In an effort to counteract student cheating, the professor of a large class crea

Question

In an effort to counteract student cheating, the professor of a large class created several versions of a midterm exam, distributing the versions equally among the students in the class. After the exam, 7 students from the class got together and petitioned to nullify the results on the grounds that the four versions were not equal in difficulty. To investigate the students' assertion, the professor performed a one-way, independent-samples ANOVA test using the 0.01level of significance. The professor looked at the scores for the different versions of the exam (the "groups") to see if, indeed, the versions were not equal in difficulty. Below is the ANOVA table that summarizes this ANOVA test. (The exam was worth 200 points.)

Fill in the missing cell in the ANOVA table (round your answer to at least two decimal places), and then answer the questions about the ANOVA test.

How many versions of the exam were looked at for the ANOVA test?

For the ANOVA test, it is assumed that each population of scores (that is, the population of scores for each version) has the same variance. What is an unbiased estimate of this common population variance based on the sample variances?

What is the /p/-value corresponding to the F statistic for the ANOVA test? Round your answer to at least three decimal places.

Can the professor conclude, based on these exam scores and using the 0.01 level of significance, that at least one of the versions was significantly different from the others in difficulty? Yes or No

Source of variation degrees of freedom sum of squares mean square F statistic Between groups 2 2016.66 1008.33 ???? Error (within groups) 258 106012.2 410.9 ------------- Total 260 108028.86 ----------------- -------------

Explanation / Answer

numebr of  versions of the exam =df for groups+1 =2+1 =3

unbiased estimate of this common population variance based on the sample variances =MSE =410.9

p/-value corresponding to the F statistic for 2 ,258 degree of freedom =0.088

as p value is greater than 0.01 level of significance ; therefore No

Source of variation degrees of freedom sum of squares mean square F statistic Between groups 2 2016.66 1008.33 2.45 Error (within groups) 258 106012.2 410.9 Total 260 108028.9

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