6. A department store manager believes that the average age of their customers is at least 60. To prove their point, the manager randomly selects a sample of customers and records their ages. The data is listed below. At = 0.01, test the manager's claim.

70 48 41 68 69 55 70 57 60 83

32 60 72 58 88 48 59 60 56 65

66 60 68 42 57 59 49 70 75 63

44

Hypothesis

H0:=60

Ha:60

Test Statistic

Z=0.1769

P VALUE

P Value = 0.4298

Conclusion

Because the p value is above the significance level, you cannot reject the null hypothesis, thus you infer that the average age of consumers is 60 years.

From the data provided , I estimated the sample mean and standard deviation.

X=nXi=311872=60.3871

=n1(X)2=3114455.354839=12.1865

Hypothesis

H0:=60

Ha:60

Test Statistic

Z=nX

Z=3112.186560.387160

Z=2.1887677190.3871

Z=0.1769

P VALUE

From the z table I found p value to be

P Value = 0.4298

Conclusion

Because the p value is above the significance level, you cannot reject the null hypothesis, thus you infer that the average age of consumers is 60 years.

HOW TO REPLY

How did you get a p-value without a mean and standard deviation? What is3112.186560.3871? and what isZ=nX? Please follow example in text and use pic of work on notebook paper for correct notation.