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A national organization has been working with utilities throughout the nation to find sites for large wind machines that generate electricity. Wind speeds must average more than 22 miles per hour (mph) for a site to be acceptable. Recently, the organization conducted wind speed tests at a particular site. Based on a sample of  n=33  wind speed recordings (taken at random intervals), the wind speed at the site averaged  22.8 mph, with a standard deviation of  4.3 mph. To determine whether the site meets the organization's requirements, perform the following hypothesis test at the 1% significance level and state your conclusion.

Question

A national organization has been working with utilities throughout the nation to find sites for large wind machines that generate electricity. Wind speeds must average more than 22 miles per hour (mph) for a site to be acceptable. Recently, the organization conducted wind speed tests at a particular site. Based on a sample of  n=33  wind speed recordings (taken at random intervals), the wind speed at the site averaged  22.8 mph, with a standard deviation of  4.3 mph. To determine whether the site meets the organization's requirements, perform the following hypothesis test at the 1% significance level and state your conclusion.

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Step 1

For a sample of 33 wind speed recordings, the mean wind speed is 22.8 mph and the standard deviation is 4.3mph. It is claimed that the average wind speed is more than 22mph. The level of significance for the test is α=0.01.

Since the population standard deviation is unknown, student’s t test can be used here.

The test hypotheses are given below:

Null hypothesis:

H0: µ = 22.

That is, the mean wind speed is 22 mph.

Alternative hypothesis:

H0: µ > 22.

That is, the mean wind speed greater than 22 mph.

Degrees of freedom:

For student’s t distribution,

df = n – 1 = 33 – 1 = 32.

Critical value:

The critical value is obtained using the Excel formula, “=T.INV (0.99, 32)”

Thus, the critical value of Student’s t is t1-α =t0.99 = 2.449.

Decision rule for Right-tailed test at α = 0.01:

If tcalc > 2.449, then reject the null hypothesis.

Test statistic:

The formula for test statistic is,

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Step 2

Where x-bar is the sample mean, µ0 is the hypothesized mean, s is the sample standard deviation and n is the sample size.

Substitute x-bar = 22.8, µ0 = 22, s = 4.3 and n = 33 in the test statistic formula.

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Step 3

Thus, the test statistic is 1.0688.

Conclusion for critical value method:

Here, the test statistic is less than the critical value.

That is, tcalc(= 1.0688) < 2.449.

Therefore, the null hypothesis is not rejected.

Hence, there is sufficient evidence to conclude that mean wind speed is 22 mph at 1% level of significance.

A site with average wind spee...

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