statistical inferences
Qualitative Data: Qualitative data cannot be explained in numbers. It is also called Categorical Data. It can be divided into groups and classes. Example: Gender, Color, Age category, etc.
Quantitative Data: Quantitative data, on the other hand is, numerical data. This gives information about the measure of something and can be used in performing mathematical operations. Example: Sales of a car company, Bitcoin Value, etc.

Probability Distribution is a function that describes the likelihood of possible outcomes of a random event. That means it tells how likely it is for an event to occur and associates a probability to it.

Nominal Data: It is a type of Qualitative Data which has no inherent order of rankings. That means this type of data does not have any numerical significance associated with them. Example: Types of Colors, Animal Species, etc.
Ordinal Data: It is a type of qualitative data which has a defined order of ranking associated with it. Some group are given more preference over others. Example: Education Level, Likert Scale in Survey response, etc.

Central Limit Theorem states that:

” The sampling distribution of a sample means approaches normal distribution as the sample size increases irrespective of the shape of Population distribution.”

This theorem holds true for sample size greater than 30. For a Sampling Distribution that follows CLT:

$\overline{x}$

Skewness in distribution refers to the distortion in the data points of distribution, making the shape asymmetric. There are two types of skewness:

Left/Negative Skewness : This is when the distribution shape is distorted towards left
Right/Positive Skewness : This is when the distribution shape is distorted towards right

Skewness happens due to the presence of outliers. Outliers in a dataset decides the direction of skewness (positive or negative).

It is a Discrete probability distribution function that models the number of successes in fixed number of Bernoulli trials, where each trial is either success or failure. The Binomial Distribution function is given as:

$P(X=k)=\binom{n}{k}p^k(1-p)^{(n-k)}$

n = number of events conducted

p = Probability of the event happening

An Outlier is a data point that is significantly different from other data points. Usually, Outliers are present in the extremes of the distribution and stand out as compared to their out data point counterparts.

Measure of Center/ Measure of Central Tendency is a part of statistics that talks about the “center” of a probability distribution (PD) /dataset. It uses 3 measures of “centers” for it, which are:

Mean : The average of all the data points present in the dataset.
Median : The middle data point of the sorted Dataset/PD.
Mode : The data point which occurs most frequently in a dataset/PD.

Measure of Dispersion / Measure of Spread talks about how much distributed the data points are with respect to a single point. Usually, Measure of Dispersion is examined around the mean of the dataset. It explains how “ spread out ” the data points are around the mean. There are few metrics which tells about the dispersion of a dataset, among which the most used ones are:

Range : The difference between the minimum and maximum value in the dataset
Standard Deviation : it is the square root of variance.
Variance : It is the average of the squared difference of each data point from the mean

The Complement Rule in Probability states that:

“The probability an event does not occur is one minus the probability of the event occurring”

( Note: The complement Rule holds true for Independent events.)

Non Probability Sampling methods is based on personal preference of the concerned people. In this type of sampling method, usually sampling is done at the person’s own convenience. Some of the methods are:

Convenience sample: A probability sampling method where the sample are chosen based on the ease to reach or contact.
Snowball sample: its a method where initially approached people are given the task to further spread the recruitment of new people, like a snowball pattern.

It is defined as the maximum expected difference between the population parameter and sample estimate.

There are many possible ways to check if a distribution is normal or not. Some of them are :

Histogram: Plot the data distribution into a histogram, if the shape of plot is like a Bell, with highest frequency at the center and melts down on both sides, it is normally distributed.
QQplot: Plot the distribution as a qqplot. If the data points mostly align along the straight diagonal line, it is normally distributed.
Measure of Central tendency: For a normal distribution, mean = median = mode.

Measure of position is used where we want to determine “where a specific data point or value falls” in a sample or distribution. It is sometimes necessary to know about the relative relation between two data points in terms of their position (like 75th percentile data, etc.). Some of the common measures of Position are:

Percentile : It is a number below which a certain percentage of data points falls.
Quartiles : Quartiles divides your data points into 4 quarters, one lower quarter, two middle quarter and one higher quarter.
Five number summary : It includes the lowest value, the 3 interquartile ranges and the highest value.

Probability sampling method is a technique of selecting a sample from a population such that each individual of the population has equal chance of getting selected into the sample. This is done by randomly selecting an individual. Some known types of Probability Sampling method are:

Simple random sampling : In this, every member of a population is selected randomly and has an equal chance of being chosen
Stratified random sampling : Here, you divide the population into groups and then randomly select a member from each group to be included in the sample.
Cluster random sampling : in this, you create cluster from the whole population and randomly choose a whole cluster and each member of that cluster as your sample.
Systematic random sampling : In this, you randomly choose a starting point from an ordered population and then choose a member at equal interval to be included in the sample.

Confidence Level : It is a statistical measure used to estimate the degree of confidence or certainty about an estimation process.

Confidence Interval : It is the range of values explaining the uncertainty surrounding an estimate.

In statistical analysis, The width confidence interval is directly proportional to the confidence level, i.e., as the confidence level increases, the width of confidence interval increases as well.

