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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.

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 Type I error is:

• 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

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 distribution to be used for this test is \(\bar{X} \sim\) ________________

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

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.

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?

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.

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 of them attended the midnight showing.

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%.

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.

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

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9.2: Exercise

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PART 1: DIFFUSION & OSMOSIS

Diffusion is the movement of molecules from an area in which they are high in concentration to an area in which they are low in concentration. Molecules move down a concentration gradient until they are equally distributed, or equilibrium is reached (Fig. 1). At equilibrium, there is no concentration gradient. Molecules still move once equilibrium is reached, but there is no net movement in any one direction.

Osmosis is a specific type of diffusion : the diffusion of water molecules across a semipermeable membrane. Like other molecules, water molecules diffuse down a concentration gradient, from an area of higher free water concentration to an area of lower free water concentration. This means that water will move across a semipermeable membrane, like the cell membrane, in the direction of the higher solute concentration. (In solution, high solute concentration = low free water concentration; conversely, low solute concentration = high free water concentration.) You will observe this concept in Part 2: Osmosis & Tonicity .

In living organisms, most substances are transported as solutes , dissolved in water, a solvent . For example, if we dissolve salt (\(\ce{NaCl}\)) in a beaker of water, salt (\(\ce{NaCl}\)) is the solute and water is the solvent. Examples of solutes in the human body include glucose, small proteins, and electrolytes like calcium and sodium ions. Waste products, such as \(\ce{CO2}\) and urea are also transported as solutes. Solutes are carried by body fluids, such as blood plasma, and pass into and out of cells through passive and active transport . In either case, the cell membrane will either inhibit or facilitate the process of diffusion: some molecules can easily diffuse across a plasma membrane and some cannot. For example, small, nonpolar molecules (such as \(\ce{CO2}\) and \(\ce{O2}\)) can cross a membrane by simple diffusion. Large molecules or polar molecules, however, cannot easily diffuse across a membrane. Cells must have specialized membrane-bound proteins that function to transport such substances across the membrane.

In Part 1: Diffusion & Osmosis , you will learn about diffusion and osmosis using dialysis membrane, a selectively permeable sheet of cellulose that permits the passage of water and small solutes, but does not allow larger molecules to diffuse across. This is because the membrane has microscopic pores that only allow small molecules through; anything larger than the size of the pores is prevented from crossing. Some of the solutes in this experiment, sucrose (\(\ce{C12H22O11}\)) and starch (\(\ce{(C6H10O5})n}\) are too large to pass through the pores of the dialysis tubing, but the solvent molecules (\(\ce{H2O}\)) and glucose (\(\ce{C12H22O11}\)), are small enough to pass easily.

Exercise 1: Molecular Weight and Diffusion Rate

Molecular weight is an indication of the mass and size of a molecule. The purpose of this experiment is to determine the relationship between molecular weight and the rate of diffusion through a semisolid gel. You will investigate two dyes, methylene blue and potassium permanganate.

Employing Steps in the Scientific Method:

• Record the Question that is being investigated in this experiment. ________________________________________________________________
• Record a Hypothesis for the question stated above. ________________________________________________________________
• Predict the results of the experiment based on your hypothesis (if/then). ________________________________________________________________
• Perform the experiment below and collect your data.
• Petri dish of agar semi-solid gel (Mueller Hinton agar plates, 150 x 15 mm) - make sure the agar has been allowed to come to room temperature
• Methylene blue solution (0.2% in 25% EtOH)
• Potassium permanganate solution (0.1% KMnO4)
• Small straws
• Small plastic metric ruler

1. Obtain a petri dish of agar

2. Take the plastic straw and gently stick it down into one side of the agar. Lift up the straw, withdrawing a small plug of agar. Repeat on the other side of the dish.

3. Using a 1mL transfer pipet, place a single drop of each dye into the appropriate agar well. (Fig. 2).

4. After 20 minutes, place a small, clear metric ruler underneath the Petri dish to measure the distance (diameter) that the dye has moved. Enter the data in Table 1.

