Question

Confidence Intervals. In Exercises 9–24, construct the confidence interval estimate of the mean.

Student Evaluations Listed below are student evaluation ratings of courses, where a rating of 5 is for “excellent.” The ratings were obtained at the University of Texas at Austin. (See Data Set 17 “Course Evaluations” in Appendix B.) Use a 90% confidence level. What does the confidence interval tell us about the population of college students in Texas?

3.8 3.0 4.0 4.8 3.0 4.2 3.5 4.7 4.4 4.2 4.3 3.8 3.3 4.0 3.8

Short answer

The mean attractiveness from the actual population will lie 90% of the time between 3.67and 4.17.

The observations were recorded from college students at the university. The sample might not be appropriate for the population of college students in Texas. Thus, the confidence interval does not tell anything about the population of students.

Step-by-step solution

Given information

The sample of 15 ratings is observed such that each rating varies from 1 to 10.

Check the requirements

The necessary conditions for using any sample data to construct confidence intervals are as follows.

The sample is collected from the population of college students in Texas that satisfies the condition of a simple random sampling. As the sample size is 15, which is less than 30, the condition for normality will only be satisfied if the data follows a normal distribution. This can be verified from the normal probability plot that the sample data points to, as shown below.

Compute the degree of freedom and the critical value

The degree of freedom is computed as follows.

$$\begin{gathered} df=n-1 \\ df=15-1 \\ df=14 \end{gathered}$$

For the 90% confidence level, the significance level is 0.10.

$$\begin{gathered} \alpha =1-0.90 \\ =0.10 \end{gathered}$$

Use the t-distribution table to obtain the critical value when $\alpha =0.10$and$df=14$ .

The critical value is obtained as 1.761 from the t-table corresponding to row 14 and column 0.10 (two-tailed).

Compute the margin of error 

Let $x$be the random variable that denotes the rating of females.

The sample mean can be obtained using the formula $\overline{x}=\frac{1}{15}\sum _{i=1}^{15}{x}_i$, where represents the data points in a sample.

Compute the sample mean as follows. So,

$$\begin{gathered} \overline{x}=\frac{\left(3.8+3+4+...+3.8\right)}{15} \\ =\frac{58.8}{15} \\ =3.92 \end{gathered}$$

.

Calculate the sample variance using the formula $s^2=\frac{1}{15-1}\sum _{i=1}^{15}{\left(x_i-\overline{x}\right)}^2$.

X	${\left(x-\overline{x}\right)}^2$
3.8	0.014
3.0	0.846
4.0	0.006
4.8	0.774
3.0	0.846
4.2	0.078
3.5	0.176
4.7	0.608
4.4	0.230
4.2	0.078
4.3	0.144
3.8	0.014
3.3	0.384
4.0	0.006
3.8	0.014
	$\sum _{i=1}^{15}{\left(x_i-\overline{x}\right)}^2=4.218$

Substitute $\sum _{i=1}^{15}{\left(x_i-\overline{x}\right)}^2=4.218$in the formula $s^2=\frac{1}{15-1}\sum _{i=1}^{15}{\left(x_i-\overline{x}\right)}^2$. So,

.$$\begin{gathered} s^2=\frac{1}{14}\times 4.218 \\ =\frac{4.218}{14} \\ =0.301 \end{gathered}$$

The square root of the sample variance is equal to the sample standard deviation. Thus, the sample standard deviation is given as follows.

$$\begin{gathered} s=\sqrt{0.301} \\ =0.5493 \end{gathered}$$

The margin of error is given by the formula $E=t_{\frac{\alpha }{2}}\times \frac{s}{\sqrt{n}}$.Substitute the respective value obtained from above in the equation and simplify to compute the margin of error. So,

$$\begin{gathered} E=1.761\times \frac{0.5493}{\sqrt{15}} \\ =0.2498 \end{gathered}$$

Construct the confidence interval 

The confidence interval is given as follows.

$$\begin{gathered} \overline{x}-E<\mu <\overline{x}-E \\ 3.92-0.2498<\mu <3.92+0.2498 \\ 3.67<\mu <4.17 \end{gathered}$$

Analyze the confidence interval   

Therefore, the mean attractiveness from the actual population will lie 90% of the time between and 4.17.

The sample is observed from one university of Texas, which is not appropriate for the population of all students in Texas. Thus, the confidence interval does not tell anything about the population of the students.