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 yardsα=0.02 (significance level)z0=2.51p=0.006Select the correct answer below:Do not reject the null hypothesis because the p-value 0.006 is less than the significance level α=0.02.Reject the null hypothesis because the p-value 0.006 is less than the significance level α=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 z is positive.

Name: Jennifer, a golfer, has a sample driving distance mean of 244.0 yards
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Description: 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

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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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

Select the correct answer below:

Do not reject the null hypothesis because the p-value 0.006 is less than the significance level α=0.02.
Reject the null hypothesis because the p-value 0.006 is less than the significance level α=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 z is positive.

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