1. ## Confidence interval question?

I'm not too sure what to do in this problem.

Consider Review Exercise 9.101. Let us assume that the data have not been collected yet. Let us also assume that previous statistics suggest that $\displaystyle \sigma_1 = \sigma_2 = \$4,000$. Are the sample sizes in Review Exercise 9.101 sufficient to produce a 95% confidence interval on$\displaystyle \mu_1 - \mu_2$having a width of only$1,000?

Here's review question 9.101.

A survey was done with the hope of comparing salaries of chemical plant managers employed in two areas of the country, the northern and western regions. Independent random samples of 300 plant managers were selected for each of the two regions. These managers were asked for their annual salaries. The results are

$\displaystyle \begin{tabular}{c c} North & West Central\\ \hline Xbar_{1} =\$102,300 & Xbar_{2} =\$98,500\\ s_1 =\$5,700 & s_2 =\$3,800\\ \end{tabular}$

I don't have much of an idea of how to solve this problem. Any help would be much appreciated. Thanks

Edit:

I'm pretty sure this is wrong, but here is my attempt to solve.

$\displaystyle 1000-1.96(\sqrt{\frac{4000^2}{x}+\frac{4000^2}{x}}<\mu_ 1 - \mu_2 <1000+1.96(\sqrt{\frac{4000^2}{x}+\frac{4000^2}{x} }$

Then I took $\displaystyle 1.96(\sqrt{\frac{4000^2}{x}+\frac{4000^2}{x})$ and set it equal to 500, so there would be a $1,000 spread. Solving for x, I got:$\displaystyle 1.96(\sqrt{\frac{4000^2}{x}+\frac{4000^2}{x}}=500\displaystyle \frac{2(4000^2)}{x}=(\frac{500}{1.96})^2\displaystyle x=\frac{32,000,000}{(\frac{500}{1.96})^2}=491.7248$So, a sample of about 492 from each of the two areas of the country being studied would be needed to produce a 95% confidence interval on$\displaystyle \mu_1 - \mu_2 $having a spread of only$1,000.