Testing Hypotheses. In Exercises 13–24, assume that a simple random sample has been and test the given claim. Unless specified by your instructor, use either the P-value method or the critical value method for testing hypotheses. Identify the null and alternative hypotheses, test statistic, P-value (or range of P-values), or critical value(s), and state the final conclusion that addresses the original claim.

Course Evaluations Data Set 17 “Course Evaluations” in Appendix B includes data from student evaluations of courses. The summary statistics are n = 93, x = 3.91, s = 0.53. Use a 0.05 significance level to test the claim that the population of student course evaluations has a mean equal to 4.00.

The summarised data for the student evaluation of the courses is as follows.

\(\begin{array}{l}n = 93\\\bar x = 3.91\\s = 0.53\end{array}\)

The significance level is 0.05 to test the claim that the mean of the course evaluations by the students is equal to 4.00.

The required conditions are verified as follows.

1. The sample, satisfies the condition of a simple random sampling, as it iss randomly collected from the student’s evaluation about the course.
2. The sample size of 93 is greater than 30. Thus, there is no need to check for normality of the sample, as the condition for the sample size is satisfied.

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As the value of \(\sigma \) is unknown, the t-test will be applied.

The null hypothesis\({H_0}\)represents the population mean of the student’s course evaluation, which is equal to 4. The alternate hypothesis\({H_1}\)represents the population mean of the student’s course evaluation, which is not equal to 4.

Let\(\mu \)be the population mean of the student’s course evaluation.

State the null and alternate hypotheses as follows.

\(\begin{array}{l}{H_0}:\mu = 4.00\\{H_1}:\mu \ne 4.00\end{array}\)

The degree of freedom is obtained by using the formula\(df = n - 1\),where\(n = 93\). So,

\(\begin{array}{c}df = 93 - 1\\ = 92\end{array}\).

The critical value can be obtained using the t-distribution table with 92 degrees of freedom and a significance level of 0.05 for a two-tailed test.

From the t-table, the value is obtained as 1.986\(\left( {{t_{\frac{{0.05}}{2}}}} \right)\)corresponding to row 92 and column 0.05 (two-tailed).

Thus, the critical values are \( \pm 1.986\).

Apply the t-test to compute the test statistic using the formula\(t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\).

Substitute the respective value in the above formula and simplify.

\(\begin{array}{c}t = \frac{{3.91 - 4}}{{\frac{{0.53}}{{\sqrt {93} }}}}\\ = - 1.638\end{array}\)

Reject the null hypothesis when the absolute value of the observed test statistic is greater than the critical value. Otherwise, fail to reject the null hypothesis.

In this case,

\(\begin{array}{c}\left| t \right| = \left| { - 1.638} \right|\\ = 1.638\\ < 1.9860\left( {{t_{\frac{{0.05}}{2}}}} \right)\end{array}\).

The absolute value of the observed test statistic is less than the critical value. This implies that the null hypothesis is failed to be rejected.

As the null hypothesis is failed to be rejected, it can be concluded that there is sufficient evidence to support the claim that the mean of the population of the student’s course evaluation is equal to 4.