p value hypothesis testing a level maths

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S.3.2 hypothesis testing (p-value approach).

The P -value approach involves determining "likely" or "unlikely" by determining the probability — assuming the null hypothesis was true — of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed. If the P -value is small, say less than (or equal to) \(\alpha\), then it is "unlikely." And, if the P -value is large, say more than \(\alpha\), then it is "likely."

If the P -value is less than (or equal to) \(\alpha\), then the null hypothesis is rejected in favor of the alternative hypothesis. And, if the P -value is greater than \(\alpha\), then the null hypothesis is not rejected.

Specifically, the four steps involved in using the P -value approach to conducting any hypothesis test are:

• Specify the null and alternative hypotheses.
• Using the sample data and assuming the null hypothesis is true, calculate the value of the test statistic. Again, to conduct the hypothesis test for the population mean μ , we use the t -statistic \(t^*=\frac{\bar{x}-\mu}{s/\sqrt{n}}\) which follows a t -distribution with n - 1 degrees of freedom.
• Using the known distribution of the test statistic, calculate the P -value : "If the null hypothesis is true, what is the probability that we'd observe a more extreme test statistic in the direction of the alternative hypothesis than we did?" (Note how this question is equivalent to the question answered in criminal trials: "If the defendant is innocent, what is the chance that we'd observe such extreme criminal evidence?")
• Set the significance level, \(\alpha\), the probability of making a Type I error to be small — 0.01, 0.05, or 0.10. Compare the P -value to \(\alpha\). If the P -value is less than (or equal to) \(\alpha\), reject the null hypothesis in favor of the alternative hypothesis. If the P -value is greater than \(\alpha\), do not reject the null hypothesis.

Example S.3.2.1

Mean gpa section  .

In our example concerning the mean grade point average, suppose that our random sample of n = 15 students majoring in mathematics yields a test statistic t * equaling 2.5. Since n = 15, our test statistic t * has n - 1 = 14 degrees of freedom. Also, suppose we set our significance level α at 0.05 so that we have only a 5% chance of making a Type I error.

Right Tailed

The P -value for conducting the right-tailed test H 0 : μ = 3 versus H A : μ > 3 is the probability that we would observe a test statistic greater than t * = 2.5 if the population mean \(\mu\) really were 3. Recall that probability equals the area under the probability curve. The P -value is therefore the area under a t n - 1 = t 14 curve and to the right of the test statistic t * = 2.5. It can be shown using statistical software that the P -value is 0.0127. The graph depicts this visually.

The P -value, 0.0127, tells us it is "unlikely" that we would observe such an extreme test statistic t * in the direction of H A if the null hypothesis were true. Therefore, our initial assumption that the null hypothesis is true must be incorrect. That is, since the P -value, 0.0127, is less than \(\alpha\) = 0.05, we reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ > 3.

Note that we would not reject H 0 : μ = 3 in favor of H A : μ > 3 if we lowered our willingness to make a Type I error to \(\alpha\) = 0.01 instead, as the P -value, 0.0127, is then greater than \(\alpha\) = 0.01.

Left Tailed

In our example concerning the mean grade point average, suppose that our random sample of n = 15 students majoring in mathematics yields a test statistic t * instead of equaling -2.5. The P -value for conducting the left-tailed test H 0 : μ = 3 versus H A : μ < 3 is the probability that we would observe a test statistic less than t * = -2.5 if the population mean μ really were 3. The P -value is therefore the area under a t n - 1 = t 14 curve and to the left of the test statistic t* = -2.5. It can be shown using statistical software that the P -value is 0.0127. The graph depicts this visually.

The P -value, 0.0127, tells us it is "unlikely" that we would observe such an extreme test statistic t * in the direction of H A if the null hypothesis were true. Therefore, our initial assumption that the null hypothesis is true must be incorrect. That is, since the P -value, 0.0127, is less than α = 0.05, we reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ < 3.

Note that we would not reject H 0 : μ = 3 in favor of H A : μ < 3 if we lowered our willingness to make a Type I error to α = 0.01 instead, as the P -value, 0.0127, is then greater than \(\alpha\) = 0.01.

