Question

The following three independent random samples are obtained from three normally distributed populations with equal variance. The dependent variable is starting hourly wage, and the groups are the types of position (internship, co-op, work study). Group 1: Internship Group 2: Co-op Group 3: Work Study 13.25 12.75 14.25 13.5 11.5 11.75 14.5 14.75 12.5 11.5 9.25 13.75 16 11 12.5 14.5 12.75 12.5 12.25 11 12 Use technology to conduct a one-factor ANOVA to determine if the group means are equal using a = 0.02 Group means (report to 2 decimal places): Group 1: Internship: Group 2: Co-op: Group 3: Work Study:

Answer 1

The Summary Statistics are given in the table below

	Group 1	Group 2	Group 3
Total	95.5	83	89.25
n	7	7	7
Mean	13.64	11.86	12.75
Sum Of Squares	13.7322	18.3572	5.0000
Variance	2.2887	3.0595	0.8333
SD	1.5128	1.7491	0.9129

Mean

Group 1 = 13.64, Group 2 = 11.86 and Group 3 = 12.75

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Conducting the test at $\alpha$ = 0.02

The Hypothesis:

H0: There is no difference between the means of the 3 groups $\mu_1 =\mu_2 =\mu_3$

Ha: The mean of at least one group is different from the others. $\mu_i \neq \mu_j$

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The ANOVA table is as below.

Source	SS	DF	Mean Square	F	Fcv	p
Between	11.16	2	5.58	2.74	4.90	0.0918
Within/Error	36.73	18	2.041
Total	47.89	20

The p value is calculated for F = 2.74 for df1 = 2 and df2 = 18

The F critical is calculated at $\alpha$ = 0.02 for df1 = 2 and df2 = 18

  The Decision Rule:

If F test is > F critical, Then Reject H0.

Also if p-value is < $\alpha$ , Then reject H0.

The Decision:

Since F test (2.74) is < F critical (4.9), We Fail to Reject H0.

Also since p-value (0.0918) is < $\alpha$ (0.02), We Fail to Reject H0.

The Conclusion: There isn't sufficient evidence at the 98% level of significance to conclude that the mean of at least one group is different from the others.

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Calculations For the ANOVA Table:

Overall Mean : Since n is equal , the overall mean = mean of means

Overall Mean = (13.64 + 11.86 + 12.75) / 3 = 12.75

SS treatment = SUM [n* ( Individual Mean - overall mean)²] = 7 * (13.64 - 12.75)² + 7 * (11.86 - 12.75)² + 7 * (12.75 - 12.75)² = 11.16

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment / df1 = 11.16 / 2 = 5.58

SS error = SUM (Sum of Squares) = 13.7322 + 18.3572 + 5 = 37.09

df2 = N - k = 21 - 3 = 18

Therefore MS error = SS error / df2 = 37.09 / 18 = 2.041

F = MSTR / MSE = 5.58 / 2.041 = 2.74

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