Question

Should you generate electricity with your own personal wind turbine? That depends
on whether you have the right amount of wind on your site. For a personal wind
turbine to work properly, your site should have an annual average wind speed of at
least 10 miles per hour, according to the Wind Energy Association. Lower winds do
not provide enough energy. One candidate site was monitored last year, with wind
speeds recorded every 8 hours. A total of 49 readings of wind speeds average 9.56
mph with a standard deviation of 2.52 mph. You've been asked to make a statistical
report to help the landowner decide whether to place a wind turbine at the site.
Test whether the average wind speed is at least 10 mph at the .025 level of
significance. Show Your Work! Formula(s), Substitutions, Answers! Round your critical
value to 3 decimal places, and your pvalue to 4 decimal places. At the .05 level of significance, the site (is / is not) suitable for a small wind turbine. If the true population mean actually is 8 mph, you have made a Type I Error               Type II Error         Correct Decision               If the level of significance had been .05, would your answer to part b change? Yes No    Circle One.

Answer 1

ANSWER

1)

(7) = 1- (0.16 + 0.2 + 0.36) = 0.28

mean = 3 * 0.16 + 6 * 0.2 + 7 * 0.28 + 12 * 0.36

= 7.96

std.deviation = summation (sqrt(x - mean)^2*P(x))
= Below are the null and alternative Hypothesis,
Null Hypothesis, H0: μ = 10
Alternative Hypothesis, Ha: μ < 10

Rejection Region
This is left tailed test, for α = 0.025 and df = 48
Critical value of t is -2.011.
Hence reject H0 if t < -2.011

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (9.56 - 10)/(2.52/sqrt(49))
t = -1.222

P-value Approach
P-value = 0.1138
As P-value >= 0.025, fail to reject null hypothesis.

2)

since p value >0.05. we fail to reject the null hypothesis

Type II Error is correct option , since we fail to reject a false null hypothesis

3)

Rejection Region
This is left tailed test, for α = 0.05 and df = 48
Critical value of t is -1.677.
Hence reject H0 if t < -1.677

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (9.56 - 10)/(2.52/sqrt(49))
t = -1.222

P-value Approach
P-value = 0.1138
As P-value >= 0.05, fail to reject null hypothesis.

At the .05 level of significance, the site ( is not) suitable for a small wind turbine.

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