Question

Twenty years ago, entering male high school students of Central High could do an average of 24 pushups in 60 seconds. To see whether this remains true today, a random sample of 36 freshmen was chosen. Suppose their average was 22.5 with a sample standard deviation of 3.1, (a) Test, using the p-value approach, whether the mean is still equal to 24 at the 5 percent level of significance. (b) Calculate the power of the test if the true mean is 23

Answer 1

a) If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is still equal to 24 at 5% of significance.

b) Power =0.48994+0.0000501=0.490

Explanation:

Part a

Data given and notation

`\bar X=22.5` represent the sample mean

`s=3.1` represent the sample standard deviation

`n=36` sample size

`\mu_o =68` represent the value that we want to test

`\alpha=0.05` represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is still equal to 24, the system of hypothesis would be:

Null hypothesis:
`\mu = 24`

Alternative hypothesis:
`\mu \\eq 24`

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

`t=(\bar X-\mu_o)/((s)/(√(n)))` (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

`t=(22.5-24)/((3.1)/(√(36)))=-2.903`

P-value

The first step is calculate the degrees of freedom, on this case:

`df=n-1=36-1=35`

Since is a bilateral test the p value would be:

`p_v =2*P(t_((35))<-2.903)=0.0064`

Conclusion

If we compare the p value and the significance level given
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is still equal to 24 at 5% of significance.

Part b

The Power of a test is the probability of rejecting the null hypothesis when, in the reality, it is false.

For this case the power of the test would be:

P(reject null hypothesis|
`\mu=23`)

If we see the nul hypothesis we reject it when we have this:

The critical values from the normal standard distribution at 5% of significance are -1.96 and 1.96. From the z score formula:

`z=(\bar x-\mu)/((s)/(√(n)))`

And if we solve for
`\bar x` we got:

`\bar X= \mu \pm z (s)/(√(n))`

Using the two critical values we have the critical values four our sampling distribution under the null hypothesis

`\bar X= 24 -1.96 (3.1)/(√(36))=22.987`

`\bar X= 24 +1.96 (3.1)/(√(36))=25.012`

So we reject the null hypothesis if
`\bar x<22.987` or
`\bar X >25.012`

So for our case:

P(reject null hypothesis|
`\mu=23`) can be founded like this:

`P(\bar X <22.987|\mu=23)=P(z<(22.987-23)/((3.1)/(√(36))))=P(Z<-0.0251)=0.48994`

`P(\bar X >25.012|\mu=23)=P(z<(25.012-23)/((3.1)/(√(36))))=P(Z>3.89)=0.0000501`

And the power on this case would be the sum of the two last probabilities:

Power =0.48994+0.0000501=0.490