Chapter 10.2, Problem 47E

(a)

To determine

To explain how can you tell the summary statistics that the distribution of earnings in each group is strongly skewed to the right and why.

Explanation of Solution

The summary statistics that the distribution of earnings in each group is strongly skewed to the right because earnings cannot take on negative values and the standard deviation is almost the same as the mean, thus distribution needs to be strongly skewed, because negative values are within two standard deviations of the mean and these values fall in the range of regular values for a symmetric distribution.

The sample sizes $675\text{ and 621}$ are at least $30$ and thus we can still assume that the distribution is approximately normal.

(b)

To determine

To construct and interpret $90\%$ confidence interval for the difference of true mean summer earnings of male and female students at the university.

Answer to Problem 47E

There is $90\%$ confidence that the mean difference is between $(412.69,635.58)$ .

Explanation of Solution

Given:

  $\begin{array}{l}{\overline{x}}_1=1884.52\\ {\overline{x}}_2=1360.39\\ n_1=376\\ n_2=621\\ s_1=1368.37\\ s_2=1037.46\\ c=0.90\end{array}$

Calculating confidence interval:

Now we will be calculating the t -value for this we need to find out the degree of freedom. Thus, the degree of freedom will be:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (675-1,621-1)\\ =620>100\\ df=100\end{array}$

Then the t -value will be as:

  $t*=1.660$

Thus the confidence interval be:

  $\begin{array}{l}({\overline{x}}_1-{\overline{x}}_2)-t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(1884.52-1360.39)-1.660\times \sqrt{\frac{{1368.37}^2}{675}+\frac{{1037.46}^2}{621}}\\ =412.69\\ ({\overline{x}}_1-{\overline{x}}_2)+t_{\alpha /2}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =(1884.52-1360.39)-1.660\times \sqrt{\frac{{1368.37}^2}{675}+\frac{{1037.46}^2}{621}}\\ =635.58\end{array}$

Thus, we conclude that there is $90\%$ confidence that the mean difference is between $(412.69,635.58)$ .

(c)

To determine

To interpret the $90\%$ confidence level in the context of this study.

Explanation of Solution

The $90\%$ confidence interval means that on average $90\%$ of all possible samples will have a confidence interval that contains the true population mean difference of earnings. This can be interpreted from this context of the study.