Question

3. Solve the following by hand, and by using either R or SAS: Let μ1-H2,H3 be, respectively, the means of three normal distributions with a common but unknown variance σ2. In order to test, at the 0-: 0.05 significance level, the hypothesis Ha : μι μ2 against all possible alternative hypotheses, we take an independent random sample of size 4 from each of these distributions. Determine whether we reject or fail to reject Ho if the observed values from these three distributions are, respectively: X1 5 96 X2 11 13 10 12 X3 10 6 9 9

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Answer 1

Answer:

Data:

	x1	x2	x3
	5	11	10
	9	13	6
	6	10	9
	8	12	9	Total
sum	28	46	34	108
sum of square	206	534	298	1038

$H_0: \mu_1 = \mu_2= \mu_3$

H₁: At least one of the mean is different from the others

Grand total = $\sum x$ =108

T1, T2 , T3 are group totals

N=12

$SST = \sum x^2 - \frac{ (\sum x)^2}{N}$

$=1038-108^2/12 =66$

$SSG = \frac {T_1^2}{n_1} +\frac {T_2^2}{n_2} +\frac {T_3^2}{n_3} - \frac{ (\sum x)^2}{N}$

$=28^2/4+46^2/4+34^2/4 - 108^2/12 =42$

SSE= SST-SSG    = 66-42 =24

ANOVA table
Source	SS	df	MS	F
Treatment	42.00	2	42/2=21.0000	21/2.6667 = 7.875
Error	24.00	9	24/9=2.6667
Total	66.00	11

Table value of F with (DF1=2, DF2=9 ) =4.26

Calculated F=7.875 > 4.26 the table value

The null hypothesis is rejected.

The data indicate there is a significant difference among the three groups

R code:

grp <- c(1,1,1,1,2,2,2,2,3,3,3,3)

y <- c(5,9,6,8,11,13,10,12,10,6,9,9)

grp<-factor(grp)

mydata <- data.frame(grp,y)

aov.1 <- aov(y~grp,data = mydata)

anova(aov.1)

R output:

Analysis of Variance Table

Response: y

          Df Sum Sq Mean Sq F value Pr(>F)

grp        2     42 21.0000   7.875 0.01054 *

Residuals 9     24 2.6667                 

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Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1