5. Repeat step 4 at 40, 60, and 80 minutes.

• What is the relationship between molecular weight and the rate of diffusion? Explain. ________________________________________________________________
• Go back and look at your initial hypothesis. Does your data support this hypothesis? Explain. _______________________________________________________________

Extension Activity: (Optional)

The results of this experiment can be presented graphically. The presentation of your data in a graph will assist you in interpreting your results. Based on your results, you can complete the final step of scientific investigation, in which you must be able to propose a logical argument that either allows you to support or reject your initial hypothesis.

• Graph your results using the data from Table 1.
• What is the dependent variable? Which axis is used to graph this data? ______________________________________________________________________
• What is your independent variable? Which axis is used to graph this data? ______________________________________________________________________

Exercise 2: Diffusion Across a Membrane

• Dialysis tubing
• Plastic clips or string
• 5 x 400 mL beakers (or plastic cups)
• Electronic balance
• Weigh boats
• 15% Sucrose solution (MW sucrose = 342 g/mol)
• 30% Sucrose solution (MW sucrose = 342 g/mol)
• 30% Glucose solution (MW glucose = 180 g/mol)
• Graduated cylinders (10 mL and 100 mL)
• Wax pencil or sharpie
• 15% starch solution (MW = variable)
• Iodine solution (MW = 166 g/mol)
• Benedict’s reagent
• Hot plate or heat block
• Cut 5 pieces of dialysis membrane approximately 10 cm long. Soak the pieces in tap water until they are soft and pliable (3-5 minutes). *This step may be done for you; check with your instructor.
• Obtain 5 beakers (plastic cups) and label them #1 - 5. Fill each beaker with 150 mL of a solution as follows:
• Beaker #1 – H2O
• Beaker #2 – H2O
• Beaker #3 – H2O
• Beaker #4 – 30% sucrose solution
• Beaker #5 - H2O and 1mL Iodine solution
• Set beakers aside.
• Remove one piece of dialysis membrane from the soaking water and open it, forming a tube. Close one end of the tube with a plastic clip, a piece of string, or simply tie it with a knot (Fig. 3)

• Fill the bag with 10 mL of H2O. Remove excess air, and close the other end of the bag with a plastic clip, a piece of string, or tie with a knot. Set aside on a paper towel.
• Repeat steps 4 and 5 for the 4 remaining dialysis tubes, filling them with 10 mL of a solution as follows:

• Bag #2 – 15% sucrose

• Bag #3 – 30% sucrose

• Bag #4 – H2O

• Bag #5 - 5 mL 30% glucose solution and 5 mL 15% starch solution

• Rinse off the outside of the bags with water and carefully blot dry.
• Weigh bags #1 - 4 to the nearest 0.5g. Record the weights in Table 1 below, in the column labeled “0 min.”
• Place each bag in the corresponding beaker (Bag #1 in Beaker #1, etc.). Make sure each bag is fully submerged in the solution.
• Set a timer for 5 minutes.
• At the end of 5 minutes, remove bags 1 - 4 from their beakers, blot excess fluid, and record the mass (in grams) in Table 1.
• Return the bags to the appropriate beaker, and wait another 5 minutes.
• Repeat steps 11 - 13 every 5 minutes and record the weights in Table 1.
• Calculate the total weight change (weight change = final weight – initial weight) for each bag. Record the values in Table 2. Calculate the rate (g/min) of osmosis for each bag by dividing the weight change by the time change. Since all 4 bags were recorded for a total of 20 minutes, the time change for all 4 bags is 20 minutes. Record the rate of osmosis for all 4 bags in Table 2.
• Make observations about bag #5 and beaker #5 in Table 3.
• Remove several mL of liquid from bag #5 and beaker #5 and add each to separate test tubes.
• Add several drops of Benedict's solution to each of the two test tubes and heat to 100 degrees Celsius in a boiling water bath or heat block for 2 - 5 minutes. Record the test results in Table 3.
• Did the weight of each bag (#1 - #4) change significantly over 20 minutes? Explain.
• In which bag(s) was there a net movement of water?
• Explain what is meant by “net movement”.
• Which carbohydrate molecules (glucose, sucrose, starch) were not able to move across the membrane? Explain.
• In terms of solvent (water) concentration, water moved from the area of _______________ concentration to the area of __________________ concentration across a selectively permeable membrane, which is defined as ________________________.
• What can you conclude about the movement of Iodine, glucose, and starch across the dialysis membrane based on your results in Table 3? Support your answers for each with the observation from bag #5 and beaker #5.
• We used the dialysis tubing to simulate a cell membrane. How is the dialysis tubing functionally the same as a cell membrane?
• We used the dialysis tubing to simulate a cell membrane. How is the dialysis tubing functionally different from a cell membrane?
• Prepare a line graph using the data from Table 2.