There two types of probability distribution:

Uniform Distribution : It is done for distribution where the likelihood of each event occurring is the same.
Binomial Distribution : It models the distribution where there are only two likely scenarios – success or failure. And those two events are mutually exclusive i.e., they cannot occur at the same time.
Bernoulli Distribution : It is also the same as Binomial except this is for a single trial of event whereas Binomial is for repeated trials
Poisson Distribution : This models the probability of a given number of events in a fixed time interval.
Normal Distribution : It is the most common distribution characterized by its iconic bell shaped curve where the mean is at the center of the shape.
Exponential Distribution : This describes the probability distribution of a Poisson process. A Poisson process determines the probability of random processes in a time period.

Hypothesis testing is a fundamental part of inferential statistics which is used to help figure out whether an assumption about a population is true of not. This is done so by testing a sample from that particular population. The possible use cases for hypothesis testing are as follows:

Inference : We can draw conclusions about a possible effect in a population
Quality testing : Hypothesis testing allows us to evaluate product features and its quality
Scientific research : We use Hypothesis testing for Scientific purposes where any assumption’s statistical significance is checked.
Policy Evaluation : We can use hypothesis testing for different policy evaluations too,

Interquartile range ( IQR ) is the range between its first quartile ( Q1 – representing 25% of data points) and third quartile ( Q3 – representing 75% of the data points). This is used as a measure of dispersion and sometimes as measure of position as well.

Conditional Probability : It is the probability of one event occurring when another (related to the first event) has already occurred. The formula for Conditional probability is:

$P(A|B) = \frac{P(A\cap B)}{P(B)}$

Bayes Theorem : It states that for any two event A and B, the probability of A given B is equal to the probability of B given A multiplied by Probability of A divided by Probability of B. It is given as:

$P(A|B) = \frac {P(B|A)\cdot{P(A)}}{P(B)}$

Bayes theorem is based on the principles of Conditional probability.

Joint Probability : It is the probability of two or more events that are happening together. It can represented as the intersection between two or more probabilities.

Marginal Probability : It is the probability of a single random variable in isolation and it is not dependent on any other event.

A Z-score or Standard score is a statistical measure which helps find out about how many standard deviation above or below the population mean is a data point situated. It is a form of Measure of Position. It is given as:

$Z = \frac {x-\mu}{\sigma}$

Z = standard score

x = data point

$\mu$

To calculate Z-score:

Subtract the mean of the population from the data point
Then divide it by the standard deviation of the distribution

based on the result, a Z-score indicates:

The data point is above the mean if Z-score is positive
The data point is below the mean if Z-score is negative
The data point is the mean if Z-score is 0

Standardization refers to the process of transforming the data into a standard scale. Standardization is done by subtracting the mean and then dividing by standard deviation. It is done so that the data is centered around 0 and has the standard deviation of 1.

Standardization is sometimes implemented on Normal Distribution so that it is transformed into a more standardized scale. This is done so that:

it is more comparable with respect to the original distribution which will further help in inferring how much the data point varies
Allows various tests like Z-test and T-test, which largely assumes that the distribution is standardized.
helps in Outlier Detection.

Axioms of Probability are foundations of probability used to assign it to an event. There are 3 axioms of probability which are:

Probability of any event is a non-negative real number.
Probability of the entire Sample Space is one

$E_{1}$

Empirical rule states that for a normal distribution:

Approximately 68% of the data falls within one standard deviation of the mean.
Approximately 95% of the data falls within two standard deviations of the mean.
Approximately 99.7% of the data falls within three standard deviations of the mean.

Standard Error is defined as the amount of variability or uncertainty associated with Sample mean. It helps us understand how much the sample mean is likely to vary from the population mean if we were to take multiple random samples from the same population It is give as:

$SE = \frac {\sigma}{\sqrt {n}}$

n = the sample size

Standard error is directly proportional to Variance, i.e., as the Variance of a dataset increases, the standard error will increase too.

Sampling bias refers to distortion in the composition of a sample collected from a population which leads to results that do not accurately represent the population. To avoid having bias in your dataset, you can implement the following:

Random Sampling method: This ensures that the sample is representative of the population by giving equal chance of selection for each member in that population.
Stratified Sampling method: Another way where you divide the population based on certain groups and randomly select a member from the groups for the sample.
Blinding: Blinding refers to the technique of keeping certain aspects of the experiment hidden from either researcher or the population. if it is hidden from both of them it is called Double Blinding.

Type I error : Type I error occurs when a null hypothesis that is actually true is rejected. This means that the researcher believed that there was a significant effect/relationship when in actuality there wasn’t. In other words, it is a false positive result.

Type II error : Type II error occurs when a null hypothesis that is actually false is not rejected. This means that the researcher believed that there was no significant effect/relationship when in actuality there was. In other words, it is a false negative result.

