Nancy, a golfer, claims that her average driving distance is 253 yards. During a practice session,...
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:μ=253 yards; Ha:μ<253 yards
- α=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 α=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 α=0.02.
Homework Answers
the p value is 0.2266
and the significance level is 0.02
the right choice is
Do not reject the null hypothesis because the p-value 0.2266 is greater than the significance level α=0.02.
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Nancy, a golfer, claims that her average driving distance is 253
yards. During a practice session,...
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