Question

In Exercises 5–16, test the given claim.

Professor Evaluation Scores Listed below are student evaluation scores of female professorsand male professors from Data Set 17 “Course Evaluations” in Appendix B. Use a 0.05 significance level to test the claim that female professors and male professors have evaluation scores with the same variation.

Female	4.4	3.4	4.8	2.9	4.4	4.9	3.5	3.7	3.4	4.8
Male	4	3.6	4.1	4.1	3.5	4.6	4	4.3	4.5	4.3

Short answer

There is enough evidence to reject the claim that female professors and male professors have evaluation scores with the same variation.

Step-by-step solution

Given information

Two samples are taken showing the student evaluation scores where one represents scores by female professors,and the other represents scores by male professors

It is claimed that the variationinthe scores by female professors is equal to the variation in the scores by male professors.

Hypotheses

Let\({\sigma _1}\)and\({\sigma _2}\)be the population standard deviationsof the scores by female professors and the scores by male professors, respectively.

Null Hypothesis:The population standard deviation of the scores by female professors is equal to the population standard deviation of scores by male professors.

Symbolically,

\({H_0}:{\sigma _1} = {\sigma _2}\)

Alternate Hypothesis: The population standard deviation of the scores by female professors is equal to the population standard deviation of scores by male professors.

Symbolically,

\({H_1}:{\sigma _1} \ne {\sigma _2}\)

Sample mean, sample size and sample variances

The sample variance has the following formula:

\({s^2} = \frac{1}{{n - 1}}\sum\limits_{i = 1}^n {{{\left( {x - \bar x} \right)}^2}} \)

The sample mean score by female professors is equal to:

\(\begin{array}{c}{{\bar x}_1} = \frac{{4.4 + 3.4 + ....... + 4.8}}{{10}}\\ = 4.02\end{array}\)

The sample variance of the scores by female professors is computed below:

\(\begin{array}{c}s_{female}^2 = \frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_i} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}\\ = \frac{{{{\left( {4.4 - 4.02} \right)}^2} + {{\left( {3.4 - 4.02} \right)}^2} + ....... + {{\left( {4.88 - 4.02} \right)}^2}}}{{10 - 1}}\\ = 0.52\end{array}\)

Thus, the sample variance of the scores by female professors is equal to 0.52.

The sample mean score by male professors is equal to:

\(\begin{array}{c}{{\bar x}_2} = \frac{{4 + 3.6 + ....... + 4.3}}{{10}}\\ = 4.1\end{array}\)

The sample variance of the scores by male professors is computed below:

\(\begin{array}{c}s_{male}^2 = \frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_i} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}\\ = \frac{{{{\left( {4 - 4.1} \right)}^2} + {{\left( {3.4 - 4.1} \right)}^2} + ....... + {{\left( {4.3 - 4.1} \right)}^2}}}{{10 - 1}}\\ = 0.12\end{array}\)

Thus, the sample variance of the scores by male professors is equal to 0.12.

Compute the test statistic

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

Here,\(s_1^2\)is the sample variance corresponding to female professors and has a value equal to 0.52.

\(s_2^2\)is the sample variance corresponding to male professors and has a value equal to 0.12.

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{0.52}}{{0.12}}\\ = 4.175\end{array}\)

State the critical value and the p-value

The value of the numerator degrees of freedom is equal to:

\(\begin{array}{c}{n_1} - 1 = 10 - 1\\ = 9\end{array}\)

The value of the denominator degrees of freedom is equal to:

\(\begin{array}{c}{n_2} - 1 = 10 - 1\\ = 9\end{array}\)

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to 9and denominator degrees of freedom equal to 9 for a right-tailed test.

The level of significance is equal to:

\(\begin{array}{c}\frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025\end{array}\)

Thus, the critical value is equal to 4.026.

The two-tailed p-value for F equal to 4.175 is equal to 0.0447.

Conclusion

Since the test statistic value is greaterthan the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

Thus, there is enough evidence to rejectthe claimthat female professors and male professors have evaluation scores with the same variation.