PART 2: OSMOSIS & TONICITY

Tonicity is the relative concentration of solute (particles), and therefore also a solvent (water), outside the cell compared with inside the cell.

• An isotonic solution has the same concentration of solute (and therefore of free water) as the cell. When cells are placed in an isotonic solution, there is no net movement of water.

• A hypertonic solution has a higher solute (therefore, lower free water) concentration than the cell. When cells are placed in a hypertonic solution, water moves out of the cell into the lower free water solution.

• A hypotonic solution has a lower solute (therefore, higher free water) concentration than the cell. When cells are placed in a hypotonic solution, water moves into the cell from the higher free water solution.

IMPORTANT NOTE: Notice that all of the above definitions have ‘solution’ as the noun. Sometimes the noun will refer to the cell instead of the solution. For example, a hypotonic cell will experience a net movement of water out of the cell. What this means is that if the ‘solution’ is hypotonic, the cell is hypertonic and vice versa.

Exercise 1: Observing Osmosis in Potato Strips

• Wax pencil or Sharpie
• Metric ruler
• 10% NaCl solution
• 0.9% NaCl solution
• Forceps and scalpel
• Obtain a potato and use a cork borer to prepare 3 cylinders of potato. Push the borer through the length of the potato. When the borer is filled, use the flat end of a wooden skewer to gently push out the potato cylinder into the petri dish. Use the scalpel to cut each potato cylinder into a length of 5 cm.
• With a wax pencil or Sharpie, label 3 test tubes (#1, #2, #3).
• Using the metric ruler, mark each tube at the 10 cm mark level from the bottom of the tube.
• Tube #1 - distilled water
• Tube #2 - 10% sodium chloride (NaCl)
• Tube #3 - 0.9% NaCl
• Place one potato cylinder into each test tube and allow them to soak for about 15 minutes in the solutions.
• You can now move on to Exercises 2 and 3 while your potatoes soak.
• After the elapsed time, observe each strip for limpness (water loss, flaccid) or stiffness (water gain,turgor).
• Which tube contained the limp (flaccid) potato strip? Explain.
• Which tube contained the stiff (turgid) potato strip? Explain.
• Which solution is isotonic to the inside of the potato cell?
• What happened to the potato strip in the isotonic solution?

Osmoregulation in Living Cells

Some organisms, known as osmoregulators , have special adaptations to keep tight control over their internal osmotic conditions while still others, known as osmoconformers , are able to live in a variety of osmotic conditions. Most living cells, however, are often at the mercy of their surrounding osmotic environment. Many freshwater plants live in an isotonic or hypotonic environment, so they have no adaptations to protect them from a hypertonic environment. Likewise, mammalian red blood cells live in the isotonic plasma inside your circulatory system so they have no protection from either a hypertonic nor a hypotonic environment.

A plant cell is surrounded by a rigid cell wall, so when the cell is placed in a hypotonic environment, the net flow of water is from the surrounding medium into the cell, and it simply expands to the cell wall and becomes turgid. When the same plant cell is placed in a hypertonic environment, water leaves the central vacuole and the cytoplasm shrinks. This causes the cell membrane to pull away from the cell wall. In this situation, the plant cell will undergo plasmolysis and die. Animal cells have no cell wall so when they are in a hypotonic environment they will expand and fill with water until they burst in a condition known as lysis. Figures 4 and 5 demonstrate these conditions.

