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 {-Espar'iol-rl the students in the class. After the exam, several students from the class got together and petitioned to nullify the results on the grounds that the four versions were not equal in difculty. To investigate the students' assertion, the professor performed a oneway, independent-samples ANOVA test using the 0.05 level of signicance. 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.) 9 (a) Fill in the missing cell in the ANOVA table (round your answer to at least two decimal places). Source of Degrees of Sum of Mean 5 uare Fstatistlc K variation freedom squares q X if) Treatments (between 5 3555.58 711.12 groups) Error (Wlthin 498 210,064.? 421.82 groups) Total 503 213,620.28 (b) How many exam scores were looked at or the ANOVA test? (c) For the ANOVA test, it is assumed t1at X 8 (d) what is the pvalue corresponding t X 8 (e) Can the professor conclude, based on t different from the others in difculty? 0 Yes ONo X 8 X b each population of scores (that is, the population of scores for each version) has the same variance. What is an unbiased estimate of thlS common population variance based on the sample variances? o the Fstatistic for the ANOVA test? Round your answer to at least three decimal places. hese exam scores and using the 0.05 level of signicance, that at least one of the versions was signicantly