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.
(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 = -2.37 (round to two decimal places as needed.)
(d) decide whether to reject or fail to reject the null hypothesis and interpret the decision in the context of the original claim.
□ h₀. there □ enough evidence at the 5% level of significance to □ the researcher’s claim that the wind speed in region a is □ the wind speed in region b.

Explanation:

Step1: Identify Hypotheses

The claim is \( \mu_A < \mu_B \), so \( H_0: \mu_A \geq \mu_B \), \( H_a: \mu_A < \mu_B \) (left - tailed test).

Step2: Rejection Region

For \( \alpha = 0.05 \) (left - tailed), critical \( z=-1.64 \). Rejection region: \( z < - 1.64 \).

Step3: Standardized Test Statistic

Formula for two - sample z - test: \( z=\frac{(\bar{x}_A-\bar{x}_B)-(\mu_A - \mu_B)}{\sqrt{\frac{\sigma_A^2}{n_A}+\frac{\sigma_B^2}{n_B}}} \)
Given \( \bar{x}_A = 13.8 \), \( \sigma_A = 2.9 \), \( n_A = 60 \); \( \bar{x}_B = 15.1 \), \( \sigma_B = 3.1 \), \( n_B = 60 \), \( \mu_A-\mu_B = 0 \) (under \( H_0 \))
\( z=\frac{(13.8 - 15.1)-0}{\sqrt{\frac{2.9^2}{60}+\frac{3.1^2}{60}}}=\frac{-1.3}{\sqrt{\frac{8.41 + 9.61}{60}}}=\frac{-1.3}{\sqrt{\frac{18.02}{60}}}=\frac{-1.3}{\sqrt{0.3003}}\approx\frac{-1.3}{0.5479}\approx - 2.37 \)

Step4: Decision Rule

Compare test statistic \( z=-2.37 \) with critical \( z=-1.64 \). Since \( - 2.37 < - 1.64 \), reject \( H_0 \).

Step5: Interpret

Reject \( H_0 \), so there is enough evidence at \( \alpha = 0.05 \) to support the claim that \( \mu_A < \mu_B \).

Answer:

Reject \( H_0 \). There is enough evidence at the 5% level of significance to support the researcher's claim that the wind speed in Region A is less than the wind speed in Region B.