Exercise 2: Observing Osmosis in Elodea Cells

• Blank slides and coverslips
• 10% salt solution
• Elodea leaf
• Place a drop of water onto a clean microscope slide.
• Using the forceps, gently tear off a small leaf from the Elodea plant.
• Prepare a wet mount by placing the Elodea leaf into the drop of water on your slide.
• Place a coverslip onto the slide.
• Use the scanning (4x) objective to bring the Elodea cells into focus. You may not be able to observe individual cells at this power.
• Switch to the low power (10x) objective. The Elodea cell walls should be visible. They will look like dark green grid lines. Use the fine focus adjustment to focus the specimen.
• Once you think you have located an Elodea cell, switch to the high power (40x) objective and refocus using the fine focus adjustment.
• Next, add several drops of 10% salt (NaCl) solution to the edge of the coverslip to allow the salt to diffuse under the coverslip. Observe what happens to the cells (this may require you to search around along the edges of the leaf). Look for cells that have been visibly altered.
• Record your observations in the following table. The cells in distilled water should look similar to the figure below.

• Which solution is hypertonic to an Elodea cell? Use your observations to support your answer.
• Would you expect pond water to be isotonic, hypotonic, or hypertonic to Elodea cells? Explain your answer.
• Explain what happens to a plant cell that undergoes plasmolysis.

Exercise 3: Observing Osmosis in Red Blood Cells (Erythrocytes)

• Sheep red blood cells
• Obtain 3 test tubes and label them (#1, #2, #3) with a wax pencil or Sharpie.
• Tube #1 - Distilled water
• Tube #2 - 10% NaCl
• Using a new transfer pipette, add 2 drops of sheep blood to each tube and swirl gently to mix the contents.
• Hold each test tube up to a sheet of paper with printed text. Attempt to read the print through each tube and record your results in the following table.
• Label 3 microscope slides (#1, #2, #3) with a wax pencil or Sharpie.
• Prepare wet mounts of each tube by placing a drop of the solution in each tube (#1 - 3) on the appropriate microscope slide (#1 -3). Add a coverslip to each slide.
• View slide #1 through the microscope using the scanning (4x) objective first. Focus the image using the coarse adjustment. Then view the blood cells under low power and then high power. Only use the fine focus adjustment to focus the specimen.
• Observe slide #2 and slide #3 in the same manner.
• Record the appearance of the red blood cells in each solution in the following table.
• Which solution allowed you to read print through the solution? Explain.
• Which solution is hypertonic to the RBCs? Use your observations to support your answer.
• Which solution is hypotonic to the RBCs? Use your observations to support your answer.
• Which solution is isotonic to the RBCs? Use your observations to support your answer.

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9.2: Exercises

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• Allen B. Downey
• Olin College via Green Tea Press

There are solutions to these exercises in the next section. You should at least attempt each one before you read the solutions.

Exercise \(\PageIndex{1}\)

Write a program that reads words.txt and prints only the words with more than 20 characters (not counting whitespace).

Exercise \(\PageIndex{2}\)

In 1939 Ernest Vincent Wright published a 50,000 word novel called Gadsby that does not contain the letter “e”. Since “e” is the most common letter in English, that’s not easy to do.

In fact, it is difficult to construct a solitary thought without using that most common symbol. It is slow going at first, but with caution and hours of training you can gradually gain facility.

All right, I’ll stop now.

Write a function called has_no_e that returns True if the given word doesn’t have the letter “e” in it.

Write a program that reads words.txt and prints only the words that have no “e”. Compute the percentage of words in the list that have no “e”.

Exercise \(\PageIndex{3}\)

Write a function named avoids that takes a word and a string of forbidden letters, and that returns True if the word doesn’t use any of the forbidden letters.

Write a program that prompts the user to enter a string of forbidden letters and then prints the number of words that don’t contain any of them. Can you find a combination of 5 forbidden letters that excludes the smallest number of words?

Exercise \(\PageIndex{4}\)

Write a function named uses_only that takes a word and a string of letters, and that returns True if the word contains only letters in the list. Can you make a sentence using only the letters acefhlo ? Other than “Hoe alfalfa”?

