Question 332017: Silver Gym’s aerobic class grew so large that the class had to be split in half and offered on two different nights. Silver also hired a new aerobic leader to handle the Tuesday night class, whereas an existing staff member continued to handle the Thursday night class. Concerned about customer satisfaction with the new leader, Silver Gym’s management decided to secure a leader-approval rating from each class after four weeks. Results are shown in the table below:
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· Tuesday Thursday
Class size 38 41
Mean Approval Rating 87.3 89.7
Standard Deviation 7.4 6.3
Did the new leader of the Tuesday night class receive a lower approval rating than the leader of the Thursday night class? Set up and test your hypotheses at a = 0.05.
Answer by jrfrunner(365)   (Show Source):
You can put this solution on YOUR website! Let Mu(n)= Population mean rating of new trainer
Let Mu(e)= Population mean rating of existing trainer
Let Xbar(n) = sample average rating for new trainer
Let Xbar(e) = sample average rating for existing trainer
Let S(n)=sample std dev for new trainer
Let S(e)=sample std dev for existing trainer
Let n(n)= sample size for new trainer
Let n(e)= sample size for exiting trainer
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to answer
Did the new leader of the Tuesday night class receive a lower approval rating than the leader of the Thursday night class?
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Ho: Mu(n)=Mu(e)
Ha: Mu(n) < Mu(e)
at alpha =0.05
lower tail test
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Find the test statistic
This is a two independent samples t test. There are two t tests, one if the population variances are equal and one if the population variances are not equal
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To determine the t test to use, you should conduct a side hypothesis test on the population variances
Ho: Variance (n) =Variance (e)
Ha: Variance (n) not equla variance (e)
test statistic: F=max(var(n),var(e))/min(var(n),var(e)) = (7.4/6.3)^2 =1.38
critical value is F(alpha, dof numerator, dof denominator) = F(.05,37,40)=1.71
since test statistic=1.38 < critical value=1.71 cannot reject Ho: variances are equal.
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Since we now established that its likely the pop variances are equal, we may pool the asmple variances
pooled sample variance =  /(38+41-2)=46.93
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test statistic: 
 =-2.4/1.54=-1.558
critical value = t(1-0.05,38+41-2)=-1.66
Since our test statistic= -1.558 > critical value = -1.66, it falls in the fail to reject Ho.
pvalue = P(t<=-1.558) = 0.062 > alpha =0.05
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Conclusion
Cannot reject the hypothesis that the mean ratings between trainers is the same.
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