Question

A study was conducted to estimate the difference in the mean salaries of elementary school teachers from two neighboring states. A sample of 10 teachers from the Indiana had a mean salary of $28,900 with a standard deviation of $2300. A sample of 14 teachers from Michigan had a mean salary of $30,300 with a standard deviation of $2100. Determine a 95% confidence interval for the difference between the mean salary in Indiana and Michigan.(Assume population variances are different.)

Answer 1

Answer:

$(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70$

$(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698$

The confidence interval would be $-3301.70 \leq \mu \leq 501.698$ and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.

Step-by-step explanation:

We have the following info given by the problem

$\bar X_1 = 28900$ the sample mean for the salaries of teachers in Indiana

$s_1 = 2300$ the sample deviation for the salary of  teachers in Indiana

$n_1 =10$ the sample size from Indiana

$\bar X_2 = 30300$ the sample mean for the salaries of teachers in Michigan

$s_2 = 2100$ the sample deviation for the salary of  teachers in Michigan

$n_2 =14$ the sample size from Michigan

We want to find a confidence interval for the difference in the two means and the formula for this case is given by;

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom for this case are:

$df = n_1 +n_2 -2= 10+14-2 =22$

The confidence is 95%so then the significance is $\alpha=0.05$ and the $\alpha/2 =0.025$, we need to find a critical value in the t distribution with 22 degrees of freedom who accumulates 0.025 of the area on each tail and we got:

$t_{\alpha/2}= 2.07$

And now replacing in the formula for the confidence interval we got:

$(28900-30300) -2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = -3301.70$

$(28900-30300) +2.07 \sqrt{\frac{2300^2}{10} +\frac{2100^2}{14}} = 501.698$

The confidence interval would be $-3301.70 \leq \mu \leq 501.698$ and since the confidence interval contains the value of 0 we don't have enough evidence to conclude that the difference between the two states for the salary of teachers are significantly different.