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 α = 0.025, answer parts (a) through (e). assume the population variances are not equal. if convenient, use technology to solve the problem. (b) find the critical value(s) and identify the rejection region(s). enter the critical value(s) below. 2.228 (type an integer or decimal rounded to three decimal places as needed. use a comma to separate answers as needed.) select the correct rejection region(s) below. a. t < -t₀, t > t₀ b. t < -t₀ c. -t₀ < t < t₀ d. t > t₀ (c) find the standardized test statistic. t = □ (type an integer or decimal rounded to three decimal places as needed.)

Explanation:

Step1: Recall the formula for the t - test statistic for two - sample t - test (unequal variances)

The formula is \(t=\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\)

Step2: Identify the values from the problem

We have \(\bar{x}_{1} = 40500\), \(s_{1}=8200\), \(n_{1}=13\), \(\bar{x}_{2}=38800\), \(s_{2}=5400\), \(n_{2}=11\)

Step3: Substitute the values into the formula

First, calculate the numerator: \(\bar{x}_{1}-\bar{x}_{2}=40500 - 38800=1700\)

Then, calculate the denominator:
\(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}=\frac{8200^{2}}{13}+\frac{5400^{2}}{11}=\frac{67240000}{13}+\frac{29160000}{11}\)
\(=5172307.692+2650909.091 = 7823216.783\)
\(\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{7823216.783}\approx2797.00\)

Finally, \(t=\frac{1700}{2797.00}\approx0.608\)

Answer:

\(t = 0.608\)