In an effort to counteract student cheating, the professor of alarge class creat

Question

In an effort to counteract student cheating, the professor of alarge class created several versions of a midterm exam,distributing the versions equally among the students in the class.After the exam, 6students from the class got together exam andpetitioned to nullify the results on the grounds that the fourversions were not equal in difficulty.

To investigate the students' assertion, the professor performeda one-way, independent-samples ANOVA test using the 0.01level ofsignificance. The professor looked at the scores for the differentversions of the exam (the "groups") to see if, indeed, the versionswere not equal in difficulty. Below is the ANOVA table thatsummarizes this ANOVA test. (The exam was worth 200points.)

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

Source of variation    degrees offreedom    sum of squares   meansquare F statistic

Treatment between groups   4                        3019.3             754.8            (answer)

Error (withingroups)              380                    152243.2         400.6           (no box)

Total                                       384                    155262.5          (nobox)         (no box)

-how many versions of the exams were looked at for the ANOVAtest?

-For the ANOVA tests it is assumed that each population ofscores (that is, the population of scores for each version) has thesame variance. What is the unbiased estimate of this commonpopulation variance based on the sample variances?

- What is the p-value corresponding to the f-statistic for theANOVA test? (round answer to at least 3 decimal places)

- Can the professor conclude, based on these exam scores, andusing the 0.01 level of significance, that at least 1 of theversions was significantly different from the others ifdifficulty? Yes or No

Explanation / Answer

Source of variation Degrees of freedom Sum of squares Mean sum squares F-statistc P-value: Treatment 4 3019.3 754.8 1.884174 0.11 Error 380 152243.2 400.6 Total 384 155262.5 Since the p-value in the above table is greaterthan 0.05, the level of significance so we reject the nullhypothesis and conclude that at least 1 of the versions wassignificantly different from the others if difficulty Source of variation Degrees of freedom Sum of squares Mean sum squares F-statistc P-value: Treatment 4 3019.3 754.8 1.884174 0.11 Error 380 152243.2 400.6 Total 384 155262.5

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