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 ?‑test of ?0:?=2.4 mm against ?1:?≠2.4 mm, where ? 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 ?‑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 ?=0.05 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 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

true mean ,    µ =    2.6
      
hypothesis mean,   µo =    2.4
significance level,   α =    0.05
sample size,   n =   195
std dev,   σ =    11.4

δ=   µ - µo =    0.2                      
                              
std error of mean,   σx = σ/√n =    0.8164                      
                              
Zα/2   = ±   1.960   (two tailed test)                  
We will fail to reject the null (commit a Type II error) if we get a Z statistic between                           -1.960   and
these Z-critical value corresponds to some X critical values ( X critical), such that                              
                              
-1.960   ≤(x̄ - µo)/σx≤   1.960                      
0.800   ≤ x̄ ≤   4.000                      
                              
now, type II error is ,ß =        P (   0.800   ≤ x̄ ≤   4.000   )      
       Z =    (x̄-true mean)/σx                  
       Z1 =   -2.205                  
       Z2 =    1.715                  
                              
   so, P(   -2.205   ≤ Z ≤   1.715   ) = P ( Z ≤   1.715   ) - P ( Z ≤   -2.205
                              
       =   0.957   -   0.014   =   0.9431  
                              
power =    1 - ß =   0.0569                      
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The researchers will reject their null hypothesis if the sample mean is

less than=___-1.96______ mm or

greater than= __1.96________mm

power = 5.69%