Statistical significance assesses whether an observed effect or relationship in data is unlikely to have occurred by random chance. This is used in Hypothesis Testing for drawing conclusions about our Null Hypothesis. The results interpreted in your testing is said to be valid and statistically significant when your p-value is less than or equal to the predetermined Significance level (i.e., you reject the null hypothesis). In other case, the The Steps involved in doing so are:

Formulate your null and alternate hypothesis
Select a statistical significance value
perform the test and find out the p-value
Interpret the result

p-value , short for probability value, is a statistical measure which is used to evaluate our hypothesis. If p-value is less than or equal to the pre-determined significance level, the null hypothesis is rejected and we conclude that there is relationship/effect between the variable. The smaller the p-value, the stronger the evidence against null hypothesis.

Confidence interval, on the other hand, is a range of values that is calculated from sample data and is used to estimate an unknown population parameter with certain level of confidence. Wider confidence intervals indicate greater uncertainty about the parameter estimate, while narrower intervals indicate greater precision.

Now, suppose you are checking your hypothesis for a value. If that value is not in the confidence interval range, you have an evidence now that you can use to reject the null hypothesis. This is similar to having a small p-value. In this way, you can use both p-value and Confidence interval to test your hypothesis.

Following are the assumptions that are made in One sample Z-test:

Normality : the true population distribution is normal.
Independence : the observations in your data set are not correlated with each other, they should be independent.
Known Standard Deviation : the true standard deviation of the population is known.

t-test is designed for normal continuous distributions or numerical data like means, score, etc. Proportions on the other hand represents categorical data. Categorical data cannot be plotted as a continuous distribution and do not follow normality. That is why t-test is not suitable for Proportions test. For that, we can use other tests like Z-test and Chi-square test.

Following are the assumptions made in Chi-square test:

Categorical : It is considered that the value is categorical in nature. (like gender, age, education level, etc.)
Independent : It is considered that the value is independent
Mutually exclusive : The samples/values should be mutually exclusive, i.e., they belong to only one category

The following assumptions are made in a t-test:

Normality : It is assumed that the data distribution is normal
Independence : The observation of the two samples are different and not dependent
Homogeneity of variance : Both have approximately same variance
Random Sampling : The samples taken are randomly sampled.

The different types of t-test can be categorized into two:

One sample t-test : t-test done where a single sample is used to compare with a known value like population mean. For e.g., Testing if the average score of students in a class differs significantly from the national average score
Two sample t-test : It is done when comparing a value for two samples. For e.g., Comparing the test scores of students who received tutoring to those who did not to assess if tutoring has an effect.
Student t-test : When two samples have equal variance, we do student’s t test. For e.g., Comparing the average weights of two groups of mice, assuming that the variances in weights are similar in both groups.
Welch t-test : When the two samples do not have equal variance, we use this method. For e.g., Comparing the test scores of students from two different schools where the variances in test scores are significantly different.

There are two possible methods to determine if two samples have equal variance:

Variance rule of Thumb : if the ratio of larger variance to smaller variance is less than 4, it is assumed that the two samples have approximately equal variance.
F-test : we can also use F-test to check whether the two variance are equal by using F-test, where we define our Null hypothesis to be both variance equal to each other.

F-test is a statistical test used to compare the variances of two or more groups or samples. This test helps in determining whether there is heteroscedasticity among two samples (they have different variance or not). To perform F-test, the following steps are taken:

Formulate your Null and Alternate Hypothesis. Here, Null hypothesis is two samples have same variance and alternate hypothesis is they have different variance.

$F = \frac {\sigma_{1}^2}{\sigma_{2}^2}$

For the degree of freedoms as calculated, find the F-value for your selected significance level.
Compare the value with the F-statistics found at step 2.
Interpret the results. If the F-value is less that the F-statistics, you reject the null hypothesis and conclude that there is heteroscedasticity between the two samples.

Box plots are visualization plots that can be considered an important part of statistics. It is helpful in giving the Measure of Center, Measure of Dispersion and Measure of Position for the data distribution. The following are some aspects in which Box plots help:

Finding out the median
Detecting Outlier
Identifying Data Skewness
Visualizing Data distribution

Lets approach the problem step by step:

First we define the proper problem statement. Here, it is calculating the height of all the trees in the world.
Create a sample of the whole population such that the sample is representative of the population
Find out the descriptive statistics (i.e., measure of central tendency, measure of position and measure of dispersion) for the sample chosen.
Formulate your hypothesis and choose a significance level. Here your hypothesis would be about the height of the trees in the sample.
Find the p-value using appropriate test metrics
compare the p-value and your significance level.
Interpret the results from it.

51. Explain different types of Sampling Biases in statistics.

Biases refer deviations from the truth that can occur at various stages of the data collection, analysis, or interpretation process. Sampling Bias occurs when a certain group of individuals get preferred/favored to be included in the research, thus making it not representative of the population as whole.

There are 3 types of Sampling Bias in Statistics:

Selection Bias : Selection bias occurs when the process of selecting individuals for a sample is not random
Survivorship Bias : Survivorship bias occurs when the analysis or study focuses only on the individuals that have “survived” or made it through a selection or filtering process.
Undercoverage bias : Undercoverage bias happens when certain segments or subgroups of the population are inadequately represented or excluded from the sample

Well, this is the end of this write-up. Here we have compiled the most-asked interview questions for statistics. To gain all the information about the interview and its related questions. Explore the whole blog post and collect all the ideas about it.

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