Exercise \(\PageIndex{5}\)

Write a function named uses_all that takes a word and a string of required letters, and that returns True if the word uses all the required letters at least once. How many words are there that use all the vowels aeiou ? How about aeiouy ?

Exercise \(\PageIndex{6}\)

Write a function called is_abecedarian that returns True if the letters in a word appear in alphabetical order (double letters are ok). How many abecedarian words are there?

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All Rotaract clubs automatically adopt the Standard Rotaract Club Constitution ( PDF ) ( Word ), and changes can be made only by the RI Board. Recommended Rotaract Club Bylaws supplement the Standard Rotaract Club Constitution and set common club practices. Clubs are welcome to adapt the Recommended Rotaract Club Bylaws, provided the changes do not conflict with RI's constitutional documents, the Standard Rotaract Club Constitution, and the Rotary Code of Policies.

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9.8: Confidence Intervals (Exercises)

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

8.1: Introduction

8.2: a single population mean using the normal distribution.

Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches.

• \(\bar{x}\) =________
• \(\sigma\) =________
• \(n\) =________
• In words, define the random variables \(X\) and \(\bar{X}\).
• Which distribution should you use for this problem? Explain your choice.
• State the confidence interval.
• Sketch the graph.
• Calculate the error bound.
• What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why?
• X is the height of a Swedish male, and is the mean height from a sample of 48 Swedish males.
• Normal. We know the standard deviation for the population, and the sample size is greater than 30.
• CI: (70.151, 71.49)

ii. E = 0.849

e. The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.

Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. Assume the underlying population is normal.

Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal.

• \(\bar{x} =\) ________
• \(\sigma =\) ________
• \(n =\) ________
• If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make?
• If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why?
• Suppose that the firm decided that it needed to be at least 96% confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why?
• \(\bar{x} = 23.6\)
• \(\sigma = 7\)
• \(n = 100\)
• \(X\) is the time needed to complete an individual tax form. \(\bar{X}\) is the mean time to complete tax forms from a sample of 100 customers.
• \(N\left(23.6, \frac{7}{\sqrt{100}}\right)\) because we know sigma.
• (22.228, 24.972)
• \(EBM = 1.372\)
• It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size.
• The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval.
• According to the error bound formula, the firm needs to survey 206 people. Since we increase the confidence level, we need to increase either our error bound or the sample size.

A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce.

• \(\bar{x} =\)________
• \(\sigma =\)________
• \(s_{x} =\)________
• In words, define the random variable \(X\).
• In words, define the random variable \(\bar{X}\).
• In complete sentences, explain why the confidence interval in part f is larger than the confidence interval in part e.
• In complete sentences, give an interpretation of what the interval in part f means.

A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8.

• \(x =\)________
• Define the random variables \(X\) and \(\bar{X}\) in words.
• What will happen to the error bound and confidence interval if 500 campers are surveyed? Why?
• \(X\) is the number of letters a single camper will send home. \(\bar{X}\) is the mean number of letters sent home from a sample of 20 campers.
• \(N 7.9\left(\frac{2.5}{\sqrt{20}}\right)\)
• CI: (6.98, 8.82)
• \(EBM: 0.92\) 
• The error bound and confidence interval will decrease.

What is meant by the term “90% confident” when constructing a confidence interval for a mean?

• If we took repeated samples, approximately 90% of the samples would produce the same confidence interval.
• If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the sample mean.
• If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean.
• If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples.

The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. During the 2012 campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Table shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is \(\sigma\) = $909,200.

• Find the point estimate for the population mean.
• Using 95% confidence, calculate the error bound.
• Create a 95% confidence interval for the mean total individual contributions.
• Interpret the confidence interval in the context of the problem.
• \(\bar{x} = $568,873\)
• \(CL = 0.95 \alpha = 1 - 0.95 = 0.05 z_{\frac{\alpha}{2}} = 1.96\) \(E = z_{0.025 } \frac{\sigma}{\sqrt{n}} = 1.96 \frac{909200}{\sqrt{40}} = $281,764\)

Alternate solution: Use the Confidence Interval For a Mean With Data calculator Click on \(\sigma\) Known n = 40, \(\bar x\) = 568873, \(sigma\) = 909200, CL = 0.95 Click Calculate The confidence interval is ($287,114, $850,632) We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637.

