Question

2. A researcher wishes to determine whether there is a difference in the average age of elementary school, high school, and community college teachers. Teachers are randomly selected from each group. Their ages are recorded below. Test the claim that at least one mean is different from the others. Use α = 0.01.
Resource: The One-Way ANOVA

Identify the null hypothesis, Ho, and the alternative hypothesis, Ha. Determine whether the hypothesis test is left-tailed, right-tailed, or two-tailed. Find the critical value(s) and identify the rejection region(s). Find the appropriate standardized test statistic. If convenient, use technology. Decide whether to reject or fail to reject the null hypothesis. Interpret the decision in the context of the original claim.

Elementary School Teachers	High School Teachers	Community College Teachers
23 28 27 25 52 37	41 36 38 47 42 31	39 45 36 61 45 35

Answer 1

(1)

H0:Null Hypothesis: ( There is no difference in the average age of elementary school, high school, and community college teachers)

Ha:Alternative Hypothesis: (At last one mean is different from the other 2 means). ( There is a difference in the average age of elementary school, high school, and community college teachers) (Claim)

(2)

two tailed

(3)

= 0.01

Degrees of Freedom for numerator = 2

Degrees of freedom for denominator = 15

From Table, critical value of F = 6.3589

Rejection Region:
Reject H0if F > 6.3589

(4)

From the given data,the following Table is calculated:

Summary of Data
	Treatments
	Elementary school teachers	High school teachers	Community college teachers	Total
N	6	6	6	18
∑X	192	235	261	688
Mean	192/6 = 32	225/6 = 39.1667	261/6 = 43.5	688/18 = 38.222
∑X2	6740	9355	11813	27908
Std.Dev.	10.9179	5.4924	9.5864	9.7351

From the above Table, ANOVA Table is calculated as follows:
Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F Stat
Between-treatments	404.7778	2	404.7778/2 = 202.3889	F = 202.3889/80,4222 = 2.51658
Within-treatments	1206.3333	15	1206.3333/15 = 80.4222
Total	1611.1111	17

The appropriate standardized test statistic is given by:

F = 202.3889/80,4222 = 2.51658

(5)

Since calculated value of F = 2.51658 is less than critical value of F = 6.3589, the difference is not significant. Fail to reject null hypothesis.

(6)

Conclusion:
The data do not support the claim that at last one mean is different from the other 2 means. .