Question

Nancy, a golfer, claims that her average driving distance is 253 yards. During a practice session, Nancy has a sample driving distance mean of 229.6 yards based on 18 drives. At the 2% significance level, does the data provide sufficient evidence to conclude that Nancy's mean driving distance is less than 253 yards? Accept or reject the hypothesis given the sample data below.

• H0:=253yards; Ha:<253yards
• =0.02 (significance level)
• z0=0.75
• p=0.2266

Select the correct answer below:

Do not reject the null hypothesis because the p-value 0.2266 is greater than the significance level =0.02.

Do not reject the null hypothesis because the value of z is negative.

Do not reject the null hypothesis because |0.75|>0.02.

Reject the null hypothesis because |0.75|>0.02.

Reject the null hypothesis because the p-value 0.2266 is greater than the significance level =0.02.