(Solved): Jennifer, a golfer, has a sample driving distance mean of 244.0 yards from 16 drives. Jennifer stil ...
Jennifer, a golfer, has a sample driving distance mean of 244.0 yards from 16 drives. Jennifer still claims that her average driving distance is 225 yards, and the high average can be attributed to chance. At the
2%significance level, does the data provide sufficient evidence to conclude that Jennifer's mean driving distance is greater than 225 yards? Given the sample data below, accept or reject the hypothesis.
H_(0):\mu =225yards;
H_(a):\mu >225yards
\alpha =0.02(significance level)
z_(0)=2.51
p=0.006Select the correct answer below: Do not reject the null hypothesis because the
p-value 0.006 is less than the significance level
\alpha =0.02. Reject the null hypothesis because the
p-value 0.006 is less than the significance level
\alpha =0.02. Reject the null hypothesis because
2.51>0.02. Do not reject the null hypothesis because
2.51>0.02. Reject the null hypothesis because the value of
zis positive.
