﻿ Two-way Anova Worksheet | Problems & Solutions Two-way Anova Worksheet

Two-way Anova Worksheet
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1.
The means of each group of a two-way analysis of variance for two independent variables are plotted and shown in the three graphs. Which of these represents a disordinal interaction? Consider the interaction to be significant.  a. Both graph 1 and graph 2 b. Graph 1 c. Graph 3 d. Graph 2

Solution:

In a disordinal interaction, the lines cross each other and an intersection occurs and the interaction is significant.

In the above 3 graphs, graph 1 satisfies this condition.

So, graph 1 represents a disordinal interaction.

2.
In a two-way ANOVA, variable A has three levels and variable B has two levels. The number of data values in each group is five. Find the degrees of freedom for factor A × B and for the within (error) factor. a. 2, 24 b. 4, 3 c. 2, 1 d. 3, 2

Solution:

The number of levels of factor A and B are 3 and 2 respectively.

The number of data values in each group is 5.

The degrees of freedom are as follows:
Interaction (A × B): d.f.N = (3 - 1)(2 - 1) = 2
Within(error): d.f.D = 3 · 2 · (5 - 1) = 24
[Interaction (A × B): d.f.N = (a - 1)(b - 1)
Within(error): d.f.D = ab(n - 1)
where a, b are levels of factors A, B respectively.]

3.
In a two-way ANOVA, factor A has six levels and factor B has five levels. The number of data values in each group is seven. Find the degrees of freedom for factor A and factor B. a. 7, 6 b. 5, 4 c. 6, 5 d. 20, 180

Solution:

The degrees of freedom for factor A and factor B having 6 and 5 levels respectively, are as follows:

Factor A: d.f.N = a - 1 = 6 - 1 = 5

Factor B: d.f.N = b - 1 = 5 - 1 = 4

4.
The means of each group of a two-way analysis of variance for two independent variables are plotted as shown in the graph. What does the graph represent about the interaction of the variables?  a. Disordinal interaction b. Ordinal interaction c. No interaction

Solution:

When there is no significant interaction effect among the variables, the lines in the graph will be parallel or approximately parallel.

The given two lines are parallel to each other, so they represent 'no interaction'.

5.
In a two-way ANOVA, factor A has $a$ levels and factor B has $b$ levels. The number of data values in each group is $n$. Find the degrees of freedom for factor A and factor B. a. ($a$ - 1)($b$ - 1), $a$$b$($n$ - 1) b. $a$ - 1, $b$ - 1 c. $a$ + 1, $b$ + 1 d. $a$, $b$

Solution:

The degrees of freedom for factor A and factor B having a and b levels respectively, are as follows:
Factor A: d.f.N = a - 1
Factor B: d.f.N = b - 1

6.
In a two-way ANOVA, factor A has $a$ levels and factor B has $b$ levels. The number of data values in each group is $n$. Find the degrees of freedom for factor A × B (Interaction) and for the Within (error) factor. a. $a$ - 1, $b$ - 1 b. $a$ + 1, $b$ + 1 c. $a$, $b$ d. ($a$ - 1)($b$ - 1), $a$$b$($n$ - 1)

Solution:

The number of levels of factor A and B are a and b respectively.

The number of data values in each group is n.

The degrees of freedom are as follows:
Interaction (A × B): d.f.N = (a - 1)(b - 1)
Within(error): d.f.D = ab(n - 1)

7.
The means of each group of a two-way analysis of variance for two independent variables are plotted and shown in the three graphs. Which of these represents an ordinal interaction? Consider the interaction to be significant.  a. Graph 2 b. Both graph 1 and graph 2 c. Graph 3 d. Graph 1

Solution:

In an ordinal interaction, the lines do not cross each other, nor are they parallel.

In the above 3 graphs, graph 2 satisfies this condition.

So, graph 2 represents an ordinal interaction.

8.
Which of the following are the assumptions for two-way ANOVA?
I. All samples are drawn from normally distributed populations.
II. All populations have a common variance.
III. The groups must be equal in sample size.
IV. Within each group, the observations were sampled randomly and independently of each other. a. I and II b. I, II, III, and IV c. I, II, and III d. Only I

Solution:

The following are the assumptions for the two-way ANOVA

The populations from which the samples were obtained must be normally distributed or approximately normally distributed.

The samples must be independent.

The variances of the populations from which the samples were selected must be equal.

The groups must be equal in sample size.

I, II, III, and IV are all the assumptions for the two-way ANOVA.

9.
In a medical school a new method of teaching in which professional actors played the roles of patients was introduced. The test scores of male and female students who were taught by either the conventional method or by a new form of training using role-play are shown in the table. Find FA , FB. Is there any difference in the mean test score under (i) Gender (ii) Teaching Method, using a two-way ANOVA at $\alpha$ = 0.05. [A : Gender and B : Teaching Method ]  a. 57.45, 0.771, (i) no, (ii)yes b. 21.04, 57.45, (i) no, (ii) no c. 21.04, 0.771, (i) yes, (ii) no d. 21.04, 57.45, (i) yes, (ii) yes

Solution:

H0: There is no difference in the mean test score depeding on the gender of the student.
H1: There is a difference in the mean test score depeding on the gender of the student.
[Hypotheses for the variable Gender.]

