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.
(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.)
a. \\(z < \square\\)
b. \\(z > 1.28\\)
c. \\(z < \square\\) and \\(z > \square\\)
calculate the standardized test statistic for \\(\mu_1 - \mu_2\\).
\\(z = \square\\)
(round to two decimal places as needed.)

Explanation:

Step1: Identify the formula for z - statistic

The formula for the standardized test statistic \( z \) for the difference between two population means (\( \mu_1-\mu_2 \)) when the population standard deviations \( \sigma_1 \) and \( \sigma_2 \) are known is:

$$ z=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} $$

Here, we are testing if the difference between the means is more than \( \$6000 \), so the null hypothesis \( H_0:\mu_1-\mu_2 = 6000 \) and the alternative hypothesis \( H_a:\mu_1-\mu_2>6000 \). So \( (\mu_1 - \mu_2)=6000 \), \( \bar{x}_1 = 67800 \), \( \bar{x}_2 = 58000 \), \( \sigma_1 = 8950 \), \( \sigma_2 = 9125 \), \( n_1 = 43 \), \( n_2 = 40 \).

Step2: Calculate the numerator

First, calculate \( (\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2) \):
\( \bar{x}_1-\bar{x}_2=67800 - 58000=9800 \)
Then, \( 9800-6000 = 3800 \)

Step3: Calculate the denominator

Calculate \( \frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2} \):
\( \frac{\sigma_1^2}{n_1}=\frac{8950^2}{43}=\frac{80102500}{43}\approx1862848.84 \)
\( \frac{\sigma_2^2}{n_2}=\frac{9125^2}{40}=\frac{83265625}{40} = 2081640.625 \)
Sum of these two: \( 1862848.84+2081640.625=3944489.465 \)
Take the square root: \( \sqrt{3944489.465}\approx1986.07 \)

Step4: Calculate the z - statistic

Now, divide the numerator by the denominator:
\( z=\frac{3800}{1986.07}\approx1.91 \)

Answer:

The standardized test statistic \( z\approx\boxed{1.91} \)