MS Mod 11 Single-Factor ANOVA Tukey’s procedure practice exam questions


MS Mod 11 Single-Factor ANOVA Tukey’s procedure practice exam...

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MS Module 11 Single-Factor ANOVA Tukey’s procedure practice exam questions

(The attached PDF file has better formatting.)

Five groups, each with 7 observations, have sample means of

    1 = 3.2, 2 = 3.1, 3 = 3.5, 4 = 3.9, 5 = 3.7

The total sum of squares (SST) is 20, and the treatment sum of squares (SSTr) is 11.

The groups have normal distributions with equal variances. We test the null hypothesis μ1 = μ2 = μ3 = μ4 = μ5 at a 5% confidence level. If we reject the null hypothesis, we test which groups means differ significantly.


Question 11.1: Error sum of squares

What is the error sum of squares SSE?

Answer 11.1: 20 – 11 = 9

(error sum of squares = total sum of squares – treatment sums of squares)


Question 11.2: Treatment mean square

What is the mean square for the groups (treatment mean square)?

Answer 11.2: 11 / (5 – 1) = 2.75

(treatment mean square = treatment sums of squares / (number of groups – 1) )


Question 11.3: Mean squared error

What is the mean squared error MSE?

Answer 11.3: 9 / (5 × 7 – 5) = 0.30

(degrees of freedom for mean squared error = (number of groups – 1) × observations per group; mean squared error = error sum of squares / its degrees of freedom)


Question 11.4: F value

What is the F value to test the null hypothesis?

Answer 11.4: 2.75 / 0.30 = 9.167


Question 11.5: Critical F value

What is the critical F value for α = 5%?

Answer 11.5: 2.69

(Table look-up: α = 5%; degrees of freedom = 4, 30)


Question 11.6: Critical Q value

What is the critical Q value for α = 5%?

Answer 11.6: 4.10

(Table look-up: α = 5%; degrees of freedom = 5, 30)


Question 11.7: W (the width of the difference) for Tukey’s honestly statistical difference?

What is w (the width of the difference) for Tukey’s honestly statistical difference?

Answer 11.7: 4.1 × (0.30 / 7)0.5 = 0.849

(Tukey’s W = critical Q value × (mean squared error / observations per group)0.5 )


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