Two-way Anova Worksheet

**Page 1**

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 |

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

So, graph 1 represents a disordinal interaction.

Correct answer : (2)

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 |

The number of data values in each group is 5.

The degrees of freedom are as follows:

Interaction (

Within(error):

[Interaction (

Within(error):

where

Correct answer : (1)

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 |

Factor

Factor

Correct answer : (2)

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 |

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

Correct answer : (3)

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. | $a$, $b$ |

Factor

Factor

Correct answer : (2)

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) |

The number of data values in each group is

The degrees of freedom are as follows:

Interaction (

Within(error):

Correct answer : (4)

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 |

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

So, graph 2 represents an ordinal interaction.

Correct answer : (1)

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.

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 |

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.

Correct answer : (2)

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 F_{A }, F_{B}. 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 |

H

[Hypotheses for the variable Gender.]

H

H

[Hypotheses for the variable Teaching Method.]

Factor

Factor

Interaction (

Within(error):

[

The critical value F

The critical value F

The critical value F

SS

The mean squares are as follows:

MS

MS

MS

MS

[Substitute and simplify.]

MS

The F values are computed as follows:

F

F

F

[Substitute and simplify.]

Since F

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.

Correct answer : (4)

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 F_{A }, F_{B}. 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 ]

At

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 |

H

[Hypotheses for the variable fertilizer.]

H

H

[Hypotheses for the variable exposure to sun.]

Factor

Factor

Interaction (

Within(error):

[

The critical value F

The critical value F

The critical value F

SS

The mean squares are as follows:

MS

MS

MS

MS

[Substitute and simplify.]

MS

The F values are computed as follows:

F

F

F

[Substitute and simplify.]

Since F

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.

Correct answer : (3)