Question 294537: 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
A 8
B 8
C 9
D 11
E 14
Total 50
Can you help me solve this problem. I am supposed to create a null and alternative hypothesis. Show how the degress of freedom are calculated, find the critical value of chi squared and interpret the p-value as well as check for small expected frequencies
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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
A 8
B 8
C 9
D 11
E 14
Total 50
Can you help me solve this problem. I am supposed to create a null and alternative hypothesis. Show how the degress of freedom are calculated, find the critical value of chi squared and interpret the p-value as well as check for small expected frequencies
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The expected values are 10,10,10,10,10
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Ho: all frequencies or propabilites are equal
Ha: at least one of the probs is different
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Determine the O-E differences for each category:
-2,-2,-1,+1,+4
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Calculate the Chi-Sq test statistic for
the data based on Chi-Sq = sim[(O-E)^2/E] = (4+4+1+1+16)/10 = 2.6
---
degrees of freedom = 5-1 = 4
----
Critical value for df=4 and alpha = 5% = 9.488
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p-value = P(Chi-Sq >2.6 when df=4) = Chi-Sqcdf(2.6,1000,4) = 0.6268
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Conclusion: Since the p-value is greater than 5% fail to reject Ho.
That is the same as saying the test stat is not in the reject interval.
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Cheers,
Stan H.
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