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At many golf clubs, a teaching professional provides a free 10-minute lesson to

ID: 3293382 • Letter: A

Question

At many golf clubs, a teaching professional provides a free 10-minute lesson to new customers. A golf magazine reports that golf facilities that provide these free lessons gain, on average, $2,300 in green fees, lessons, or equipment expenditures. A teaching professional believes that the average gain is less than $2,300. Complete parts a through c below. a. In order to support the claim made by the teaching professional, what null and alternative hypotheses should you test? b. Suppose you select alpha = 0.05. Interpret this value in the words of the problem. The probability that the null hypothesis is _____ when the average gain ___ $2,300 is 0.05. c. For alpha = 0.05, specify the rejection region of a large-sample test. Choose the correct answer below. A. z 1.645 C. -2.575

Explanation / Answer

                ? = $2300                             Population average

a.            In order to support claim made by the teaching professional, the null                    

                and alternative hypothesis that should be tested are                     

                                               

                Ho : ? = 2300                     

                Ha : ? < 2300                     

                                               

b.            ? = 0.05                               

                Interpretation is                              

                The probability that the null hypothesis is rejected when the average                   

                gain = $2300 is 0.05                         

                                               

c.             For ? = 0.05, the rejection region of a large sample test is given by                          

                We find from the z-tables the value of Z such that                           

                P(z < Z) = 0.05                    This is because the alternative hypothesis is Ha : ? < 2300

                From the z-tables, we get the value of Z = -1.645                                                    

                Answer :                             

                E.     z < -1.645                   

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