Question

a personnel director in a particular state claims that the mean annual income is the same in one of the counties (county a) as it is in another county (county b). in county a, a random sample of 18 residents has a mean annual income of $41,100 and a standard deviation of $8400. in county b, a random sample of 8 residents has a mean annual income of $38,400 and a standard deviation of $5700. at α=0.10, answer parts (a) through (e). assume the population variances are not equal. assume the samples are random and independent, and the populations are normally distributed.

c. \the mean annual incomes in counties a and b are equal.\
d. \the mean annual income in county a is greater than in county b.\

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 alternative hypothesis, hₐ
the null hypothesis, h₀

(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.)

Explanation:

Step1: Identify Test Type

This is a two - sample t - test for means with unequal variances. The degrees of freedom for the Welch - Satterthwaite approximation is calculated as \(df=\frac{(\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2})^{2}}{\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}+\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}}\)

Given \(n_1 = 18\), \(s_1=8400\), \(n_2 = 8\), \(s_2 = 5700\)

First, calculate \(\frac{s_1^{2}}{n_1}=\frac{8400^{2}}{18}=\frac{70560000}{18}=3920000\)

\(\frac{s_2^{2}}{n_2}=\frac{5700^{2}}{8}=\frac{32490000}{8}=4061250\)

Then, \((\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2})^{2}=(3920000 + 4061250)^{2}=(7981250)^{2}=6.3699390625\times10^{13}\)

\(\frac{(s_1^{2}/n_1)^{2}}{n_1 - 1}=\frac{3920000^{2}}{17}=\frac{15366400000000}{17}\approx9.0390588235\times10^{11}\)

\(\frac{(s_2^{2}/n_2)^{2}}{n_2 - 1}=\frac{4061250^{2}}{7}=\frac{16493751562500}{7}\approx2.3562502232\times10^{12}\)

\(df=\frac{6.3699390625\times10^{13}}{9.0390588235\times10^{11}+2.3562502232\times10^{12}}=\frac{6.3699390625\times10^{13}}{3.2601561055\times10^{12}}\approx19.54\approx20\) (we can also use a calculator or software for more accurate calculation, but for simplicity, we can approximate the degrees of freedom. Alternatively, we can use the smaller of \(n_1 - 1=17\) and \(n_2 - 1 = 7\), but the Welch - Satterthwaite is more appropriate. However, since \(\alpha=0.10\) and it's a two - tailed test, we can also use the t - distribution table with a reasonable approximation of degrees of freedom. Let's use \(df\approx19\) (a common approximation) or we can calculate more accurately.

Step2: Find Critical Values

For a two - tailed test with \(\alpha = 0.10\), the significance level in each tail is \(\alpha/2=0.05\)

Using a t - distribution table or a calculator (e.g., in R, qt(0.95, df = 19)), the critical values are \(t_{\alpha/2}\) and \(-t_{\alpha/2}\)

For \(df = 19\), \(t_{0.05,19}=1.729\) (from t - table)

So the critical values are \(\pm1.729\)

Answer:

The critical values are \(- 1.729,1.729\)