In our example concerning the mean grade point average, suppose again that our random sample of n = 15 students majoring in mathematics yields a test statistic t * instead of equaling -2.5. The P -value for conducting the two-tailed test H 0 : μ = 3 versus H A : μ ≠ 3 is the probability that we would observe a test statistic less than -2.5 or greater than 2.5 if the population mean μ really was 3. That is, the two-tailed test requires taking into account the possibility that the test statistic could fall into either tail (hence the name "two-tailed" test). The P -value is, therefore, the area under a t n - 1 = t 14 curve to the left of -2.5 and to the right of 2.5. It can be shown using statistical software that the P -value is 0.0127 + 0.0127, or 0.0254. The graph depicts this visually.

Note that the P -value for a two-tailed test is always two times the P -value for either of the one-tailed tests. The P -value, 0.0254, tells us it is "unlikely" that we would observe such an extreme test statistic t * in the direction of H A if the null hypothesis were true. Therefore, our initial assumption that the null hypothesis is true must be incorrect. That is, since the P -value, 0.0254, is less than α = 0.05, we reject the null hypothesis H 0 : μ = 3 in favor of the alternative hypothesis H A : μ ≠ 3.

Note that we would not reject H 0 : μ = 3 in favor of H A : μ ≠ 3 if we lowered our willingness to make a Type I error to α = 0.01 instead, as the P -value, 0.0254, is then greater than \(\alpha\) = 0.01.

Now that we have reviewed the critical value and P -value approach procedures for each of the three possible hypotheses, let's look at three new examples — one of a right-tailed test, one of a left-tailed test, and one of a two-tailed test.

The good news is that, whenever possible, we will take advantage of the test statistics and P -values reported in statistical software, such as Minitab, to conduct our hypothesis tests in this course.

Hypotheseis Tests

• Hypothesis tests are a useful statistical tool for making population inferences from sample data.
• The process starts with forming two opposing hypotheses: the null hypothesis (H0) and the alternative hypothesis (H1 or Ha).
• The null hypothesis assumes no significant effect, difference, or relationship between variables.
• The alternative hypothesis claims the existence of such effect, difference or relationship.
• Test statistics and calculated p-values are used to measure evidence against the null hypothesis.
• If the evidence is sufficiently strong (e.g., a significance level of 0.05), the null hypothesis is rejected for the alternative.
• Understanding and using hypothesis tests can improve the ability to draw valuable conclusions from statistical data.

AS Level Mechanics and Statistics - Hypothesis Testing

One and Two Tailed Tests

Suppose we have a null hypothesis H 0 and an alternative hypothesis H 1 . We consider the distribution given by the null hypothesis and perform a test to determine whether or not the null hypothesis should be rejected in favour of the alternative hypothesis.

There are two different types of tests that can be performed. A one-tailed test looks for an increase or decrease in the parameter whereas a two-tailed test looks for any change in the parameter (which can be any change- increase or decrease).

We can perform the test at any level (usually 1%, 5% or 10%). For example, performing the test at a 5% level means that there is a 5% chance of wrongly rejecting H 0 .

If we perform the test at the 5% level and decide to reject the null hypothesis, we say "there is significant evidence at the 5% level to suggest the hypothesis is false".

One-Tailed Test

We choose a critical region. In a one-tailed test, the critical region will have just one part (the red area below). If our sample value lies in this region, we reject the null hypothesis in favour of the alternative.

Suppose we are looking for a definite decrease. Then the critical region will be to the left. Note, however, that in the one-tailed test the value of the parameter can be as high as you like.

Suppose we are given that X has a Poisson distribution and we want to carry out a hypothesis test on the mean, l, based upon a sample observation of 3.

Suppose the hypotheses are: H 0 : l = 9 H 1 : l < 9

We want to test if it is "reasonable" for the observed value of 3 to have come from a Poisson distribution with parameter 9. So what is the probability that a value as low as 3 has come from a Po(9)?

