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

Answer:

The critical value for this hypothesis test is 6.571.

Step-by-step 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:

$\chi ^{2}_{cal.}=\frac{ns^{2}}{\sigma^{2}}$

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.

Answer 2

Answer:

Critical value for this hypothesis test is 6.571

Step-by-step 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. = $\frac{(n-1) s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, s = sample standard deviation = 4.0 minutes

            n = sample size = 15

So, test statistics = $\frac{(15-1) 4^{2} }{5^{2} }$ = 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.