1. Describe in your own words what an ANOVA test is and what it can be used for in business research.

ANOVA stands for “Analysis Of Variance. The purpose of conducting ANOVA is to see if there is any difference between two or more groups on some variable. ANOVA is a general technique that can be used to test the hypothesis that the means among two or more groups are equal, under the assumption that the sampled populations are normally distributed.
Example: The production of three varieties of wheat sown in 12 plots is recorded and ANOVA is used to test if there are significant differences in the production of the three varieties.

2. If we have three or more population means to compare, why can’t we test them two at a time until we exhaust all possible combinations?

In a two-sample t- test, we have only two groups, and just one t- test will be able to show if or not the two groups differ significantly.
If we have to compare three or more groups, a single t- test will not work. For example if we have to compare 4 groups to see if they are significantly different, we will need to conduct 12 separate t- tests (there are 12 combination-pairs of the 4 groups possible). This is tedious and unnecessary. This is precisely where ANOVA comes in.

3. Tedko Associates is a marketing research firm that specializes in comparative shopping. Tedko is hired by General Motors to compare the selling prices of Pontiac Sunbird with Chevy Cavalier. Posing as a potential customer, a representative of Tedko visited 8 Pontiac dealerships and 6 Chevrolet dealerships and obtained quotes on comparable cars. The standard deviation for the selling prices of 8 Pontiac Sunbirds is $350 and on six Cavaliers, $290. At the 0.01 significance level do Pontiac Sunbirds exhibit greater variance in quotes than Cavaliers?

n1 = 8
n2 = 6
s1^2 = 122500
s2^2 = 84100
(1) Formulate the hypotheses:
Ho: σ1^2 ≤ σ2^2
Ha: σ1^2 σ2^2
(2) Decide the test statistic and the level of significance:
F Test
α = 0.01
Numerator DOF = 7
Denominator DOF = 5
Critical F- score = 10.456
(3) State the decision Rule:
Reject Ho if F >10.456
(4) Calculate the value of test statistic:
F = s1^2 / s2^2 = 1.4566
(5) Compare with the critical value and make a decision:
Since 1.456614.2004we reject Ho and accept Ha
Decision: There is no sufficient evidence that the Pontiac Sunbird has greater variance than Cavaliers.

4. A physician who specializes in weight control has three different diets she recommend. As an experiment, she randomly selected 15 patients and then assigned 5 to each diet. After three weeks, the following weight losses in pounds were noted. At the 0.05 significance level, can she conclude that there is a difference in the mean amount of weight loss among the three diets? Plan A Plan B Plan C 5 6 7 7 7 8 4 7 9 5 5 8 4 6 9

Plan A / Plan B / Plan C
5 / 8 / 5
6 / 4 / 8
7 / 7 / 4
7 / 9 / 6
7 / 5 / 9
Ho: There is no difference in the mean weight loss figures of the three plans, that is μ1 = μ2 = μ3
Ha: At least one of the three plans has a mean weight loss figure different from the other two, that is at least one of μ1, μ2 and μ3 is different from the other two
MegaStat ANOVA Output:
One factor ANOVA
Mean / n / Std. Dev
6.466666667 / 6.4 / 5 / 0.89 / Plan A
6.466666667 / 6.6 / 5 / 2.07 / Plan B
6.466666667 / 6.4 / 5 / 2.07 / Plan C
6.5 / 15 / 1.64 / Total
ANOVA table
Source / SS / df / MS / F / p-value
Treatment / 0.13 / 2 / 0.067 / 0.021 / .9790
Error / 37.60 / 12 / 3.133
Total / 37.73 / 14
Since the test F- value (0.021) < the critical F- value (3.8853), we fail to reject Ho (Alternatively, since the test
p- value > 0.05, we fail to reject Ho)
Conclusion: There is no sufficient evidence that at least one of the three plans has a mean weight loss figure different from the other two.