Quantitative Analysis in Social Science Research

QUESTION 1 (30 points): Select three different examples that demonstrate an ANOVA test. For two of these examples, you should be able to reject the null hypothesis. For one of these examples, you should fail to reject the null hypothesis. Set up the research questions and hypotheses. Describe the main components of the ANOVA and interpret the findings. Create a new fourth example in which you would use an ANOVA test and explain whether or not you think you`d be able to reject the null hypothesis (and why). This example must not already appear as one of the associations in the tables. Your answer should make specific references to the labels and values of the variables.
ANOVA
ANOVA also known as Analysis Of Variance, is commonly be used in cases where there are more than two groups. When thee are only two samples, then we can use the t-test to compare the means of the different samples however, it may become unreliable when we have more than two samples. If we are comparing only two means, then the t-test also known as the independent samples will give the same results as the case of ANOVA. A question therefore arise on whether why not to use multiple t-tests. The t-test can only be used to test differences between two means. When there are more than two means, it is possible to compare each mean with each other mean using many t-tests. However, conducting such multiple t-tests can lead to severe complications and in such circumstances we use ANOVA. This technique is commonly used when an alternative procedure is required for testing the hypotheses concerning means especially in places when there are several populations.
In summary, we can say that ANOVA is used to compare the means of more than two samples. This can best be demonstrated by the use of examples as shown below:
Example 1:
Testing the effect of five different exercises
To make the test, we recruit ten men and assign one type of exercise to two men. Therefore we will have a total of five groups. After several weeks of training, their weights are recorded. From the results, we can therefore be able to establish whether the effect of these exercises on them was significant or not. This can be done comparing the average weight of the men in the different categories. This forms an example of one-way balanced ANOVA. This is because there is only one category that its effects have been considered for the study and balanced as the same number of men has been assigned on each exercise. Therefore the basic idea in the test is to test whether the samples are all alike or not. In this case we will accept the null hypothesis since there was a significant difference in the weights of the men based on the types of exercises that they went through.
Example 2: difference in resistance for different three temperature levels.
The data below resulted from measuring the difference in resistance resulting from subjecting identical resistors to three different temperatures for a period of 24 hours. The sample size of each group was 5.



Level 1

Level 2

Level 3




6.9

8.3

8.0


5.4

6.8

10.5


5.8

7.8

8.1


4.6

9.2

6.9


4.0

6.5

9.3



means

5.34

7.72

8.56



The resulting ANOVA table is
Example ANOVA table

Source

SS

DF

MS

F


Treatments

27.897

2

13.949

9.59

Error

17.452

12

1.454



Total (corrected)

45.349

14



Correction Factor

779.041

1



Interpretation of the ANOVA table The test statistic is the F value of 9.59. Using an α of .05, we have that F.05; 2, 12 = 3.89. Since the test statistic is much larger than the critical value, we reject the null hypothesis of equal population means and conclude that there is a (statistically) significant difference among the population means. The p-value for 9.59 is .00325, so the test statistic is significant at that level.
Example 3:
A manager wishes to determine whether the mean times required to complete a certain task differ for the three levels of employee training. He randomly selected 10 employees with each of the three levels of training (Beginner, Intermediate and Advanced). Do the data provide sufficient evidence to indicate that the mean times required to complete a certain task differ for at least two of the three levels of training? The data is summarized in the table.  
Level of Training

n

 

s2

Advanced

10

24.2

21.54

Intermediate

10

27.1

18.64

Beginner

10

30.2

17.76

Ha: The mean times required to complete a certain task differ for at least two of the three levels of training. 
Ho: The mean times required to complete a certain task do not differ the three levels of training. ( µB = µI = µA)
In this case, we are going to reject the null hypothesis since the given interval does not contain a 0 as required.
Example 4:
A teacher predicts that students will learn most effectively with a constant background sound, as opposed to an unpredictable sound or no sound at all. The teacher therefore randomly divides twenty-four students into three groups of eight. All students study a passage of text for 30 minutes. Those in group 1 study with background sound at a constant volume in the background. Those in group 2 study with noise that changes volume periodically. Those in group 3 study with no sound at all. After studying, all students take a 10 point multiple choice test over the material. Their scores follow:
group

test scores

1) constant sound

7

4

6

8

6

6

2

9

2) random sound

5

5

3

4

4

7

2

2

3) no sound

2

4

7

1

2

1

5

5

The following table is therefore obtained:
x1

x12

x2

x22 

x3

x32

7

49

5

25

2

4

4

16

5

25

4

16

6

36

3

9

7

49

8

64

4

16

1

1

6

36

4

16

2

4

6

36

7

49

1

1

2

4

2

4

5

25

9

81

2

4

5

25

x1 = 48

x12 = 322

x2 = 32

x22 = 148

x3 = 27

x32 = 125

(x1)2 = 2304

 

(x2)2 = 1024

 

(x3)2 = 729

 

M1 = 6

 

M2 = 4

 

M3 = 3.375

 

(according to the F sig/probability table with df = (2,21) F must be at least 3.4668 to reach p < .05, so F score is statistically significant)