Question

. Given the following information obtained from three normally distributed populations, construct an ANOVA table. (Round intermediate calculations to at least 4 decimal places. Round "SS" to 2 decimal places, "MS" to 4 decimal places, and "F" to 3 decimal places.)

SSTR = 286.3; SSE = 2,785.8; c = 3; n₁ = n₂ = n₃ = 10

b. At the 5% significance level, what is the conclusion to the ANOVA test of mean differences?

• Do not reject H₀; we cannot conclude that some means differ.

• Reject H₀; we can conclude that some means differ.

• Do not reject H₀; we can conclude that some means differ.

• Reject H₀; we cannot conclude that some means differ.

Answer 1

Solution:

Given:

SSTR = 286.3;

SSE = 2,785.8;

c = 3;

n₁ = n₂ = n₃ = 10

Part a)

N = 10 +10+ 10 = 30

df_treatement = c - 1

df_treatement = 3 - 1

df_treatement = 2

df_total = N - 1

df_total = 30 - 1

df_total = 29

and

df_error = N - c

df_error = 30 - 3

df_error = 27

SST = SSTR + SSE

SST = 286.3 + 2785.8

SST = 3072.1

MSTR = SSTR / df_treatement

MSTR = 286.3 / 2

MSTR =  143.1500

and

MSE = SSE / df_error  

MSE = 2785.8 / 27

MSE =  103.1778

thus

F = MSTR / MSE

F = 143.1500 / 103.1778

F = 1.387

Thus we get:

Source	SS	df	MS	F
Treatment	286.3	2	143.1500	1.387
Error	2785.8	27	103.1778
Total	3072.1	29

Part b) At the 5% significance level, what is the conclusion to the ANOVA test of mean differences?

df_numerator = df_treatement = 2

df_denominator = df_error = 27

Level of significance = 0.05

F critical value = 3.35

Decision Rule:

Reject null hypothesis H0, if F test statistic value  > F  critical value = 3.35, otherwise we fail to reject H0.

Since F test statistic value =  F = 1.387 < F  critical value = 3.35, we fail to reject H0.

Thus correct answer is:

Do not reject H₀; we cannot conclude that some means differ.