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 17 residents has a mean annual income of $42,400 and a standard deviation of $8900. in county b, a random sample of 8 residents has a mean annual income of $38,000 and a standard deviation of $5100. at α = 0.05, answer parts (a) through (e). assume the population variances are not equal. if convenient, use technology to solve the problem.

○ c. \the mean annual income in county a is less than in county b.\
○ d. \the mean annual incomes in counties a and b are not equal.\

what are h₀ and hₐ?
the null hypothesis, h₀, is μ₁ ≤ μ₂. the alternative hypothesis, hₐ, is μ₁ > μ₂.
which hypothesis is the claim?
○ the null hypothesis, h₀
✓ the alternative hypothesis, hₐ

(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: Identify Test Type

This is a two - sample t - test (unequal variances) for comparing means. We have \(n_1 = 17\), \(\bar{x}_1=42400\), \(s_1 = 8900\); \(n_2=8\), \(\bar{x}_2 = 38000\), \(s_2=5100\). The degrees of freedom for unequal variances is calculated using the formula \(df=\frac{(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2})^{2}}{\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}+\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}}\)

First, calculate \(\frac{s_1^{2}}{n_1}=\frac{8900^{2}}{17}=\frac{79210000}{17}\approx4659411.76\)

\(\frac{s_2^{2}}{n_2}=\frac{5100^{2}}{8}=\frac{26010000}{8} = 3251250\)

Then, \((\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2})^{2}=(4659411.76 + 3251250)^{2}=(7910661.76)^{2}\approx6.258\times10^{13}\)

\(\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}=\frac{(4659411.76)^{2}}{16}\approx\frac{2.171\times10^{13}}{16}\approx1.357\times10^{12}\)

\(\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}=\frac{(3251250)^{2}}{7}\approx\frac{1.057\times10^{13}}{7}\approx1.510\times10^{12}\)

\(df=\frac{6.258\times10^{13}}{1.357\times10^{12}+1.510\times10^{12}}=\frac{6.258\times10^{13}}{2.867\times10^{12}}\approx21.83\), we can round down to \(df = 21\) (or use more precise calculation).

Step2: Determine Critical Value

Since the alternative hypothesis is \(H_a:\mu_1>\mu_2\), it is a right - tailed test with \(\alpha = 0.05\) and \(df\approx21\) (using the rounded - down value). Looking up in the t - distribution table, the critical value \(t_{\alpha,df}\) for \(\alpha = 0.05\) and \(df = 21\) is \(t_{0.05,21}\approx1.721\)

Answer:

1.721