7 Driving distance on the PGA tour in 2012 was determined to

7. Driving distance on the PGA tour in 2012 was determined to a mean of 282 yards. With even better technology in drivers and golf balls, observers of the PGA tour believe that the driving distance in 2013 has a larger mean value than in 2012. A study of 50 drives taken randomly in a 2013 PGA tournament recorded an Xbar of 286 yards with a sample standard deviation of 7 yards. (11 pts) a. Is this study looking to see simply if the mean is different than 2012? Yes / No? b. What is an appropriate null hypothesis? c. What is the value of the t test statistic? d. What is the critical value for t? e. What is the p value? f. At a 95% confidence level, what is your conclusion concerning the null hypothesis? Reject / Not Reject? g. Using the Critical Value Test (test statistic vs. critical value), explain your conclusion concerning the null hypothesis? h. Using the p value test (p value vs. value), explain your conclusion concerning the null hypothesis? i. If you were in the golf equipment business, why might you be interested in this study?

Solution

a) Yes

b) Ho : miu = 282

H1 : miu =/ 282

c) T = 286 - 282 / 7/srqt(50)

T = 4.04

d) critical value for alpha = 0.05

alpha/2 = 0.025

df=49

critical value : +/- 2.021

f)

we reject Ho because T is greater than critical value