In Exercises 5–20, assume that the two samples are independent random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of n1−1 and n2−1).

Are male and female professors rated differently? According to Data Set 17 “Course Evaluations” Appendix B, given below are student evaluation scores of female professors and male professors. The test claims that female and male professors have the same mean evaluation ratings. Does there appear to be a difference?

Females	4.4	3.4	4.8	2.9	4.4	4.9	3.5	3.7	3.4	4.8
Males	4.0	3.6	4.1	4.1	3.5	4.6	4.0	4.3	4.5	4.3

Null hypothesis:female professors and male professors have the same mean evaluation ratings.

\({H_0}\):\({\mu _1} = {\mu _2}\).

Alternative hypothesis:female professors and male professors do not have the same mean evaluation ratings.

\({H_1}\):\({\mu _1} \ne {\mu _2}\).

Here, \({\mu _1},{\mu _2}\)are population mean evaluation ratings of female and male professors respectively

The sample mean of female professors is given by,

\(\begin{array}{c}{{\bar x}_1} = \frac{{\sum\limits_{i = 1}^{{n_1}} {{x_i}} }}{{{n_1}}}\\ = \frac{{4.4 + 3.4 + 4.8 + .... + 4.8}}{{10}}\\ = 4.02\end{array}\)

Therefore, the mean value of \({\bar x_1} = 4.02\)

The sample mean of male professors is given by,

\(\begin{array}{c}{{\bar x}_2} = \frac{{\sum\limits_{i = 1}^{{n_2}} {{x_i}} }}{{{n_2}}}\\ = \frac{{4.0 + 3.6 + 4.1 + .... + 4.3}}{{10}}\\ = 4.1\end{array}\)

The standard deviation for female professors is given by,

\(\begin{array}{c}{s_1} = \sqrt {\frac{{\sum\limits_{i = 1}^{{n_1}} {{{({x_i} - {{\bar x}_1})}^2}} }}{{{n_1} - 1}}} \\ = \sqrt {\frac{{{{\left( {4.4 - 4.02} \right)}^2} + {{\left( {3.4 - 4.02} \right)}^2} + .... + {{\left( {4.8 - 4.02} \right)}^2}}}{{10 - 1}}} \\ = 0.7208\end{array}\)

Therefore, the standard deviation is\({s_1} = 0.7208\)

The standard deviation for male professors is given by,

\(\begin{array}{c}{s_2} = \sqrt {\frac{{\sum\limits_{i = 1}^{{n_2}} {{{({x_i} - {{\bar x}_2})}^2}} }}{{{n_2} - 1}}} \\ = \sqrt {\frac{{{{\left( {4.0 - 4.1} \right)}^2} + {{\left( {3.6 - 4.1} \right)}^2} + .... + {{\left( {4.3 - 4.1} \right)}^2}}}{{10 - 1}}} \\ = 0.3528\end{array}\)

Therefore, the standard deviation is \({s_2} = 0.3528\)

When two samples are independent and have an equal but unknown standard deviation, use a t-test for two samples.

The test statistic is given by,

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{\left( {4.02 - 4.1} \right) - \left( 0 \right)}}{{\sqrt {\frac{{{{0.7208}^2}}}{{40}} + \frac{{{{0.3528}^2}}}{{40}}} }}\\ = - 0.3152\end{array}\)

Calculated value of test statistic is -0.315.

Degrees of freedom:

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

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

Degrees of freedom is the minimum of two values, which is 9.

The two-tailed test has two critical values.

Refer to the t-table for two-tailed test with a 0.05 level of significance and 9 degrees of freedom.

The critical value is given by,\({t_{\frac{{0.05}}{2}}} = 2.262\)

Thus, the critical value are \( \pm 2.262\)

The decision rule,

If\(\left| t \right| > {t_{\frac{\alpha }{2}}}\), reject the null hypothesis. Otherwise fail to reject the null hypothesis.

In this case, 0.315 < 2.262, which implies that the null hypothesis is failed to be rejected.

Thus, there is enough evidence to support the null hypothesis that both female and male professors have the same mean evaluation ratings.

Therefore, there does not appear to be any difference between the evaluation scores of two professors.