The American Community Survey (ACS), part of the United States Census Bureau, conducts a yearly census similar to the one taken every ten years, but with a smaller percentage of participants. The most recent survey estimates with 90% confidence that the mean household income in the U.S. falls between $69,720 and $69,922. Find the point estimate for mean U.S. household income and the error bound for mean U.S. household income.

The average height of young adult males has a normal distribution with standard deviation of 2.5 inches. You want to estimate the mean height of students at your college or university to within one inch with 93% confidence. How many male students must you measure?

Use the formula for \(E\), solved for \(n\):

\(n = \frac{z^{2}\sigma^{2}}{E^{2}}\)

From the statement of the problem, you know that \(\sigma\) = 2.5, and you need \(E = 1\).

\(z = z_{0.035} = 1.812\)

(This is the value of \(z\) for which the area under the density curve to the right of \(z\) is 0.035.)

\(n = \frac{z^{2}\sigma^{2}}{EBM^{2}} = \frac{1.812^{2}2.5^{2}}{1^{2}} \approx 20.52\)

You need to measure at least 21 male students to achieve your goal.

8.3: A Single Population Mean using the Student t Distribution

In six packages of “The Flintstones® Real Fruit Snacks” there were five Bam-Bam snack pieces. The total number of snack pieces in the six bags was 68. We wish to calculate a 96% confidence interval for the population proportion of Bam-Bam snack pieces.

• Define the random variables \(X\) and \(P′\) in words.
• Which distribution should you use for this problem? Explain your choice
• Calculate \(p′\).
• Do you think that six packages of fruit snacks yield enough data to give accurate results? Why or why not?

A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal.

• \(\bar{x} =\) __________
• \(s_{x} =\) __________
• \(n =\) __________
• \(n – 1 =\) __________
• What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why?
• \(CI: (6244, 11,014)\)
• \(EB = 2385\)
• It will become smaller

Suppose that a committee is studying whether or not there is waste of time in our judicial system. It is interested in the mean amount of time individuals waste at the courthouse waiting to be called for jury duty. The committee randomly surveyed 81 people who recently served as jurors. The sample mean wait time was eight hours with a sample standard deviation of four hours.

• Explain in a complete sentence what the confidence interval means.

A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.

• Define the random variable \(X\) in words.
• Define the random variable \(\bar{X}\) in words.
• What does it mean to be “95% confident” in this problem?
• \(\bar{x} = 2.51\)
• \(s_{x} = 0.318\)
• \(n - 1 = 8\)
• the effective length of time for a tranquilizer
• the mean effective length of time of tranquilizers from a sample of nine patients
• We need to use a Student’s-t distribution, because we do not know the population standard deviation.
• \(CI: (2.27, 2.76)\)
• Check student's solution.
• \(EBM: 0.25\)
• If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time.

Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal.

• Why would the error bound change if the confidence level were lowered to 90%?

The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns.

The FEC has reported financial information for 556 Leadership PACs that operating during the 2011–2012 election cycle. The following table shows the total receipts during this cycle for a random selection of 20 Leadership PACs.

\(\bar{x} = $251,854.23\)

\(s = $521,130.41\)

Use this sample data to construct a 96% confidence interval for the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle. Use the Student's \(t\)-distribution.

Note that we are not given the population standard deviation, only the standard deviation of the sample.

There are 30 measures in the sample, so \(n = 30\), and \(df = 30 - 1 = 29\)

\(CL = 0.96\), so \(\alpha = 1 - CL = 1 - 0.96 = 0.04\)

\(\frac{\alpha}{2} = 0.02 t_{0.02} = t_{0.02} = 2.150\)

\(EBM = t_{\frac{\alpha}{2}}\left(\frac{s}{\sqrt{n}}\right) = 2.150\left(\frac{521,130.41}{\sqrt{30}}\right) - $204,561.66\)

\(\bar{x} - E = $251,854.23 - $204,561.66 = $47,292.57\)

\(\bar{x} + E = $251,854.23+ $204,561.66 = $456,415.89\)

We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle lies between $47,292.57 and $456,415.89.

