Question

is the difference between the mean annual salaries of statisticians in region 1 and region 2 more than $6000? to decide, you select a random sample of statisticians from each region. the results of each survey are shown to the right. at α = 0.10, what should you conclude?
region 1:
\\(\bar{x}_1 = \\$67,800\\)
\\(\sigma_1 = \\$8950\\)
\\(n_1 = 43\\)
region 2:
\\(\bar{x}_2 = \\$58,000\\)
\\(\sigma_2 = \\$9125\\)
\\(n_2 = 40\\)
click here to view page 1 of the standard normal distribution table.
click here to view page 2 of the standard normal distribution table.

choose the correct null and alternative hypotheses below.
a. \\(h_0: \mu_1 - \mu_2 \
eq 6000\\)
\\(h_a: \mu_1 - \mu_2 = 6000\\)
b. \\(h_0: \mu_1 - \mu_2 = 6000\\)
\\(h_a: \mu_1 - \mu_2 \
eq 6000\\)
c. \\(h_0: \mu_1 - \mu_2 \leq 6000\\)
\\(h_a: \mu_1 - \mu_2 > 6000\\)
d. \\(h_0: \mu_1 - \mu_2 \geq 6000\\)
\\(h_a: \mu_1 - \mu_2 < 6000\\)
e. \\(h_0: \mu_1 - \mu_2 > 6000\\)
\\(h_a: \mu_1 - \mu_2 \geq 6000\\)
f. \\(h_0: \mu_1 - \mu_2 > 6000\\)
\\(h_a: \mu_1 - \mu_2 \leq 6000\\)

determine the critical value(s).
\\(z_0 = \square\\)
(round to two decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Identify Test Type and Significance

This is a right - tailed z - test for the difference between two population means (since we know the population standard deviations \(\sigma_1\) and \(\sigma_2\)) with \(\alpha = 0.10\). For a right - tailed test, the critical value \(z_0\) is the value such that the area to the right of \(z_0\) under the standard normal curve is \(\alpha=0.10\).

Step2: Find the Critical Value

The area to the left of the critical value \(z_0\) is \(1 - \alpha=1 - 0.10 = 0.90\). We look up the \(z\) - score in the standard normal distribution table (or use a calculator with a normal - distribution function) that corresponds to an area of \(0.90\) to the left. Looking at the standard normal table, the \(z\) - score corresponding to a cumulative probability of \(0.90\) is approximately \(1.28\) (because \(P(Z\leq1.28)\approx0.90\)).

Answer:

\(z_0 = 1.28\)