Question

a personnel director in a particular state claims that the mean annual income is greater in one of the states counties (county a) than it is in another county (county b). in county a, a random sample of 13 residents has a mean annual income of $40,500 and a standard deviation of $8200. in county b, a random sample of 11 residents has a mean annual income of $38,800 and a standard deviation of $5400. at \\( \alpha = 0.025 \\), answer parts (a) through (e). assume the population variances are not equal. if convenient, use technology to solve the problem.
a. \the mean annual income in county a is greater than in county b.\
b. \the mean annual incomes in counties a and b are not equal.\
c. \the mean annual incomes in counties a and b are equal.\
d. \the mean annual income in county a is less than in county b.\
what are \\( h _ { 0 } \\) and \\( h _ { a } \\) ?
the null hypothesis, \\( h _ { 0 } \\), is \\( \mu _ { 1 } \leq \mu _ { 2 } \\). the alternative hypothesis, \\( h _ { a } \\), is \\( \mu _ { 1 } > \mu _ { 2 } \\)
which hypothesis is the claim?
the alternative hypothesis, \\( h _ { a } \\)
the null hypothesis, \\( h _ { 0 } \\)
(b) find the critical value(s) and identify the rejection region(s).
enter the critical value(s) below.
(type an integer or decimal rounded to three decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Determine the type of test

Since the alternative hypothesis is \(H_{a}:\mu_{1}>\mu_{2}\), this is a right - tailed test.

Step2: Calculate the degrees of freedom

The formula for degrees of freedom when variances are not equal is \(df=\frac{(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}})^{2}}{\frac{(s_{1}^{2}/n_{1})^{2}}{n_{1}-1}+\frac{(s_{2}^{2}/n_{2})^{2}}{n_{2}-1}}\)

Given \(n_{1} = 13\), \(s_{1}=8200\), \(n_{2}=11\), \(s_{2}=5400\)

\(\frac{s_{1}^{2}}{n_{1}}=\frac{8200^{2}}{13}=\frac{67240000}{13}\approx5172307.69\)

\(\frac{s_{2}^{2}}{n_{2}}=\frac{5400^{2}}{11}=\frac{29160000}{11}\approx2650909.09\)

\(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}\approx5172307.69 + 2650909.09=7823216.78\)

\((\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}})^{2}\approx(7823216.78)^{2}\)

\(\frac{(s_{1}^{2}/n_{1})^{2}}{n_{1}-1}=\frac{(5172307.69)^{2}}{12}\)

\(\frac{(s_{2}^{2}/n_{2})^{2}}{n_{2}-1}=\frac{(2650909.09)^{2}}{10}\)

\(df=\frac{(7823216.78)^{2}}{\frac{(5172307.69)^{2}}{12}+\frac{(2650909.09)^{2}}{10}}\approx19.97\approx20\) (using technology for a more accurate calculation)

Step3: Find the critical value

For a right - tailed test with \(\alpha = 0.025\) and \(df\approx20\), using the t - distribution table or technology, the critical value \(t_{\alpha}\) is \(t_{0.025,20}= 2.086\)

Answer:

\(2.086\)