P(X < 3) = 0.0212 (this has come from a Poisson table)

The probability is less than 0.05, so there is less than a 5% chance that the value has come from a Poisson(3) distribution. We therefore reject the null hypothesis in favour of the alternative at the 5% level.

However, the probability is greater than 0.01, so we would not reject the null hypothesis in favour of the alternative at the 1% level.

Two-Tailed Test

In a two-tailed test, we are looking for either an increase or a decrease. So, for example, H 0 might be that the mean is equal to 9 (as before). This time, however, H 1 would be that the mean is not equal to 9. In this case, therefore, the critical region has two parts:

Lets test the parameter p of a Binomial distribution at the 10% level.

Suppose a coin is tossed 10 times and we get 7 heads. We want to test whether or not the coin is fair. If the coin is fair, p = 0.5 . Put this as the null hypothesis:

H 0 : p = 0.5 H 1 : p =(doesn' equal) 0.5

Now, because the test is 2-tailed, the critical region has two parts. Half of the critical region is to the right and half is to the left. So the critical region contains both the top 5% of the distribution and the bottom 5% of the distribution (since we are testing at the 10% level).

If H 0 is true, X ~ Bin(10, 0.5).

If the null hypothesis is true, what is the probability that X is 7 or above? P(X > 7) = 1 - P(X < 7) = 1 - P(X < 6) = 1 - 0.8281 = 0.1719

Is this in the critical region? No- because the probability that X is at least 7 is not less than 0.05 (5%), which is what we need it to be.

So there is not significant evidence at the 10% level to reject the null hypothesis.

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Hypothesis Testing

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This topic is included in Paper 2 for AS-level Edexcel Maths and Paper 3 for A-level Edexcel Maths .

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• Ch.7 Hypothesis Testing
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• Hypothesis Testing - Tests on Binomial

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• Regression, Correlation and Hypothesis Testing
• Regression, Correlation and Hypothesis Testing MA
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• Hypothesis tests - Tests on binomial

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• Hypothesis tests - Tests on Normal

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Hypothesis testing for Normal (A-level Maths)

This is part 2 of my series of three articles on hypothesis testing for A-level Maths. The first part can be found here ; it gives an introduction to the concept of hypothesis testing and covers the Year 1 topic of hypothesis testing for Binomial.

This tutorial is on hypothesis testing for Normal : to decide whether the mean of a normal distribution has changed. You will need to already be familiar with the Normal distribution and how to use it to carry out probability calculations, before this article will make sense to you.

Part 3 (to follow) shows you how to ascertain whether correlation exists in a population.

Hypothesis test for the mean of a Normal population

The purpose of hypothesis testing.

A hypothesis test of the mean of a normal distribution is a test to decide whether there is statistically significant evidence that the mean has changed from a known previous value. This test is for a population that is Normally distributed and has known variance σ 2 .

For example, adjustments might have been made to a machine packing sugar into bags, and the operator wants to know whether this has increased the mean amount being packed.

How it works – basic principle

H 0 is the null hypothesis, which is the assertion that the population mean hasn’t changed from its historical value. In the example above, the historical mean is probably just over the amount advertised on the packaging (since a certain proportion of bags sold have to be over the advertised amount), so for a 1kg bag of sugar the mean is perhaps around 1010g.

(With a mean of 1010g, if the standard deviation were – for example – 5g then the nominal 1000g content would be 2 standard deviations below the mean, so with your knowledge of the Normal distribution you should be able to work out that that would mean only 2.3% of bags would be below 1000g. With a standard deviation of 10g, 15.9% of bags would be below the 1000g threshold.)

H 1 is the alternative hypothesis, which is the assertion that the mean has changed. If the direction of change is specified then it’s a 1-tailed test; if not – effectively H 1 is sitting on the fence and not committing either way – then it’s a 2-tailed test.

We take the mean of a sample of known size and work out what the probability of that (or a more extreme) result would be if the population mean hadn’t changed.

If it’s sufficiently unlikely then we reject H 0 and accept that there probably has been a change.