Alternate Solution

Alternate solution: Use the Confidence Interval For a Mean With Data calculator CL = 0.95 Click Calculate The confidence interval is ($47262, $456447)

Forbes magazine published data on the best small firms in 2012. These were firms that had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. The Table shows the ages of the corporate CEOs for a random sample of these firms.

Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution.

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats.

• \(n-1 =\) __________
• \(\bar{x} =\)
• \(s_{x} =\)
• \(X\) is the number of unoccupied seats on a single flight. \(\bar{X}\) is the mean number of unoccupied seats from a sample of 225 flights.
• We will use a Student’s \(t\)-distribution, because we do not know the population standard deviation.
• \(CI: (11.12 , 12.08)\)
• \(E\): 0.48

In a recent sample of 84 used car sales costs, the sample mean was $6,425 with a standard deviation of $3,156. Assume the underlying distribution is approximately normal.

• Explain what a “95% confidence interval” means for this study.

Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal.

• If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done?
• Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies.
• Calculate the mean.
• Is the mean within the interval you calculated in part a? Did you expect it to be? Why or why not?
• CI: (7.64 , 9.36)
• \(EBM: 0.86\)
• The sample should have been increased.
• Answers will vary.

A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal.

• If many random samples were taken of size 14, what percent of the confidence intervals constructed should contain the population mean worth of coupons? Explain why.

Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed.

Find the 95% Confidence Interval for the true population mean for the amount of soda served.

• (12.42, 14.18)
• (12.32, 14.29)
• (12.50, 14.10)
• Impossible to determine

What is the error bound?

8.4: A Population Proportion

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car.

• When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03?
• If it were later determined that it was important to be more than 95% confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why?
• The sample size would need to be increased since the critical value increases as the confidence level increases.

Suppose that the insurance companies did do a survey. They randomly surveyed 400 drivers and found that 320 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.

• \(x =\) __________
• \(p′ =\) __________
• Define the random variables \(X\) and \(P′\), in words.
• If this survey were done by telephone, list three difficulties the companies might have in obtaining random results.

According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job.

\(P′ =\) the proportion of people in a sample who feel that the president is doing an acceptable job.

• \(N\left(0.61, \sqrt{\frac{(0.61)(0.39)}{1200}}\right)\)
• \(CI: (0.59, 0.63)\)
• Check student’s solution
• \(E: 0.02\)

An article regarding interracial dating and marriage recently appeared in the Washington Post. Of the 1,709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this survey, 86% of blacks said that they would welcome a white person into their families. Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.

• We are interested in finding the 95% confidence interval for the percent of all black adults who would welcome a white person into their families. Define the random variables \(X\) and \(P′\), in words.

Refer to the information in Exercise .

• percent of all Asians who would welcome a white person into their families.
• percent of all Asians who would welcome a Latino into their families.
• percent of all Asians who would welcome a black person into their families.
• Even though the three point estimates are different, do any of the confidence intervals overlap? Which?
• For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions?
• For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions?
• (0.72, 0.82)
• (0.65, 0.76)
• (0.60, 0.72)
• Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.
• We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.
• We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families.

Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period.

• Explain what a “97% confidence interval” means for this study.

A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine. One of the questions asked was “What is the main problem facing the country?” Twenty percent answered “crime.” We are interested in the population proportion of adult Americans who feel that crime is the main problem.

• Suppose we want to lower the sampling error. What is one way to accomplish that?
• The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is \(\pm 3%\). In one to three complete sentences, explain what the ±3% represents.
• \(X =\) the number of adult Americans who feel that crime is the main problem; \(P′ =\) the proportion of adult Americans who feel that crime is the main problem
• Since we are estimating a proportion, given \(P′ = 0.2\) and \(n = 1000\), the distribution we should use is \(N\left(0.61, \sqrt{\frac{(0.2)(0.8)}{1000}}\right)\).
• \(CI: (0.18, 0.22)\)
• Check student’s solution.
• One way to lower the sampling error is to increase the sample size.
• The stated “\(\pm 3%\)” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%.

