MS Mod 12 ANOVA unequal group sizes practice exam questions


MS Mod 12 ANOVA unequal group sizes practice exam questions

Author
Message
NEAS
Supreme Being
Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)Supreme Being (5.9K reputation)

Group: Administrators
Posts: 4.3K, Visits: 1.3K

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    mean
group 1    21    51
group 2    34    76
group 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 observations

What 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 factor

What 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 squares


What 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 squares

What 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 squares

What 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 freedom

What 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 groups

What are the degrees of freedom for the groups?

Answer 12.7: 3 – 1 = 2


Question 12.8: Degrees of freedom for the error sum of squares

What are the degrees of freedom for the error sum of squares (SSE)?

Answer 12.8: 85 – 2 = 83


Question 12.9: Mean squared deviation for the groups

What is MSTr, the mean squared deviation for the groups (treatment mean square)?

Answer 12.9: 8,362.85 / 2 = 4,181.425


Question 12.10: Mean squared error

What is MSE, the mean squared error?

Answer 12.10: 187,740.00 / 83 = 2,261.928


Question 12.11: F value

What is the F value for testing the null hypothesis?

Answer 12.11: 4,181.425 / 2,261.928 = 1.849



Attachments
GO
Merge Selected
Merge into selected topic...



Merge into merge target...



Merge into a specific topic ID...





Reading This Topic


Login
Existing Account
Email Address:


Password:


Social Logins

  • Login with twitter
  • Login with twitter
Select a Forum....











































































































































































































































Neas-Seminars

Search