Question

Suppose that for the past several decades, daily precipitation in Seattle, Washington has had a mean of 2.4 mm and a standard deviation of 11.4 mm. Researchers suspect that in recent years, the mean amount of daily precipitation has changed, so they plan to obtain data for a random sample of 195 days over the past five years and use this data to conduct a one-sample ?z‑test of ?0:?=2.4H0:μ=2.4 mm against ?1:?≠2.4H1:μ≠2.4 mm, where ?μ is the mean daily precipitation for the last five years. Although they realize that rainfall does not follow a normal distribution, they feel safe using a ?z‑test because the sample size is large.

The researchers want to know what the power of this test is to reject the null hypothesis at significance level ?=0.05α=0.05 if the actual mean daily precipitation is 2.6 mm or more. Computing power by hand requires two steps.

The first step is to use a significance level of 0.05 to determine the values of the sample mean for which they will reject their null hypothesis. Give your answer to the nearest 0.1 mm.

The researchers will reject their null hypothesis if the sample mean is

less than_____mm or

greater than_____mm

In the second step, find the power of the test by first assuming that the the actual mean is 2.6 mm. Then, compute the probability of getting a sample mean in the rejection region found in the first step. Leave the boundaries of the critical region rounded to one decimal place in your calculation, and give your answer as a percentage rounded to two decimal places.

Power = ______%

Answer 1

Given the sample size = n = 195, population standard deviation = , to test the null hypothesis v/s the alternative hypothesis , at alpha = 0.05, the test statistic, reject H0 if , where, ,

therefore, for H0, , Therefore, , therefore, reject H0 if

, or

The actual mean = , therefore, power of the test =

P[Z>1.71] = 0.0436

Therefore, Power = 4.36%