Question

In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and 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\({n_1} - 1\)and\({n_2} - 1\).)

Are Male Professors and Female Professors Rated Differently?

a. Use a 0.05 significance level to test the claim that two samples of course evaluation scores are from populations with the same mean. Use these summary statistics: Female professors:

n = 40, \(\bar x\)= 3.79, s = 0.51; male professors: n = 53, \(\bar x\) = 4.01, s = 0.53. (Using the raw data in Data Set 17 “Course Evaluations” will yield different results.)

b. Using the summary statistics given in part (a), construct a 95% confidence interval estimate of the difference between the mean course evaluations score for female professors and male professors.

c. Example 1 used similar sample data with samples of size 12 and 15, and Example 1 led to the conclusion that there is not sufficient evidence to warrant rejection of the null hypothesis.

Do the larger samples in this exercise affect the results much?

Short answer

a.There is sufficient evidence to reject the claim that thepopulation evaluation score of courses taught by female professors is equal to the population mean evaluation score of courses taught by male professors.

b. The 95% confidence interval of the difference in the two population means is equal to (-0.4397, -0.0003).

c. There is an effect of the large sample size on the conclusion of the claim.

Step-by-step solution

Given information

For a sample of 40 course evaluation scores of courses taught by female professors, the mean score is equal to 3.79, and the standard deviation is equal to 0.51. In another sample of 53 course evaluation scores of courses taught by male professors, the mean score is equal to 4.01, and the standard deviation is equal to 0.53.

It is claimed that the mean evaluation score of courses taught by female professors is equal to the mean evaluation score of courses taught by male professors.

Hypotheses

Let\({\mu _1}\)and\({\mu _2}\)be the population mean evaluation scores of courses taught by female professors and male professors, respectively.

Null hypothesis: The population mean evaluation score of courses taught by female professors is equal to the population mean evaluation score of courses taught by male professors.

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Symbolically,

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

Alternate hypothesis: The population mean evaluation score of courses taught by female professors is not equal to the population mean evaluation score of courses taught by male professors.

Symbolically,

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

Compute the test statistic

The sample mean score of courses taught by female professors\(\left( {{{\bar x}_1}} \right)\)is equal to 3.79.

The sample mean score of courses taught by male professors\(\left( {{{\bar x}_2}} \right)\)is equal to 4.01.

The sample standard deviation\(\left( {{s_1}} \right)\)of scores of courses taught by female professors is equal to 0.51.

The sample standard deviation\(\left( {{s_2}} \right)\)of scores of courses taught by female professors is equal to 0.53.

The sample size\(\left( {{n_1}} \right)\)is equal to 40, and the sample size\(\left( {{n_2}} \right)\)is equal to 53.

Apply the t-test to compute the test statistic as shown below.

\(\begin{array}{c}t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }}\\ = \frac{{\left( {3.79 - 4.01} \right) - 0}}{{\sqrt {\frac{{{{\left( {0.51} \right)}^2}}}{{40}} + \frac{{{{\left( {0.53} \right)}^2}}}{{53}}} }}\\ = - 2.025\end{array}\).

State the critical value and the p-value

The degree of freedom is the smaller of the two values\(\left( {{n_1} - 1} \right)\)and\(\left( {{n_2} - 1} \right)\).

The values are computed below.

\(\begin{array}{c}\left( {{n_1} - 1} \right) = 40 - 1\\ = 39\end{array}\)

\(\begin{array}{c}\left( {{n_2} - 1} \right) = 53 - 1\\ = 52\end{array}\)

Thus, the value of the degrees of freedom is equal to the smaller of the two values 39 and 52, which is 39.

The critical values can be obtained using the t-distribution table with degrees of freedom equal to 39 and the significance level equal to 0.05 for a two-tailed test.

Thus, the critical values are -2.0227 and 2.0227.

The p-value for t equal to -2.025 is equal to 0.0498.

Conclusion of the test

a.

As the test statistic value does not lie within the two critical values and the p-value is less than 0.05, the null hypothesis is rejected.

Therefore,there is sufficient evidence to reject the claim that thepopulation evaluation score of courses taught by female professors is equal to the population mean evaluation score of courses taught by male professors.

Confidence interval

b.

The confidence interval has the following expression:

\(CI = \left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E < \left( {{\mu _1} - {\mu _2}} \right) < \left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E\).

The margin of error is given by the following formula:
\(E = {t_{\frac{\alpha }{2}}}\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \).

Substitute the respective values in the above formula to compute the margin of error.

\(\begin{array}{c}E = 2.0227 \times \sqrt {\frac{{{{0.51}^2}}}{{40}} + \frac{{{{0.53}^2}}}{{53}}} \\ = 0.219745\end{array}\).

Substituting the value of E and the sample means, the following confidence interval is obtained:

\(\begin{array}{c}\left( {3.79 - 4.01} \right) - 0.219745 < \left( {{\mu _1} - {\mu _2}} \right) < \left( {3.79 - 4.01} \right) + 0.219745\\ - 0.4397 < \left( {{\mu _1} - {\mu _2}} \right) < - 0.0003\end{array}\)

Thus, the 95% confidence interval of the difference in the two population means is equal to (-0.4397, -0.0003).

Interpretation on the basis of the confidence interval

It can be observed that 0 does not lie within the interval. This implies that there is a significant difference between the two population means.

Change in conclusion due to an increase in sample sizes

c.

The values of\({n_1}\)and\({n_2}\)used in Example 1 are 12 and 15, respectively.

The value of\({n_1}\)and\({n_2}\)used for this exercise are 40 and 53, respectively.

The conclusion of the claim obtained in the example is thatthere is not sufficient evidence to reject the claim that thepopulation mean evaluation score of courses taught by female professors is equal to the population mean evaluation score of courses taught by male professors.

The conclusion of the claim obtained in the exercise is thatthere is sufficient evidence to reject the claim that thepopulation mean evaluation score of courses taught by female professors is equal to the population mean evaluation score of courses taught by male professors.

Thus, there is an effect of the large sample size on the conclusion of the claim.