H0: There is no difference in the mean test score depending on the teaching methods.
H1: There is a difference in the mean test score depending on the teaching methods.
[Hypotheses for the variable Teaching Method.]

a = 2, b = 2, and n = 5. The degrees of freedom(d.f.) are
Factor A: d.f.N = a - 1 = 2 - 1 = 1
Factor B: d.f.N = b - 1 = 2 - 1 = 1
Interaction (A × B): d.f.N = (a - 1)(b - 1) = (2 - 1)(2 - 1) = 1
Within(error): d.f.D = ab(n - 1) = 2·2 (5 - 1) = 16

[a, b are levels of factors A, B respectively and n is the number of data values in each group.]

The critical value FA corresponding to α = 0.05, d.f.N = 1, and d.f.D = 16 is 4.49

The critical value FB corresponding to α = 0.05, d.f.N = 1, and d.f.D = 16 is 4.49

The critical value FA × B corresponding to α = 0.05, d.f.N = 1, and d.f.D = 16 is 4.49

SSA = 994.05, SSB = 2714.5, SSA × B = 36.45 and SSW = 756

The mean squares are as follows:
MSA = SS / Aa-1 = 994.05
MSB = SS / Bb-1 = 2714.5
MSA × B = SS / A×B(a-1)(b-1) = 36.45
MSW = SS / Wab(n-1) = 47.25

[Substitute and simplify.]

MSA = 994.05, MSB = 2714.5, MSA × B = 36.45 and MSW = 47.25

The F values are computed as follows:
FA = MS / AMS / W = 21.04
FB = MS / BMS / W = 57.45
FA × B = MS / A×BMS / W = 0.771
[Substitute and simplify.]

Since FA = 21.04 and FB = 57.45 are greater than the critical value 4.49, the null hypotheses concerning the gender of the student and teaching methods should be rejected.

So, it can be concluded that there is a difference in the mean test score of students under gender and also the teaching method. The ANOVA summary table is shown below.

10.
There are three groups of plants that were exposed to 8 hours, 12 hours and 16 hours of sunlight per day during a given growing period and two kinds of fertilizers (P and Q) were used. The heights of the plants(in feet) after six month are tabulated as follows.
At $\alpha$ = 0.05, find the test values FA , FB. Is there any difference in the mean height of plants depending on the (i) type of fertilizer (ii) duration of exposure to sun using a two-way ANOVA. [ A : Fertilizer and B : Exposure to sun ]  a. 4.7, 3.9, (i) no, (ii) no b. 30.08, 64.03, (i) no, (ii) no c. 30.08, 64.03, (i) yes, (ii) yes d. 4.7, 3.9, (i) yes, (ii) yes

Solution:

H0: There is no difference in the mean height depending on the type of fertilizer used.
H1: There is a difference in the mean height depending on the type of fertilizer used.
[Hypotheses for the variable fertilizer.]

H0: There is no difference in the mean height depending on the duration of exposure to sun.
H1: There is a difference in the mean height depending on the duration of exposure to sun.
[Hypotheses for the variable exposure to sun.]

a = 2, b = 3, and n = 3. The degrees of freedom(d.f.) are
Factor A: d.f.N = a - 1 = 2 - 1 = 1
Factor B: d.f.N = b - 1 = 3 - 1 = 2
Interaction (A × B): d.f.N = (a - 1)(b - 1) = (2 - 1)(3 - 1) = 2
Within(error): d.f.D = ab(n - 1) = 2·3 (3 - 1) = 12
[a, b are levels of factors A, B respectively and n is the number of data values in each group.]

The critical value FA corresponding to α = 0.05, d.f.N = 1, and d.f.D = 12 is 4.7

The critical value FB corresponding to α = 0.05, d.f.N = 2, and d.f.D = 12 is 3.9

The critical value FA × B corresponding to α = 0.05, d.f.N = 2, and d.f.D = 12 is 3.9

SSA = 20.1, SSB = 85.8, SSA × B = 0.444 and SSW = 8

The mean squares are as follows:
MSA = SS / Aa-1 = 20.1
MSB = SS / Bb-1 = 42.89
MSA × B = SS / A×B(a-1)(b-1) = 0.222
MSW = SS / Wab(n-1) = 0.667
[Substitute and simplify.]

MSA = 20.1, MSB = 42.89, MSA × B = 0.222 and MSW = 0.667

The F values are computed as follows:
FA = MS / AMS / W = 30.08
FB = MS / BMS / W = 64.33
FA × B = MS / A×BMS / W = 0.333
[Substitute and simplify.]

Since FA = 30.08 > 4.7, FB = 64.33 > 3.9, the null hypothesis concerning the type of fertilizer, and the duration of exposure to sun should be rejected.

So, it can be concluded that the type of fertilizer does affect the mean height of the plants and also the duration of exposure to sun does affect the mean height of the plants. The ANOVA summary table is shown below.