Question

1. (8 pts) The superintendent of a large school district speculated that high school students involved in extracurricular activities had a lower mean number of absences per year than high school students not in extracurricular activities. She generated a random sample of students from each group and recorded the number of absences each student had in the most recent school year. The data are listed below. Test the superintendent’s claim at the α=.01 significance level. [To receive full credit, your response should be sure to state your hypotheses, check the relevant requirement(s), find a test value, find a critical value or p-value (either is OK), make your decision, and state your conclusion.]

Students in EC Activities: 4 1 2 0 5 6 2 1 0 3 0 1 1 4 8 6 9 2 0 4 2 10 5 6 1

Students not in EC Activities: 5 7 0 9 4 3 12 8 4 2 0 5 5 4 9 14 6 10 9 6 3

Answer 1

1 represents sample data of Students in EC Activities.

2 represents sample data of Students not in EC Activities.

.

The sample means are shown below:

Also, the sample standard deviations are:

and the sample sizes are n1​=25 and n2​=21.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: μ1​ < μ2​

This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.327 and FU​=3.222, and since F = 0.637, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region

The significance level is α = 0.01, and the degrees of freedom are df = 44.

In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this left-tailed test is tc​=−2.414, for α=0.01 and df = 44.

The rejection region for this left-tailed test is R = { t : t < −2.414}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that t = −0.825 ≥ tc ​= −2.414, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.207, and since p = 0.207 ≥ 0.01, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is less than μ2​, at the 0.01 significance level.

Therefore, there is not enough evidence to claim that the high school students involved in extracurricular activities had a lower mean number of absences per year than high school students not in extracurricular activities at the 0.01 significance level.