BE 251 SP17 Final Ex Database ATT Ex1 Ex2 Ex3 Ex Tot HW Height Gender Class Majo

Question

BE 251 SP17 Final Ex Database ATT Ex1 Ex2 Ex3 Ex Tot HW Height Gender Class Major 9 15 13 13 40 19 75 M JR ACCT 8 10 9 12 31 8 67 F JR ACCT 8 8 9 14 32 14 63 F SO ACCT 8 14 13 12 39 20 66 F SO ACCT 10 16 14 12 42 20 67 M SO ACCT 10 12 12 11 35 20 73 M SO ACCT 8 12 11 10 33 17 66 F SO ACCT 6 13 13 14 41 15 76 M SO FINC 10 12 13 13 38 19 74 M SO FINC 8 14 13 10 37 14 68 M JR MGMT 8 13 7 9 30 11 65 M JR MGMT 8 14 11 13 37 13 71 M SO MGMT 8 14 15 14 43 16 65 F JR MGMT 8 10 10 11 30 9 62 F JR MGMT 8 13 9 12 34 14 64 F SO MKTG 8 11 12 9 32 12 69 M SO MKTG 10 13 11 9 34 18 71 M SO MKTG 10 14 12 12 38 16 72 M JR MKTG 10 14 13 12 39 18 69 M SO OTH 10 15 13 13 41 16 62 F SO OTH 10 13 9 9 31 19 68 M SO OTH 8 13 13 9 35 14 67 F JR OTH 8 9 6 7 21 4 69 F JR OTH 10 14 8 11 33 12 65 F JR OTH 8 9 8 8 25 16 74 M SO OTH 14 31 BE 251 SP17 Final Exam Database Height Gender Class Question 7: Hypothesis Testing and ANOVA Use the data on Exam Grades to respond to the following questions: (be sure to use the Excel functions to answer) Using the estimates of the mean exam grade (one for each exam) and a population standard deviation o of 4.34 1. Plot the point estimates of the mean grade for each of the three exams. Is there anything apparent from the graph? 2. Create a null and alternate hypothesis to test that: a). The mean grade on an exam was greater for Exam 2 than for Exam 1. b) The mean grade on an exam was no greater for Exam 3 than for Exam 2 3. Using ANOVA, test the hypothesis that all three exam means are What do you conclude? Conclusions from your analysis: Analyze the results above to discuss the performance of the students on each of the 3 exams s the performance notably different among the exams?

Explanation / Answer

used excel data analysis--->one way anova to answer yoiur question.

For question1:

point estimates are emans of three groups

mean(exam1 ) is more than others

Solutionc:

Anova:

Null hypothesis:

Ho:

there is no significance difference in means of exam1,2 and 3

Alternative Hypothesis:

Atleast one of the means of the exams are different

alpha=0.05

test statistic F=3.952

p=0.0235

Decsion:

if p<0.05 reject null hypothesis

if p>0.05 accept null hypothesis

here p=0.02354<0.05

Rejecty Null hypothesis

Accept Alternative Hypothesis

Conclusion:

there is sufficient statistical evidence at 5% level of significance to conclude that the performance notable different among exams.

Anova: Single Factor SUMMARY Groups Count Sum Average Variance Ex1 25 315 12.6 4.25 Ex2 25 277 11.08 5.66 Ex3 25 279 11.16 3.973333

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