Question

Student Evaluations of Professors Use the sample data given in Exercise 1 and test the claim that evaluation ratings of female professors have the same median as evaluation ratings of male professors. Use a 0.05 significance level.

Short answer

There is not enough evidence to conclude that there is a difference in the median values between the evaluation ratings of male and female professors.

Step-by-step solution

Given information

Two samples show the evaluation ratings of female and male professors.

The significance level is 0.05.

Identify the hypotheses

The Wilcoxon rank-sum test is performed to test the difference betweenthe medians of the given samples.

The null hypothesis is as follows:

There is no difference in the median values of the evaluation ratings for female and male professors.

The alternative hypothesis is as follows:

There is a difference in the median values of the evaluation ratings for female and male professors.

It is a two-tailed test.

Assign ranks to the values after combining the samples

The ranks are computed by combining the two samples and tagging each observation with the sample name/number it comes from.

Rank 1 is assigned to the smallest observation; rank 2 is assigned to the next smallest observation, and soon until the largest value.

If the two or more observations have the same value, then theaverage (or mean) of the ranks is assigned to them.

The following table shows the ranks:

Student Evaluation	Sample Name	Ranks
3.9	F	11.5
3.4	F	4
3.7	F	8.5
4.1	F	14.5
3.7	F	8.5
3.5	F	6
4.4	F	21
3.4	F	4
4.8	F	24
4.1	F	14.5
2.3	F	1
4.2	F	17.5
3.6	F	7
4.4	F	21
3.8	M	10
3.4	M	4
4.9	M	25.5
4.1	M	14.5
3.2	M	2
4.2	M	17.5
3.9	M	11.5
4.9	M	25.5
4.7	M	23
4.4	M	21
4.3	M	19
4.1	M	14.5

Calculate the sum of the ranks

Compute the sum of ranks of females as shown below:

\(11.5 + 4 + 8.5 + .... + 21 = 163\)

Compute the sum of ranks ofmales as shown:

\(10 + 4 + 25.5 + .... + 14.5 = 188\)

Thus, the sum of the ranks offemales is 163 and is denoted by R.

Calculate the mean and the standard deviation

Let \({n_1}\) be the sample sizeoffemale professors.

Let \({n_2}\) be the sample sizeofmale professors.

Here,

\(\begin{array}{l}{n_1} = 14\\{n_2} = 12\end{array}\)

Compute the mean value \(\left( {{\mu _R}} \right)\).

\(\begin{array}{c}{\mu _R} = \frac{{{n_1}\left( {{n_1} + {n_2} + 1} \right)}}{2}\\ = \frac{{14\left( {14 + 12 + 1} \right)}}{2}\\ = 189\end{array}\)

Compute the standard deviation \(\left( {{\sigma _R}} \right)\).

\(\begin{array}{c}{\sigma _R} = \sqrt {\frac{{{n_1}{n_2}\left( {{n_1} + {n_2} + 1} \right)}}{{12}}} \\ = \sqrt {\frac{{14 \times 12\left( {14 + 12 + 1} \right)}}{{12}}} \\ = 19.44\end{array}\)

Calculate the test statistic

The test statistic is obtained as shown below:

\(\begin{array}{c}z = \frac{{R - {\mu _R}}}{{{\sigma _R}}}\; \sim N\left( {0,1} \right)\\ = \frac{{163 - 189}}{{19.44}}\\ = - 1.34\end{array}\)

Thus, the absolute value of the z-score is 1.34.

Determine the critical value andthe conclusion of the test

The critical value of z from the standard normal table for a two-tailed test, with \(\alpha = 0.05\), is 1.96.

As the test statistic value is less than the critical value, the null hypothesis fails to reject.

There is not enough evidence to conclude that there is a difference in the median values between the evaluation ratings of male and female professors. In other words, there is not enough evidence to reject the claim that the median values of the evaluation ratings for female and male professors are the same.