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 \\( \alpha = 0.10 \\), what should you conclude?
\\( \

$$\begin{array}{|c|c|} \\hline \\text{region 1} & \\text{region 2} \\\\ \\hline \\bar{x}_1 = \\$67,800 & \\bar{x}_2 = \\$58,000 \\\\ \\sigma_1 = \\$8950 & \\sigma_2 = \\$9125 \\\\ n_1 = 43 & n_2 = 40 \\\\ \\hline \\end{array}$$

click here to view page 1 of the standard normal distribution table.
click here to view page 2 of the standard normal distribution table.
\\( \bigcirc \\) d. \\( h_0: \mu_1 - \mu_2 \geq 6000 \\) \\( h_a: \mu_1 - \mu_2 < 6000 \\) \\( \bigcirc \\) e. \\( h_0: \mu_1 - \mu_2 > 6000 \\) \\( h_a: \mu_1 - \mu_2 \geq 6000 \\) \\( \bigcirc \\) f. \\( h_0: \mu_1 - \mu_2 > 6000 \\) \\( h_a: \mu_1 - \mu_2 \leq 6000 \\)
determine the critical value(s).
\\( z_0 = 1.28 \\)
(round to two decimal places as needed. use a comma to separate answers as needed.)
determine the rejection region.
select the correct choice below, and, if necessary, fill in any answer boxes to complete your choice.
(round to two decimal places as needed.)
\\( \bigcirc \\) a. \\( z < \square \\)
\\( \bigcirc \\) b. \\( z > \square \\)
\\( \bigcirc \\) c. \\( z < \square \\) and \\( z > \square \\)

Explanation:

Step1: Identify Hypotheses

We test if \(\mu_1 - \mu_2> 6000\). So \(H_0: \mu_1 - \mu_2\leq6000\), \(H_a: \mu_1 - \mu_2 > 6000\) (one - tailed test, right - tailed as we care about difference > 6000).

Step2: Determine Critical Value

For a right - tailed test with \(\alpha = 0.10\), we look at the standard normal distribution. The critical value \(z_0\) is such that \(P(Z > z_0)=\alpha = 0.10\). From standard normal tables, \(z_0 = 1.28\) (since \(P(Z\leq1.28)=0.90\), so \(P(Z > 1.28)=0.10\)).

Step3: Rejection Region

In a right - tailed test, we reject \(H_0\) when the test statistic \(z>z_0\). Since \(z_0 = 1.28\), the rejection region is \(z > 1.28\).

Step4: Calculate Test Statistic (Optional for this part, but to confirm)

The formula for the test statistic in two - sample z - test (when \(\sigma_1,\sigma_2\) are known) is:

$$ z=\frac{(\bar{x}_1-\bar{x}_2)-d_0}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}} $$

where \(d_0 = 6000\), \(\bar{x}_1 = 67800\), \(\bar{x}_2 = 58000\), \(\sigma_1 = 8950\), \(\sigma_2 = 9125\), \(n_1 = 43\), \(n_2 = 40\)

First, calculate \(\bar{x}_1-\bar{x}_2=67800 - 58000=9800\)

Then, calculate the denominator:

$$ \sqrt{\frac{8950^{2}}{43}+\frac{9125^{2}}{40}}=\sqrt{\frac{80102500}{43}+\frac{83265625}{40}}=\sqrt{1862848.84 + 2081640.625}=\sqrt{3944489.465}\approx1986.07 $$

The test statistic \(z=\frac{(9800 - 6000)}{1986.07}=\frac{3800}{1986.07}\approx1.91\)

Since \(1.91>1.28\) (test statistic is in rejection region), we reject \(H_0\). But for the rejection region part of the question:

Answer:

The critical value is \(z_0 = 1.28\). The rejection region is \(z>1.28\), so the correct choice for rejection region is B. \(z > 1.28\)