Question

Describe a situation where you would use ANOVA by stating

1. the quantity of populations that are to be investigated,

2. a quantitative variable on these populations,

3. the sizes of samples from these Populations,

4. your null hypothesis,

5. your alternative hypothesis, and

6. a significance level.

Then find 7. The degrees of freedom of the numerator of your F-statistic, p

8. the degrees of freedom of the denominator of your F-statistic, and state

9. what the P-value for your test statistic being less than your significance level would imply. (you do not need to list the values of the variable for individuals in either the samples or the population, and that the values for 3 and 6 do not need to be calculated, only stated.)

Answer 1

1. Total number of students in each of five classes in a school.

2. Mean number of students in each class.

3. Sample of 6 students from each class.

4. Null hypothesis: No significant difference between mean number of students of each class.

5. Alternative hypothesis: There is a significant difference between mean number of students of five classes, that is, at least two classes differ significantly regarding their mean number of students.

6. 0.05

7. Df = C-1 = 5-1 = 4 (S1 square = SSC/C-1)

8. Df = N-C = 30-5 = 25 (S2 square= SSE/N-C)

where Df = degrees of freedom; C= No. of classes (samples) = 5; N = goal students from all five samples= 5*6= 30. S1 square greater than S2 square.

9. P value less than the significance level of 0.05 would imply that the null hypothesis is to be rejected and conclude that at least two classes will significantly differ regarding their mean number of students.