Question

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. h0:μ=225 yards; ha:μ>225 yards α=0.02 (significance level) z0=2.51 p=0.006

Answer 1

The hypothesis test results, with a p-value of 0.006 which is lower than the significance level of 0.02, suggest that Jennifer's driving distance mean is indeed greater than 225 yards, leading to the rejection of the null hypothesis.

Step-by-step explanation:

When assessing whether Jennifer, a golfer, has a sample driving distance mean greater than 225 yards, we utilize the provided sample data to perform a hypothesis test. Her mean of 244.0 yards from 16 drives is being tested against her claim that her average driving distance is actually 225 yards. With H0: μ=225 yards and Ha: μ>225 yards, we look at the significance level (α=0.02) and the calculated test statistic z0=2.51 with a corresponding p-value=0.006.

To decide whether to accept or reject the null hypothesis, we compare the p-value to the significance level. Since the p-value (0.006) is less than the significance level (0.02), there is sufficient evidence to reject the null hypothesis at the 2% significance level. Therefore, we conclude that Jennifer's mean driving distance is greater than 225 yards.