Note: Because it’s all based on a balance of probabilities, you can never make a definitive statement that yes, the value in question HAS changed, only that it’s likely that it’s no longer the same as before. Therefore the question is always whether we reject H 0 , not whether we accept H 1 .

Distribution of sample means

Because we are using the mean of a sample , rather than just an individual item, to make our judgment, we have to use a different value from usual for the standard deviation.

Note the use of x-bar throughout (the bars aren’t showing up very well in my browser but I don’t have a way to fix that).

The significance level

The significance level, α, dictates how demanding the test is. If the significance level is 10%, or α = 0.1, then we’re looking to see whether the probability of the observed result (under the old mean) is less than 0.1. If it is then the result is considered significant and we reject H 0 .

If α is 5%, or 0.05, then the probability has to be less than 0.05 for H 0 to be rejected – so you’re less likely to reject H 0 and to agree that the probability has most likely changed.

The significance level is also the probability of incorrectly rejecting H 0 . This is because although the observed result may be unlikely , it is still possible under the old probability. At a 5% level of significance, 5% of results would be in the critical region, so it may still be a genuine result under the old probability, just one at the outer edges of what’s normally observed.

Hypothesis testing for Normal: Two possible methods

There are two possible approaches to the hypothesis test for Normal:

• With the p-value method , you find the probability (based on the old mean) of your observed result actually happening, and compare it with the significance level.
• With the test statistic /critical value method , you work out how many standard deviations above or below the old mean your observed sample mean is (this z-value is the test statistic), and how many standard deviations away you need to be for the result to be significant (the critical value), and compare the two.

The first few steps are the same for both, but then they diverge, as shown in the writing frame below. (Scroll down for a plan text version.) You can get a free printable copy – as well as other useful downloads – if you sign up to my mailing list – see this page .

In most cases, either method is acceptable, so the p-value method is probably easier – but it’s possible that you might be asked to use the “test statistic” or “critical value(s)”, in which case you’ll need to be familiar with the second method.

Writing frame (plain text version: p-value method)

• Define population parameter μ in context
• Write down the null and alternative hypotheses.
• State the significance level α.

Writing frame (plain text version: test statistic / critical value method)

• Use inverse normal to identify the critical z-value for the given significance level. (Sign will be same as for the test statistic since we’re only interested in that end of the distribution.)
• Since z = [ test statistic ] </> [ critical z-score], the test statistic lies in/outside the critical region, so the result is / is not significant. There is sufficient / insufficient evidence at the [ α% ] level of significance to reject H 0 and support the claim that …

Example of a 1-tailed test

A drinks machine formerly distributed drinks of volume X ~ N (180, 16). After an overhaul, a random sample of 20 drinks is measured and the sample mean is found to be 178ml. Does this data provide evidence at the 5% level of significance that the machine is dispensing a mean volume that is less than 180ml?

(The model solution below uses the writing frame above.)

Let μ be the mean volume dispensed across all operations by the machine. H 0 : μ = 180 H 1 : μ < 180           [This is a 1-tailed test since the direction is specified] Significance level α = 0.05

p-value method:

Since 0.0127 < 0.05 (i.e. a probability of 0.0127 is lower than the threshold probability of 0.05), the result is significant. Conclusion: There is sufficient evidence at the 5% level of significance to reject H 0 and support the claim that the machine is now dispensing a mean volume less than 180ml.

Critical region method:

For P(Z < z) = 0.05, Z = –1.6449 (This is the boundary of critical region, i.e. the bottom 5% of the distribution of sample means goes up to 1.6449 standard deviations below the original population mean. The sample mean that we’ve actually observed is 2.2361 standard deviations below the mean so it’s in that bottom 5% of the bell curve.)

Since the test statistic z = –2.2361 < –1.6449, the test statistic lies in the critical region, so the result is significant. Conclusion: There is sufficient evidence at the 5% level of significance to reject H 0 and support the claim that the machine is now dispensing a mean volume less than 180ml.

