91影视

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

Let\(\mu \)be the population mean of the student鈥檚 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鈥檚 course evaluation is equal to 4.