Question

A one-way analysis of variance experiment produced the following ANOVA table. (You may find it useful to reference the q table).

SUMMARY
Groups	Count				Average
Column 1	6				0.71
Column 2	6				1.43
Column 3	6				2.15
Source of Variation	SS		df	MS	F	p-value
Between Groups	10.85		2	5.43	20.88	0.0000
Within Groups	3.86		15	0.26
Total	14.71		17

SUMMARY
Groups	Count	Average
Column 1	6	0.71
Column 2	6	1.43
Column 3	6	2.15
ANOVA
Source of Variation	SS	df	MS	F	p-value
Between Groups	10.85	2	5.43	20.88	0.0000
Within Groups	3.86	15	0.26
Total	14.71	17

a. Conduct an ANOVA test at the 1% significance level to determine if some population means differ.

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

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

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

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

b. Calculate 99% confidence interval estimates of μ₁ − μ_2,μ₁ − μ_3, and μ₂ − μ₃ with Tukey’s HSD approach. (If the exact value for n_T – c is not found in the table, then round down. Negative values should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places. Round your answers to 2 decimal places.)

Population Mean Differences	Confidence Interval
μ1 − μ2	[		,		]
μ1 − μ3	[		,		]
μ2 − μ3	[		,		]

c. Given your response to part b, which means significantly differ?

Population Mean Differences	Can we conclude that the population means differ?
μ1 − μ2
μ1 − μ3
μ2 − μ3

Answer 1

a)since p value <0.01

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

b)

MSE=			0.2600
df(error)=			15
number of treatments =			3
pooled standard deviation=Sp =√MSE=			0.510

critical q with 0.01 level and k=3, N-k=15 df=			4.84
Tukey's (HSD) =(q/√2)*(sp*√(1/ni+1/nj)         =			1.01

			Lower bound	Upper bound
	(xi-xj )	ME	(xi-xj)-ME	(xi-xj)+ME
μ1-μ2	-0.72	1.01	-1.73	0.29
μ1-μ3	-1.44	1.01	-2.45	-0.43
μ2-μ3	-0.72	1.01	-1.73	0.29

c)

μ1-μ2	no
μ1-μ3	YEs
μ2-μ3	No