Question

A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes. A random sample of 15 students was selected and the sample standard deviation for the time needed to complete the exam was found to be 4.0 minutes. Using α=0.05, determine the critical value for this hypothesis test. Round to three decimal places.

Answer 1

Critical value for this hypothesis test is 6.571

Explanation:

We are given that a random sample of 15 students was selected and the sample standard deviation for the time needed to complete the exam was found to be 4.0 minutes.

A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes.

So, Null Hypothesis,
`H_0` :
`\sigma` = 5.0 minutes

Alternate Hypothesis,
`H_1` :
`\sigma` < 5.0 minutes

Now, the test statistics used here for testing population standard deviation is;

T.S. =
`((n-1) s^(2) )/(\sigma^(2) )` ~
`\chi^(2) __n_-_1`

where, s = sample standard deviation = 4.0 minutes

n = sample size = 15

So, test statistics =
`((15-1) 4^(2) )/(5^(2) )` = 8.96

At, 5% level of significance the chi-square table gives critical value of 6.571 at 14 degree of freedom. Since our test statistics is more than the critical value as 8.96 > 6.571 so we have insufficient evidence to reject null hypothesis.

Answer 2

The critical value for this hypothesis test is 6.571.

Explanation:

In this case the professor wants to determine whether the average number of minutes that a student needs to complete a statistics exam has a standard deviation that is less than 5.0 minutes.

Then the variance will be,
`\sigma^(2)=(5.0)^(2)=25`

The hypothesis to determine whether the population variance is less than 25.0 minutes or not, is:

H₀: The population variance is not less than 25.0 minutes, i.e. σ² = 25.

Hₐ: The population variance is less than 25.0 minutes, i.e. σ² < 25.

The test statistics is:

`\chi ^(2)_(cal.)=(ns^(2))/(\sigma^(2))`

The decision rule is:

If the calculated value of the test statistic is less than the critical value,
`\chi^(2)_(n-1)` then the null hypothesis will be rejected.

Compute the critical value as follows:

`\chi^(2)_((1-\alpha), (n-1))=\chi^(2)_((1-0.05),(15-1))=\chi^(2)_(0.95, 14)=6.571`

*Use a chi-square table.

Thus, the critical value for this hypothesis test is 6.571.