Question

Three students, Linda, Tuan, and Javier, are given laboratory rats for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 0.05, test the hypothesis that the three formulas produce the same mean weight gain.

H₀: μ₁ = μ₂ = μ₃
H_a: At least two of the means differ from each other

Forumla A	Forumla B	Forumla C
45	35.3	35.3
40.1	34.3	45
54.4	13.9	43.6
38.1	32.9	41.8
32	38.1	51
45.9	27.1	48.5
48.7	11.3	37.3
37	50	44.4
50.7	53.6	40.5

Run a single-factor ANOVA with α=0.05

. Round answers to 4 decimal places.

Test Statistic =
p-value =

Based on the p-value, what is the conclusion

Reject the null hypothesis: at least one of the group means is different

Fail to reject the null hypothesis: not sufficient evidence to suggest the group means are different

Answer 1

The Following data has been found from the given data:

	A	B	C
Total	391.9	297.1	387.3996
n	9	9	9
Mean	43.544	33.0111	43.0444
SS	420.90	1623.35	200.02
Variance	52.6128	202.9186	25.0028

The Hypothesis:

H0: There is no difference between the mean of the three treatments. ()

Ha: At least 2 of the means differ from each other.

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The Test Statistic:

.The ANOVA table is as below.

The p value is calculated for F = 3.399 for df1 = 2 and df2 = 24; p value = 0.0501

Source	SS	DF	Mean Square	F	p
Between	635.6026	2	317.8013	3.3990	0.0501
Within/Error	2244.2736	24	93.5114
Total	2879.8762	26

F observed = 3.399

The Decision Rule: If p-value is , Then reject H0.

The Decision: Since p-value (0.0501) is > (0.05), We Fail to Reject H0.

The Conclusion: Fail to reject the null hypothesis. There is insufficient evidence to suggest that the group means are different.

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Calculations For the ANOVA Table:

Overall Mean = [(9 * 43.5444) + (9 * 33.011) + (9 * 43.0444)]/27 = 39.867

SS treatment = SUM n* ( x̅_i - overall mean)² = 9 * (43.544 - 39.867)² + 9 * (33.011 - 39.867)² + 9 * (43.044 - 39.867)² = 635.6026

df1 = k - 1 = 3 - 1 = 2

MSTR = SS treatment/df1 = 635.602 / 2 = 317.8013

SSerror = SUM (SS) = 420.9 + 1623.35 + 200.02 = 2244.2736

df2 = N - k = 27 - 3 = 24

Therefore MS error = SSerror/df2 = 2244.27/24 = 93.5114

F = MSTR/MSE = 317.8/93.511 = 3.3990

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