Question

The following data are given for a two-factor ANOVA with two treatments and three blocks. Treatment Block 1 2 A 42 32 B 33 20 C 48 39 Using the 0.05 significance level conduct a test of hypothesis to determine whether the block or the treatment means differ.

State the null and alternate hypotheses for treatments.

State the decision rule for treatments. (Round your answer to 1 decimal place.)

State the null and alternate hypotheses for blocks. (Round your answer to 1 decimal place.)

Also, state the decision rule for blocks.

Compute SST, SSB, SS total, and SSE and complete an ANOVA table. (Round your SS, MS values to 3 decimal places and F value to 2 decimal places.)

Give your decision regarding the two sets of hypotheses.

Answer 1

By using the MS-Excel two-factor ANOVA:

Summary of the data,

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
treatments 1	3	123	41	57
treatments 2	3	91	30.333	92.333
blocks 1	2	74	37	50
blocks 2	2	53	26.5	84.5
blocks 3	2	87	43.5	40.5
ANOVA Table
Source of Variation	SS	df	MS	F	P-value	F crit
treatments	170.667	1	170.667	78.769	0.012	18.513
blocks	294.333	2	147.167	67.923	0.015	19
Error	4.333	2	2.167
Total	469.333	5

From above ANOVA table:

SStotal = 469.334

SSB = 294.334

SST = 170.667

SSE = 4.334

Testing of Hypothesis:

H0: The mean effect of all treatments is the same.

against,

H1: The mean effect of treatment is different.

Test Statistics:

MST = SST/DF = 170.667

MSB = SSB/DF = 141.167

MSE = SSE/DF = 2.167

F = MST / MSE = 78.77 ( for treatment)

p-value = 0.012

F = MSB / MSE = 67.93 ( for block)

p-value = 0.015

Decision Rule:

If the p-value is greater than 0.05 level of significance then accept the null hypothesis.

for treatment p-value 0.012 < 0.05

so we reject the null hypothesis.

i.e.The mean effect of treatment is different.

for block p-value 0.015 < 0.05

so we reject the null hypothesis.

i.e.The mean effect of the block is different.

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