Question

Is there any difference among population means? Please show steps to completion.

How do you compute SS, MS, and F values by hand?

Including critical and computed values show steps of how to obtain.

alpha = 0.05
set 1
27
23
17
21
xbar= 22
s = 4.16

set 2
21
28
25
18
xbar = 23
s= 4.40

set 3
33
28
23
20
xbar = 26
s = 5.71

Answer 1

The variance

Set A = 4.16 * 4.16 = 17.3056, Set B = 4.4 * 4.4 = 19.36, Set C = 5.71 * 5.71 = 32.6041

k = number of columns = 3, N = Total Observations = 4 * 3 = 12

df1 = df treatments = k - 1 = 3 - 1= 2

df2 = df error = N - k = 12 - 3 = 9

The Hypothesis:

H0: There is no difference between the means of the three Sets

Ha: At least 1 mean of the three sets is different from the others.

The ANOVA table is as below. the p value is calculated for F = 0.75 for df1 = 2 and df2 = 9

The Fcritical is calculated at = 0.05 for df1 = 2 and df2 = 9

Source	SS	DF	Mean Square	F	Fcv	p
Between	34.67	2.0	17.33	0.75	4.2565	0.4993
Within/Error	207.81	9.0	23.09
Total	242.5	11.0

The Decision Rule:

If Ftest is > F critical, Then Reject H0.

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

The Decision:

Since Ftest (0.75) is > F critical (4.2565), We Fail to Reject H0.

Also since p-value (0.4993) is < (0.05), We Fail to Reject H0.

The Conclusion: There isn't sufficient evidence at the 95% level of significance to conclude that there is a difference between the means of the three sets.

______________________________________________________

The Calculations for ANOVA are below:

The overall mean = [(22 * 4) + (23 * 4) + (26 * 4)] / 12 = 23.67

SS treatment = SUM n* ( - overall mean)² = 4 * (22 - 23.67)² + 4 * (23 - 23.67)² + 4 * (26 - 23.67)² = 34.67

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 34.67 / 2 = 17.33

SSerror = SUM [(n - 1) * Variance] = 3 * 17.3056 + 3 * 19.36 + 3 * 32.6041 = 207.81

df2 = N - k = 12 - 3 = 9

Therefore MS error = SSerror/df2 = 207.81 / 9 = 123.09

F = MSTR/MSE = 17.33 / 123.09 = 0.75