Chapter 10, Problem 14CQ

To determine

To find: The $90\%$ confidence interval for the difference ${\mu }_1-{\mu }_2$ .

Answer to Problem 14CQ

The $90\%$ confidence interval for the difference ${\mu }_1-{\mu }_2$ is $\left(1832.7054,3253.2946\right)$ .

Explanation of Solution

Given information:

The sample is $76$ , the mean annual income is $\$34214$ and the standard deviation is $\$2171$ . The sample in neighboring town is $88$ , the mean income is $\$31671$ with standard of deviation of $\$3279$ .

Concept used:

Excel formula is used.

Calculation:

The point estimate for the difference is,

  $\begin{array}{c}{\mu }_1-{\mu }_2=\overline{x_1}-\overline{x_2}\\ =34214-31671\\ =\$2543\end{array}$

The formula for degree of freedom is,

  $\begin{array}{c}dof=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}}\\ =\frac{{\left(\frac{{\left(2171\right)}^2}{76}+\frac{{\left(3279\right)}^2}{88}\right)}^2}{\frac{{\left(\frac{{\left(2171\right)}^2}{76}\right)}^2}{76-1}+\frac{{\left(\frac{{\left(3279\right)}^2}{88}\right)}^2}{76-1}}\\ =152.2363\\ =152\end{array}$

Assume level of significance as $0.10$ .

For $152$ degree of freedom and $0.10$ level of significance, the critical value is,

  $\begin{array}{c}{t}_c=\left[=\text{TINV}\left(\text{probability,deg\_freedom}\right)\right]\\ =\left[=\text{TINV}\left(\text{0}\text{.1,152}\right)\right]\\ =1.6549\\ =1.655\end{array}$

The formula for margin of error is,

  $\begin{array}{c}E=t_c\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =1.655\sqrt{\frac{{\left(2171\right)}^2}{76}+\frac{{\left(3279\right)}^2}{88}}\\ =710.294573\\ =710.3\end{array}$

The confidence interval is given as,

  $\begin{array}{c}CI=\left(\overline{x_1}-\overline{x_2}\right)\pm t_c\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\left(\overline{x_1}-\overline{x_2}\right)\pm E\\ =2543\pm 710.294573\\ =1832.7054,3253.2946\end{array}$

Therefore, the $90\%$ confidence interval for the difference ${\mu }_1-{\mu }_2$ is $\left(1832.7054,3253.2946\right)$ .