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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.)

Respuesta :

Answer:

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

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

The confidence interval would be [tex] -3301.70 \leq \mu \leq 501.698[/tex] 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

[tex]\bar X_1 = 28900[/tex] the sample mean for the salaries of teachers in Indiana

[tex]s_1 = 2300[/tex] the sample deviation for the salary of  teachers in Indiana

[tex] n_1 =10[/tex] the sample size from Indiana

[tex]\bar X_2 = 30300[/tex] the sample mean for the salaries of teachers in Michigan

[tex]s_2 = 2100[/tex] the sample deviation for the salary of  teachers in Michigan

[tex] n_2 =14[/tex] 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;

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

The degrees of freedom for this case are:

[tex] df = n_1 +n_2 -2= 10+14-2 =22[/tex]

The confidence is 95%so then the significance is [tex]\alpha=0.05[/tex] and the [tex]\alpha/2 =0.025[/tex], 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:

[tex] t_{\alpha/2}= 2.07[/tex]

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

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

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

The confidence interval would be [tex] -3301.70 \leq \mu \leq 501.698[/tex] 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.

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