Refer to Exercise . Another question in the poll was “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools.

Use the following information to answer the next three exercises: According to a Field Poll, 79% of California adults (actual results are 400 out of 506 surveyed) feel that “education and our schools” is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education and the schools is one of the top issues facing California.

A point estimate for the true population proportion is:

A 90% confidence interval for the population proportion is _______.

• (0.761, 0.820)
• (0.125, 0.188)
• (0.755, 0.826)
• (0.130, 0.183)

The error bound is approximately _____.

Use the following information to answer the next two exercises: Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed met the minimum recommendations for earthquake preparedness, and 338 did not.

Find the confidence interval at the 90% Confidence Level for the true population proportion of southern California community homes meeting at least the minimum recommendations for earthquake preparedness.

• (0.2975, 0.3796)
• (0.6270, 0.6959)
• (0.3041, 0.3730)
• (0.6204, 0.7025)

The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______.

On May 23, 2013, Gallup reported that of the 1,005 people surveyed, 76% of U.S. workers believe that they will continue working past retirement age. The confidence level for this study was reported at 95% with a \(\pm 3%\) margin of error.

• Determine the estimated proportion from the sample.
• Determine the sample size.
• Identify \(CL\) and \(\alpha\).
• Calculate the error bound based on the information provided.
• Compare the error bound in part d to the margin of error reported by Gallup. Explain any differences between the values.
• Create a confidence interval for the results of this study.
• A reporter is covering the release of this study for a local news station. How should she explain the confidence interval to her audience?

A national survey of 1,000 adults was conducted on May 13, 2013 by Rasmussen Reports. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education.

• Find the point estimate and the error bound for this confidence interval.
• Can we (with 95% confidence) conclude that more than half of all American adults believe this?
• Use the point estimate from part a and \(n = 1,000\) to calculate a 75% confidence interval for the proportion of American adults that believe that major college sports programs corrupt higher education.
• Can we (with 75% confidence) conclude that at least half of all American adults believe this?
• \(p′ = \frac{(0.55+0.49)}{2} = 0.52; EBP = 0.55 - 0.52 = 0.03\)
• No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.

Alternate Solution Confidence Interval forr a Proportion calculator: \(x = (0.52)(1,000) = 520, n = 1000, CL = 0.75\)

Answer is (0.502, 0.538)

• Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence.

Public Policy Polling recently conducted a survey asking adults across the U.S. about music preferences. When asked, 80 of the 571 participants admitted that they have illegally downloaded music.

• Create a 99% confidence interval for the true proportion of American adults who have illegally downloaded music.
• This survey was conducted through automated telephone interviews on May 6 and 7, 2013. The error bound of the survey compensates for sampling error, or natural variability among samples. List some factors that could affect the survey’s outcome that are not covered by the margin of error.
• Without performing any calculations, describe how the confidence interval would change if the confidence level changed from 99% to 90%.

You plan to conduct a survey on your college campus to learn about the political awareness of students. You want to estimate the true proportion of college students on your campus who voted in the 2012 presidential election with 95% confidence and a margin of error no greater than five percent. How many students must you interview?

\(CL = 0.95 \alpha = 1 - 0.95 = 0.05 \frac{\alpha}{2} = 0.025 z_{\frac{\alpha}{2}} = 1.96.\) Use \(p′ = q′ = 0.5\).

\(n = \frac{z_{\frac{\alpha}{2}}^{2}p'q'}{EPB^{2}} = \frac{1.96^{2}(0.5)(0.5)}{0.05^{2}} = 384.16\)

You need to interview at least 385 students to estimate the proportion to within 5% at 95% confidence.

In a recent Zogby International Poll, nine of 48 respondents rated the likelihood of a terrorist attack in their community as “likely” or “very likely.” Use the “plus four” method to create a 97% confidence interval for the proportion of American adults who believe that a terrorist attack in their community is likely or very likely. Explain what this confidence interval means in the context of the problem.

8.5: Confidence Interval (Home Costs)

8.6: confidence interval (place of birth), 8.7: confidence interval (women's heights).