Question

Jacqueline, a golfer, has a sample driving distance mean of 202.0 yards from 12 drives. Jacqueline still claims that her average driving distance is 223 yards, and the low average can be attributed to chance. At the 2% significance level, does the data provide sufficient evidence to conclude that Jacqueline's mean driving distance is less than 223 yards? Given the sample data below, accept or reject the hypothesis.

• H0:=223yards; Ha:<223yards
• =0.02 (significance level)
• z0=1.26
• p=0.1038

Select the correct answer below:

Reject the null hypothesis because the p-value 0.1038 is greater than the significance level =0.02.

Do not reject the null hypothesis because the p-value 0.1038 is greater than the significance level =0.02.

Reject the null hypothesis because |1.26|>0.02.

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

Do not reject the null hypothesis because |1.26|>0.02.