Question

an energy company wants to choose between two regions in a state to install energy-producing wind turbines. a researcher claims that the wind speed in region a is less than the wind speed in region b. to test the regions, the average wind speed is calculated for 60 days in each region. the mean wind speed in region a is 13.8 miles per hour. assume the population standard deviation is 2.9 miles per hour. the mean wind speed in region b is 15.1 miles per hour. assume the population standard deviation is 3.1 miles per hour. at α = 0.05, can the company support the researcher’s claim? complete parts (a) through (d) below.

h₀: μ₁ ≥ μ₂
hₐ: μ₁ < μ₂

(b) find the critical value(s) and identify the rejection region.
the critical value(s) is/are z₀ = -1.64.
(round to two decimal places as needed. use a comma to separate answers as needed.)
what is the rejection region? select the correct choice below and fill in the answer box(es) within your choice.
(round to two decimal places as needed.)
a. z < or z >
b. z < -1.64
c. z >

(c) find the standardized test statistic z.
z = (round to two decimal places as needed.)

Explanation:

Step1: Recall the formula for z - test statistic for two - sample means

The formula for the z - test statistic when comparing two population means (with known population standard deviations) is:

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

In our case, the null hypothesis \(H_0:\mu_1\geq\mu_2\) and the alternative hypothesis \(H_a:\mu_1 < \mu_2\), so \((\mu_1-\mu_2) = 0\) (under the null hypothesis). Let \(\bar{x}_1 = 13.8\) (mean of Region A), \(\sigma_1=2.9\) (population standard deviation of Region A), \(n_1 = 60\) (sample size of Region A), \(\bar{x}_2=15.1\) (mean of Region B), \(\sigma_2 = 3.1\) (population standard deviation of Region B), and \(n_2=60\) (sample size of Region B).

Step2: Substitute the values into the formula

First, calculate the denominator:

$$\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}=\sqrt{\frac{(2.9)^{2}}{60}+\frac{(3.1)^{2}}{60}}$$

$$=\sqrt{\frac{8.41 + 9.61}{60}}=\sqrt{\frac{18.02}{60}}$$

$$=\sqrt{0.3003}\approx0.548$$

Then, calculate the numerator: \((\bar{x}_1-\bar{x}_2)-0=(13.8 - 15.1)=- 1.3\)
Now, calculate the z - statistic:

$$z=\frac{-1.3}{0.548}\approx - 2.37$$

Answer:

\(z\approx - 2.37\)