Two-way analysis of variance is used when we want to know the following pieces of information.

• The means of the blocks or subpopulations

• The means of the treatment groups

• The means of the interaction of the subpopulation and treatment groups

Now you are probably confused but remember that two-way analysis of variance is an extension of randomized block designed. With randomized block design, there were two hypotheses one for the treatment groups and one for the blocks or subpopulations. What we are doing for two-analysis is assessing the interaction effect, which is the amount of the variation of

subpopulation and treatment group). The assessment of the interaction effect gives us the third hypothesis. To put it in simple words when both the subpopulation and the treatment are present combined they have some sort of influence just as they do when one or the other is present. Therefore, two-way analysis of variance is randomized block designed plus an interaction effect hypothesis.

Another important difference is the use of repeated measures. In a two-way analysis of variance, at least one of the groups received the treatment more than once. In a randomized block design, each group receives the treatment only one time. Your research questions determine if any group needs to experience the treatment more than once.

Below are the assumptions

• Sample randomly selected

• Populations have homogeneous standard deviations

• Population distributions are normal

• Population covariances are equal.

Here are the steps

1. Set up hypotheses (there will be three of them)

a.Treatment means (AKA factor A)

i. H0: There is no difference in the treatment means

ii. H1: H0 is false

b. Block means (AKA factor B)

i. H0: There is no difference in the block means

ii. H1: is false

c. Interaction between Factor A and B

i. H0: There is no interacting effect between factor A & B

ii. H1: There is an interacting effect between factor A & B

2. Determine your level of statistical significance

3. Determine F critical (there will be three now and the computer does this)

4. Calculate the F-test values (there will be three now and the computer does this)

5. Test hypotheses

6. State conclusion

Here is an example

A music teacher wants to study the effect of instrument type and service center on the repair time measured in minutes. Four instruments (sax, trumpet, clarinet, flute) were picked for the analysis. Each service center was assigned to perform the particular repair on two instruments in each category

Instrument

Service centers Sax Trumpet Clarinet Flute

1 60 50 58 60

70 56 62 64

2 50 53 48 54

54 57 64 46

3 62 54 46 51

64 66 52 49

Here are your research questions

• Is there a difference in the means of the repair time between service centers?

• Is there a difference in the means of the repair time between instrument type?

• Is there an interaction due to service center and type of instrument on the mean of the repair time

Let us go through each of our steps

Step 1: State the hypotheses

• Treatment means (AKA factor A)

a. H0: There is no difference in the means of the service centers

b. H1: H0 is false

• Block means (AKA factor B)

a. H0: There is no difference in means of the instrument types

b. H1: is false

• Interaction between Factor A and B

a. H0: There is no interacting effect between service center and instrument type

b. H1: There is an interacting effect between service center and instrument type

Step 2: Significance level

• Set at 0.1

Step 3: Determine F-Critical

For the instruments, F-critical is 2.81

For the service centers, F-critical is 2.61

For the interaction effect, F-critical is 2.33

Step 4: Calculate F-values

Service centers 3.2

Instrument type 1.4

Interaction 2.1

Step 5: Make decision

Since the F-value of 3.2 is greater than the F-critical of 2.8 we reject the null hypothesis for the service centers

Since the F-value of 1.4 is less than the F-critical of 2.61 we do not reject the null hypothesis for the instrument types

Since the F-value of 2.1 is less than the F-critical of 2.3 we do reject the null hypothesis for the interaction effect of service center and instrument type.

Step 6: Conclusion

Since we reject the null hypothesis that there is no difference in the means of the repair time of the service centers, we conclude that there is evidence of a difference in the repair times between service centers. This means that one service center is faster than the others are. To find out, do a posthoc test.

Since we do not reject the null hypothesis that there is no difference in the means of the repair time of the instrument types, we conclude that there is no evidence of a difference in the repair time between instrument types. In other words, it does not matter what type of instrument is being fixed as they will all take about the same amount of time.

Since we do not reject the null hypothesis that there is no interaction effect of service center and instrument type on the mean of the repair time, we conclude that there is no evidence of an interaction effect of service center and instrument type on repair time. In other words, if service center and instrument type are considered at the same time there is no difference in how fast the instruments are repaired.