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, 4 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 0.01 level 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.)             (a)Fill in the missing cell in the ANOVA table (round your answer to at least two decimal places).   Source of variation Degrees of freedom Sum of squares Mean square F statistic Treatments (between groups) 4 6300.54 1575.14     Error (within groups) 430 157,285.4 365.78   Total 434 163,585.94

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
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Author:Carter
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Chapter10: Statistics
Section10.6: Summarizing Categorical Data
Problem 30PPS
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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, 4 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 0.01 level 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.)

 

 

 

 

 

 

(a)Fill in the missing cell in the ANOVA table (round your answer to at least two decimal places).
 
Source of variation Degrees of freedom Sum of squares Mean square F statistic
Treatments
(between groups)
4
6300.54
1575.14
 
 
Error
(within groups)
430
157,285.4
365.78
 
Total
434
163,585.94
 
(d) Using the 0.01 level of significance, what is the critical value of the F statistic for the ANOVA test? Round your answer to at least two decimal places.
(e) 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?
OYes
ONo
Transcribed Image Text:(d) Using the 0.01 level of significance, what is the critical value of the F statistic for the ANOVA test? Round your answer to at least two decimal places. (e) 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? OYes ONo
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