Question

Bob is worried that students’ scores on examinations differ significantly based on the paper colors of the examinations, on average. Bob conducts an Analysis of Variance in which he selects four examinations at random for each of four paper colors.

a. If the value for Sum of Squares Total equals 2,144 and the value for Mean Squares Within equals 135.5, what does the calculated value for the associated test statistic equal?

(a) 1.2743

(b) 0.7819

(c) 1.6238

(d) 15.8229

b. If the level of significance equals 0.05, can Bob conclude students’ scores on examinations differ significantly based on the paper colors of the examinations, on average?

(a) not w/out Turkey's HSD

(b) Not w/out Hypotheses

(c) yes

(d) no

Answer 1

A. In ANOVA the test statistics is given by the formula:

And also,

There are 4 papers of 4 colours each so the total number of observations are n= 4x4=16

And k=4; as there are 4 different colours of examination papers

Total sum of squares is 2144

Mean Sum of Squares Between is 135.5

So, total mean sum of squares within is 135.5 x (16-4) = 1626

Thus, the Total sum of squares Between can be calculated as: 2144-1626 = 518

And the mean sum of squares between is 518/(4-1)=172.67

So the F-statistic is 172.67/135.5 = 1.2743

So the correct option is (a) 1.2743

B. The test statistic follows an F-distribution with degrees of freedom (3,12). At 0.05 level of significance the value of the F-critical value is 3.490. Since the calculated value of the F-test statistic is less that the F-critical value we do not sufficient evidence to reject the null hypothesis.

The correct option is (d) no