Question

Answer the one-way ANOVA questions using the data below. Use α = 0.01.

1	2	3	4
53 49 47 42 48 34 44	44 39 36 34 24 30 12	46 42 39 43 36 34 47	25 35 30 27 33 28 22

b) Compute the appropriate test statistic(s) to make a decision about H₀.
critical value = ___________ ; test statistic = _____________

Decision:  ---Select--- Reject H0 Fail to reject H0

e) Regardless of the H₀ decision in b), conduct Tukey's post hoc test for the following comparisons:
3 vs. 4: difference =   ; significant:  ---Select--- Yes No
1 vs. 2: difference =   ; significant:  ---Select--- Yes No

f) Regardless of the H₀ decision in b), conduct Scheffe's post hoc test for the following comparisons:
1 vs. 3: test statistic =   ; significant:  ---Select--- Yes No
2 vs. 3: test statistic =   ; significant:  ---Select--- Yes No

Answer 1

The Completed summary table is below

	Group 1	Group 2	Group 3	Group 4
Total	317	219	287	200
n	7	7	7	7
Mean	45.29	31.29	41.00	28.57
Sum Of Squares	223.43	677.43	144.00	121.71
Variance	37.2381	112.9048	24.0000	20.2857
SD	6.1023	10.6257	4.899	4.504

The Hypothesis:

H0: There is no difference between the means of the 4 groups

Ha: The mean of at least one group is different from the others.

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The ANOVA table is as below.

The p value is calculated for F = 9 for df1 = 3 and df2 = 24

The F critical is calculated at = 0.01 for df1 = 3 and df2 = 24

Source	SS	DF	Mean Square	F	Fcv	p
Between	1312.41	3	437.47	9.00	4.72	0.0004
Within/Error	1166.57	24	48.607
Total	2478.98	27

  The Decision Rule:

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

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

The Decision:

Since F test () is > F critical (), We Reject H0.

Also since p-value (0.) is < (0.01), We Reject H0.

The Conclusion: There is sufficient evidence at the 95% level of significance to conclude that the mean of at least one treatment 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 = (45.29 + 31.29 + 41 + 28.57) / 4 = 36.54

SS treatment = SUM [n* ( Individual Mean - overall mean)²] = 7 * (45.29 - 36.54)² + 7 * (31.29 - 36.54)² + 7 * (41 - 36.54)² + 7 * (28.57 - 36.54)² = 1312.41

df1 = k - 1 = 4 - 1 = 3

MSTR = SS treatment / df1 = 1312.41 / 3 = 437.47

SS error = SUM (Sum of Squares) = 223.43 + 677.43 + 144 + 121.21 = 1166.57

df2 = N - k = 28 - 4 = 24

Therefore MS error = SS error / df2 = 1167.57 / 24 = 48.607

F = MSTR / MSE = 437.47 / 48.607 = 9.00

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Tukeys Post Hoc

q = Tukeys Critical Value

n = number of replicates in each sample. Here n = 7, MSerror = 48.607

Tukeys critical is found from the critical value tables for = 0.01, for k (# of columns) = 4 on the horizontal and df error = 24 on the vertical.is 4.907

q = 4.907
Therefore:

The Rule is that if |M_i – M_j| is > Tukeys HSD, then there is a significant difference between groups.

3 vs 4: Mean difference = 41 - 28.57 = 12.43, which is > HSD, therefore YES the difference is significant

1 vs 2: Mean difference = 45.29 - 31.29 = 14, which is > HSD, therefore YES the difference is significant

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Scheffes Post HOC

Sceffes Value is calculated as Sqrt[ (k-1) * f value * MSE * (1/ni + 1/nj)]

ni = nj = 7, MSE = 48.607, f = 9 and k - 1 = 3

Therefore Scheffes Value = Sqrt[3 * 9 * 48.607 * (2/7)] = 19.36

The the difference is > Scheffs, then the difference is significant

1 vs 3: 45.29 - 41 = 4.29 which is < 19.36. Therefore NO, Not significant.

2 vs 3: 31.29 - 41 = 9.71 which is < 19.36. Therefore NO, Not significant.

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