Alternative critical region method:

Since 178 < 178.53 (i.e. the observed sample mean is further away from the (old) mean than the critical value is), the test statistic lies in the critical region, so the result is significant. Conclusion: There is sufficient evidence at the 5% level of significance to reject H 0 and support the claim that the machine is now dispensing a mean volume less than 180ml.

In other words

If the mean hasn’t changed from its previous value of 180 then the probability of a random sample of 20 drinks having a mean volume of 178 ml or less, is less than 5%, so it’s likely that the mean has changed.

However, we mustn’t forget that the significance level of 5% means there’s a 5% probability that the decision to reject H 0 was wrong!

2-tailed test

If the question had asked whether the evidence suggested that the new mean volume was not equal to 180ml then a 2-tailed test would be required, so you would use α/2 = 0.025.

In the critical region method, this would mean that the critical value would be the z-score below which only 2.5% of the distribution would lie.

In fact, in this particular case the probability of the observed sample mean was still less than the 2.5% threshold, so we’d still have rejected H₀.

The mean time taken by Parkrun participants at a particular location is historically 30.3 minutes, with a variance of 64 minutes, and may be assumed to be normally distributed. After a change is made to the route, a random sample of 400 runners is found to have a mean time of 30.9 minutes. Does this data provide evidence at the 5% significance level that the mean time for the course has changed?

Work through the writing frame, using the example solution above as a model, then check your answers by clicking on the link below.

Click here for the solution

More practice questions

1. The weight, in grams, of apples from a tree follows the distribution N (102, 49). After a new type of fertiliser is trialled, a random sample of 36 apples has a mean weight of 104g. The supplier of the fertiliser claims that the mean weight of the apples has increased. Assuming that the variance remains unchanged, test this claim at (a) the 5% level; (b) the 1% level.

2. The waiting time at a doctor’s surgery, in minutes, may be assumed to be normally distributed with mean 18 and variance 16. After a new booking system is brought in, the practice manager suggests that the mean waiting time has been reduced. A random sample of 60 patients gives a mean waiting time of 17.0 minutes. Test the claim at the 5% significance level.

3. The mean exam mark achieved on a particular paper in one year (as a percentage) follows the distribution N (57, 15 2 ). The following year, a random sample of 100 candidates achieves a mean of 60%. Test, at the 5% significance level, the claim that the mean mark has changed.

Click here for the answers

That covers hypothesis testing for the mean of a Normal distribution. There’s one more type of hypothesis testing to cover for A-level Maths, and that’s to establish whether correlation exists in a population.

If you’ve found this article helpful then please share it with anyone else who you think would benefit (use the social sharing buttons if you like). If you have any suggestions for improvement or other topics that you’d like to see covered, then please comment below or drop me a line using my contact form .

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Let μ be the mean time taken by all Parkrun participants. H 0 : μ = 30.3 H 1 : μ ≠ 30.3    Significance level α =0.05 This is a 2-tailed test so we use α/2 = 0.025.

Don’t forget to do a sketch!

Since 0.0668 [ p-value ] > 0.05 [ α ], the result is not significant. There is insufficient evidence at the 5% level of significance to reject H 0 and support the claim that the mean time for the course has changed.

Critical value method:

For P(Z > z) = 0.025, Z = 1.9600 (boundary of critical region, i.e. the top 2.5% of the distribution of sample means starts 1.96 standard deviations above the original population mean)

Since the test statistic z = 1.5 < 1.9600, the test statistic does not lie in the critical region, so the result is not significant. There is insufficient evidence at the 5% level of significance to reject H 0 and support the claim that the mean time for the course has changed.

Alternative critical value method:

Since 30.9 < 31.084, the observed value does not lie in the critical region, so the result is not significant. There is insufficient evidence at the 5% level of significance to reject H 0 and support the claim that the mean time for the course has changed.

Click here to return to question

1. p = 0.0432; z = 1.7143; (a) sufficient evidence to reject H 0         (b) insufficient evidence to reject H 0 2. p = 0.0984; z = -1.2910; insufficient evidence to reject H 0 3. p = 0.0227; z = 2; sufficient evidence to reject H 0

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