Chapter 13.1, Problem 5E

a:

To determine

The difference between the two population means.

Explanation of Solution

The difference between the two population means (DP) can be calculated as follows:

$\begin{array}{c}\text{DP}=({\overline{x}}_1-{\overline{x}}_2)\pm t_{\frac{\alpha }{2}}\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ =({\overline{x}}_1-{\overline{x}}_2)\pm t_{\frac{95\%}{2}}\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ =(524-469)\pm 2.009\sqrt{\left(\frac{(25-1){129}^2+(25-1){131}^2}{25+25-2}\right)\left(\frac{1}{25}+\frac{1}{25}\right)}\\ =\text{ -18}\text{.87 and 73}\text{.87}\end{array}$

The difference between the two population means lies between -18.87 and 73.87.

b:

To determine

The new difference between the two population means.

Explanation of Solution

The difference between the two population means (DP) can be calculated as follows:

$\begin{array}{c}\text{DP}=({\overline{x}}_1-{\overline{x}}_2)\pm t_{\frac{\alpha }{2}}\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ =({\overline{x}}_1-{\overline{x}}_2)\pm t_{\frac{95\%}{2}}\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ =(524-469)\pm 2.009\sqrt{\left(\frac{(25-1){255}^2+(25-1){260}^2}{25+25-2}\right)\left(\frac{1}{25}+\frac{1}{25}\right)}\\ =-\text{91}\text{.33 and 201}\text{.33}\end{array}$

The difference between the two population means lies between -91.33 and 201.33.

c:

To determine

The impact of increasing standard deviation.

Explanation of Solution

When the standard deviation increases, it increases the difference in the interval.

d:

To determine

The difference between the two population means with sample size 100.

Explanation of Solution

The difference between the two population means (DP) can be calculated as follows:

$\begin{array}{c}\text{DP}=({\overline{x}}_1-{\overline{x}}_2)\pm t_{\frac{\alpha }{2}}\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ =({\overline{x}}_1-{\overline{x}}_2)\pm t_{\frac{95\%}{2}}\sqrt{s_p^2\left(\frac{1}{n_1}+\frac{1}{n_2}\right)}\\ =(524-469)\pm 1.972\sqrt{\left(\frac{(100-1){129}^2+(100-1){131}^2}{100+100-2}\right)\left(\frac{1}{100}+\frac{1}{100}\right)}\\ =\text{ 18}\text{.74 and 91}\text{.26}\end{array}$

The difference between the two population means lies between -18.87 and 91.26.

e:

To determine

The impact of increasing sample size.

Explanation of Solution

When the sample size increases, it reduces the difference in the interval.