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.

○ 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 need

Explanation:

Step1: Determine the test type and degrees of freedom

This is a two - sample t - test with unequal variances. The formula for the degrees of freedom \(df\) when variances are unequal is \(df=\frac{(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}})^{2}}{\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}+\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}}\)

For County A: \(n_{1} = 17\), \(\bar{x}_{1}=42400\), \(s_{1}=8900\)

For County B: \(n_{2}=8\), \(\bar{x}_{2} = 38000\), \(s_{2}=5100\)

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, \((\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}})^{2}=(4659411.76 + 3251250)^{2}=(7910661.76)^{2}\approx6.258\times10^{13}\)

Next, \(\frac{(\frac{s_{1}^{2}}{n_{1}})^{2}}{n_{1}-1}=\frac{(4659411.76)^{2}}{16}\approx\frac{2.171\times10^{13}}{16}\approx1.357\times10^{12}\)

\(\frac{(\frac{s_{2}^{2}}{n_{2}})^{2}}{n_{2}-1}=\frac{(3251250)^{2}}{7}=\frac{1.057\times10^{13}}{7}\approx1.510\times10^{12}\)

\(df=\frac{6.258\times10^{13}}{1.357\times10^{12}+1.510\times10^{12}}=\frac{6.258\times10^{13}}{2.867\times10^{12}}\approx21.83\approx21\) (we can also use a calculator or software for more accurate calculation. Alternatively, some methods use the smaller of \(n_1 - 1\) and \(n_2 - 1\), but the formula above is more accurate for unequal variances)

Step2: Find the critical value

We are performing a right - tailed test with \(\alpha = 0.05\) and \(df\approx21\) (using the t - distribution table or a calculator like the t - inv function in Excel. The t - inv function in Excel for a right - tailed test with \(\alpha=0.05\) and \(df = 21\) is \(t_{0.05,21}\))

Using a t - table or calculator, the critical value \(t_{\alpha,df}=t_{0.05,21}\approx1.721\)

The rejection region is \(t>1.721\)

Answer:

The critical value is \(1.721\) and the rejection region is \(t > 1.721\)