Mathematics, 27.05.2020 23:08 whitefang3153

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 90 days in each region. The mean wind speed in Region A is 13.9 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 3.3 miles per hour. At alphaequals0.05, can the company support the researcher's claim? Complete parts (a) through (d) below.

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We conclude that the wind speed in Region A is less than the wind speed in Region B which means researcher's claim was correct.

Step-by-step explanation:

We are given that 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 90 days in each region.

The mean wind speed in Region A is 13.9 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 3.3 miles per hour.

Let $\mu_1$ = mean wind speed in Region A.

$\mu_2$ = mean wind speed in Region B.

SO, Null Hypothesis, H_0 : $\mu_1 \geq \mu_2$ {means that the wind speed in Region A is more than or equal to the wind speed in Region B}

Alternate Hypothesis, H_A : $\mu_1 < \mu_2$ {means that the wind speed in Region A is less than the wind speed in Region B}

The test statistics that would be used here Two-sample z-test statistics as we know about population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }$ ~ N(0,1)

where, $\bar X_1$ = sample mean wind speed in Region A = 13.9 miles per hour

$\bar X_2$ = sample mean wind speed in Region B = 15.1 miles per hour

$\sigma_1$ = population standard deviation in Region A = 2.9 miles per hour

$\sigma_2$ = population standard deviation in Region B = 3.3 miles per hour

n_1 = sample of days speed calculated in Region A = 90

n_2 = sample of days speed calculated in Region A = 90

So, the test statistics = $\frac{(13.9-15.1)-(0)}{\sqrt{\frac{2.9^{2} }{90}+\frac{3.3^{2} }{90} } }$

= -2.591

The value of z test statistics is -2.591.

Now, at 0.05 significance level the z table gives critical value of -1.645 for left-tailed test.

Since our test statistic is less than the critical value of z as -2.591 < -1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the wind speed in Region A is less than the wind speed in Region B which means researcher's claim was correct.

Answer from: Quest

no i do not : ( what is it?

Answer from: Quest

answer: what is the question you're trying to ask?

step-by-step explanation: