Question

A random sample of five observations from three normally distributed populations produced the following data: (You may find it useful to reference the F table.)

Treatments
A				B				C
	24				18				31
	26				21				27
	19				27				21
	24				23				16
	30				19				30
x−Ax−A	=	24.6		x−Bx−B	=	21.6		x−Cx−C	=	25.0
s2AsA2	=	15.8		s2BsB2	=	12.8		s2CsC2	=	40.5

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a. Calculate the grand mean. (Round intermediate calculations to at least 4 decimal places and final answer to 4 decimal places.)

b. Calculate SSTR and MSTR. (Round intermediate calculations to at least 4 decimal places and final answers to 4 decimal places.)

c. Calculate SSE and MSE. (Round intermediate calculations to at least 4 decimal places and final answers to 4 decimal places.)

d. Specify the competing hypotheses in order to determine whether some differences exist between the population means.

• H₀: μ_A = μ_B = μ_C; H_A: Not all population means are equal.

• H₀: μ_A ≤ μ_B ≤ μ_C; H_A: Not all population means are equal.

• H₀: μ_A ≥ μ_B ≥ μ_C; H_A: Not all population means are equal.

e-1. Calculate the value of the F_{(df1, df2)} test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

e-2. Find the p-value.

• p-value 0.10
• 0.05 p-value < 0.10
• 0.025 p-value < 0.05
• 0.01 p-value < 0.025

• p-value < 0.01

f. At the 1% significance level, what is the conclusion to the test?

• Do not reject H₀ since the p-value is not less than significance level.

• Reject H₀ since the p-value is less than significance level.

• Reject H₀ since the p-value is not less than significance level.

• Do not reject H₀ since the p-value is less than significance level.

g. Interpret the results at αα = 0.01.

• We conclude that some means differ.

• We cannot conclude that some means differ.

• We cannot conclude that all means differ.

• We cannot conclude that population mean C is greater than population mean A.

Answer 1

Overall total = 356
Grand mean Y̅.. = 356 / 15 = 23.73

Y̅ .. is overall mean
y̅i . is treatment mean
SS total = ΣΣ(Yij - & Y̅..)2 = 310.92

SS treatment = ΣΣ(Yij - & y̅i.)2 = 34.5333

MS treatment = ΣΣ(Yij - & y̅i.)2 / a - 1 = 17.2667

SS error = Σ(y̅i. - & Y̅..)2 = 276.4000
MS error = Σ(y̅i. - & Y̅..)2 / N - a = 23.0333

ANOVA table
Source	SS	df	MS	F	p-value
Treatment	34.53	2	17.267	0.75	.4934
Error	276.40	12	23.033
Total	310.93	14

Part d)

H0: μA = μB = μC; HA: Not all population means are equal.

Part e)

Test Statistic :-
f = MS treatment / MS error = 0.7496

Part e-2)

P value = 0.4934

p-value > 0.10

Reject null hypothesis if P value < α = 0.01
Since P value = 0.49 > 0.01, hence we fail to reject the null hypothesis
conclusion :- Treatment means are same

Do not reject H0 since the p-value is not less than significance level.

We cannot conclude that all means differ.