Question 138909
Ages for the 2005 Boston Red Sox pitchers are shown below. 
(a) Assuming this is a random sample of major league pitchers, at the 5 percent level of significance does this sample show that the true mean age of all American League pitchers is over 30 years? State your hypotheses and decision rule and show all work. 
Ho: u = 30
Ha: u > 30
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n=12  ; x-bar= 33.9167
I ran a T-test with df=11, alpha= 5% and go the following:
test statistic = -36.4565
p-value = 1
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(b) If there is a difference, is it important?
A slight difference but not important.
Since p-value > 5%, Fail to reject Ho
Average age is statistically 30. 
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(c) Find the p-value and interpret it.
p-value = 1 means 100% of test results would give stronger evidence
for rejecting Ho.

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Ages of Boston Red Sox Pitchers, October 2005
Arroyo 28 
Foulke 33 
Mantei 32 
Timlin 39
Clement 31 
Gonzalez 30 
Miller 29 
Wakefield 39
Embree 35 
Halama 33 
Myers 36 
Wells 42 
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2. In Utica, Michigan, 205 of 226 school buses passed the annual safety inspection. In Detroit, Michigan, only 151 of 296 buses passed the inspection. 
(a) State the hypotheses for a right-tailed test.
Ho: p(Ut)-p(Dt)= 0
Ha: p(Ut)-p(Dt)>0
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(b) Obtain a test statistic and p-value.
p-hat(Ut) = 205/226= 0.9071;p-hat(Dt)=0.5101; p(pooled)=(431/522)=0.8257
z = (9071-0.5101)/[ 0.8257*0.1743/226 + 0.8257*0.1743/296]= 9.6491
p-value: 2.52x10^-22
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(c) Is normality assured? 
I'll let you check out those conditions.
They should be listed in your text.
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(d) If significant, is the difference also large enough to be important?
Since p-value is so small the difference is very significant.
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3. Prof. Green’s multiple-choice exam had 50 questions with the distribution of correct answers shown below. Research question: At α = .05, can you reject the hypothesis that Green’s exam answers came from a uniform population 
Correct Answer Frequency.......Expected
A ................8................10
B ................8................10
C ................9................10
D ...............11................10
E ...............14................10
Total 50 
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Ho: Observed Frequencies are equal
Ha; Observed Frequencies are not equal
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I ran a Chi-Sq Test on a 2x5 matrix with observed in the 1st row and expected in the 2nd row and got the following results:
Test statistic: Chi-Sq = 1.2114
p-value: 0.8762
Meaning: 87.6% of test results could have given stronger evidence for
rejecting Ho.
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Conclusion: Since p-value is greater than alpha=5%, Fail to reject Ho.
The grades follow a uniform distribution.
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Cheers,
Stan H.