Question

We're interested in the amount of time spent at work by college graduates employed full-time. The standard amount of time spent at work by full-time employees is 40 hours per week. We suspect that the mean number of hours worked per week by college graduates is less than 40 hours and wish to do a statistical test. We select a random sample of 100 college graduates employed full-time and compute the mean number of hours worked per week by the graduates in the sample. Suppose that the population of hours worked per week by college graduates has a standard deviation of 5 hours and that we perform our hypothesis test using the 0.1 level of significance. Based on this information, answer the questions below. Carry your intermediate computations to at least four decimal places, and round your responses as indicated. (If necessary, consult a list of formulas.) H is What are the null and alternative hypotheses that we should use for the test? What is the probability that we commit a Type I error? Round your response to at least two decimal places Assuming that the actual value of wis 39.2 hours, what is the probability that we accept the null hypothesis? Round your response to at least two decimal places. The power of the second test is greater than the power of the original test Suppose that we decide to perform another statistical test using the same population, the same null and alternative hypotheses, and the same sample sire, but for this second test we use a significance level of 0.01 Instead of a significance level of 0.1. Assuming that the actual value of wis 39.2 hours, how does the power of this second test compare to the power of the original The power of the second test is less than the power of the original test The powers of the two tests are equal

Answer 1

Ho:µ is equal to 40
Ha: µ is less than 40

P(type I error)=0.10

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true mean ,    µ =    39.2              
                      
hypothesis mean,   µo =    40              
significance level,   α =    0.1              
sample size,   n =   100           0.3125  
std dev,   σ =    5              
                      
δ=   µ - µo =    -0.8              
                      
std error of mean,   σx = σ/√n =    5.0000   / √    100   =   0.50000

Zα =       -1.2816   (left tail test)
P(type II error) , ß =   P(Z > Zα - δ/σx)                  
= P(Z >    -1.282   - (   -0.8   /   0.5000   ))
=P(Z>    0.318   ) =   0.3751 (answer) [ Excel function: =1-normsdist(z) ]      

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power of second test is less than the power of the original test