Question

A two-factor ANOVA was perofrmed with a-2, b-2, and r-3. The following are the data Male Female Less than bachelor's degree 15 10 6 12 10 At least one bachelor's degree 10 (a) Complete the ANOVA table F-statistics F-MS(A)/MSE F-MS(B)/MSE Variation df Mean squares SS(A) SS(B) (a-1) (b-1) SS(AB) SSE MS(A) MS(B) Factor A (Gender) a Factor B (Education)a-1 Interaction MS(AB) F-MS(AB)/MSE rror l-a MSE Total n- SS(Total) (b) Write down Ho and H and determine whether there are differences between men and women. (Use a 0.05.) (c) Write down Ho and Hi and determine whether there are differences in number of jobs between education levels. (Use a 0.05.) (d) Write down Ho and H and determine whether there is interaction between gender and education. (Use α = 0.05.)

Answer 1

Answer a

Given that the experiment has two factors (Factor Gender and Factor Education level ) at a = 2(Male and Female) and b = 2(less than bachelor degree and at least one bachelor degree) levels. Thus there are ab = 2× 2 = 4 different combinations of Gender and Education level.

With each combination r = 3 loads. r is called the number of replicates.This sums up to n = abr = 12 loads in total.

The Yijk denotes the no of observations for k (k = 1, 2, 3) with gender i (i = 1, 2) at education level j (j = 1, 2, 3) are recorded as follows

From the given data we can write

	Less than bachelor's degree			Mean	At least one bachelor's degree			Mean	mGender
Male	15	8	6	9.67	12	10	9	10.33	10.00
Female	10	5	7	7.33	10	6	9	8.33	7.83
meducation				8.50				9.333	8.92

2 23 i-1j-lk-1 (15-9.67)2 +(9-8.33)2 -70.67

DF error is n-ab = 12 - 4 = 8

MSE = SSE/ DF = 8.83

6 x[(10.00-8.92)^2 +(7.83-8.92)^2] = 2.35

In similar way we can fill the ANOVA table

Variation	df	SS	Mean squares	F - Statistics	P-value
Factor A(Gender)	1	2.35	2.35	0.27	0.6200
Factor B(Education)	1	0.35	0.35	0.04	0.8472
Interaction	1	4.36	4.36	0.49	0.5023
Error	8	70.67	8.83
Total	11	77.72667

Answer b

Main Effect of Factor A(Gender):

H0:μ_M.=μ_F.
HA: not all μi. are equal where μi. is the mean for ith gender.

Now form ANOVA table we got the p value for Factor A is greater than 0.05.

Therefore we are unable to reject the null hypothesis at 0.05 level of significance and conclude that the no significant effect of gender .

Answer c

Main Effect of Factor B(Education level):

H0:μ.1=μ.2
HA: not all μ.j are equal where μi.is the mean for ith education level

Again form ANOVA table we got the p value for Factor B is greater than 0.05.

Therefore we are unable to reject the null hypothesis at 0.05 level of significance and conclude that the there is not significant difference in education level as well.

Answer c

A × B Interaction:

H0: there is no interaction
HA: an interaction exists

ANOVA table says that p-value for interaction is also greater than 0.05 and we are unable to reject the null hypothesis.

Hence we can conclude that there no significant difference in education and gender together.