## MS Mod 12 ANOVA unequal group sizes practice exam questions

 Author Message NEAS Supreme Being         Group: Administrators Posts: 4.2K, Visits: 1.2K MS Module 12 ANOVA unequal group sizes practice exam questions(The attached PDF file has better formatting.) An experiment has three groups; the number of observations per group and the group means are    size    meangroup 1    21    51group 2    34    76group 3    31    70●    The sum of the squares of the observations is 590,645●    The observations have normal distributions in each group, and the variance in each group is the same.●    The null hypothesis is that the means of the groups are equal: H0: μ1 = μ2 = μ3 [μj = mean of Group j]Question 12.1: Square of sum of observationsWhat is the square of the sum of all the observations, or x..2 ?Answer 12.1: (21 × 51 + 34 × 76 + 31 × 70)2 = 33,930,625(square of the sum of the observations = ( (observations in group × mean of group) )2 )Question 12.2: Correction factorWhat is the correction factor for SST and SSTr?Answer 12.2: 33,930,625 / (21 + 34 + 31) = 394,542.15(correction factor = square of the sum of the observations / total observations)Question 12.3: Total sum of squaresWhat is SST, the total sum of squares?Answer 12.3: 590,645 – 394,542.15 = 196,102.85(total sum of squares = sum of squares of observations – correction factor) Question 12.4: Treatment sums of squaresWhat is SSTr, the treatment sum of squares?Answer 12.4: (21 × 512 + 34 × 762 + 31 × 702) – 394,542.15 = 8,362.85(total sum of squares =  (observations by group × square of mean by group) – correction factor)Question 12.5: Error sum of squaresWhat is SSE, the error sum of squares?Answer 12.5: 196,102.85 – 8,362.85 = 187,740.00(error sum of squares = total sum of squares – treatment sums of squares)Question 12.6: Total degrees of freedomWhat are the total degrees of freedom?Answer 12.6: (21 + 34 + 31 – 1) = 85(total degrees of freedom = number of observations – 1)Question 12.7: Degrees of freedom for the groupsWhat are the degrees of freedom for the groups?Answer 12.7: 3 – 1 = 2Question 12.8: Degrees of freedom for the error sum of squaresWhat are the degrees of freedom for the error sum of squares (SSE)?Answer 12.8: 85 – 2 = 83Question 12.9: Mean squared deviation for the groupsWhat is MSTr, the mean squared deviation for the groups (treatment mean square)?Answer 12.9: 8,362.85 / 2 = 4,181.425Question 12.10: Mean squared error What is MSE, the mean squared error?Answer 12.10: 187,740.00 / 83 = 2,261.928Question 12.11: F valueWhat is the F value for testing the null hypothesis?Answer 12.11: 4,181.425 / 2,261.928 = 1.849 Attachments MS Module 12 ANOVA unequalsizes practice exam questions.pdf (229 views, 45.00 KB)
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