The table below contains three samples obtained from three different populations
ID: 3159645 • Letter: T
Question
The table below contains three samples obtained from three different populations. Please conduct an ANOVA test for the equality of the three population means and state if the test rejects the null hypothesis of the equality of the three means at the 90% confidence level?
Sample 1
Sample 2
Sample 3
7
11
6
5
8
5
7
11
4
6
7
5
6
12
5
6
11
6
8
8
8
9
7
6
7
6
14
9
7
4
8
8
8
8
9
13
9
6
10
8
11
14
8
9
14
9
12
8
6
11
8
12
Question 1 options:
The test fails to reject the null hypothesis that the three means are equal.
The test rejects the null hypothesis that the three means are equal
The test is inconclusive
Sample 1
Sample 2
Sample 3
7
11
6
5
8
5
7
11
4
6
7
5
6
12
5
6
11
6
8
8
8
9
7
6
7
6
14
9
7
4
8
8
8
8
9
13
9
6
10
8
11
14
8
9
14
9
12
8
6
11
8
12
Explanation / Answer
confidence level is 90%
hence level of significance is alpha=0.1
here there are three samples from three different populations.
let u1 , u2 , u3 be the respective means of the three populations.
our interest lies is testing
H0: u1=u2=u3 vs H1: not H0
now this is similar in testing for differential effects in a one way anova where the factors are the three different populations.
let the linear model be yij=u+ai+eij
where yij is the value of the respeonse for the jth member of the ith population
so here i=1,2,3 and j=r1,r2,r3
where r1=number of observations for the first sample=18
r2=number of observations for the second sample=18
r3=number of observations of the third sample=16
u=mean effect
ai=additional effect due to ith population
eij=error associated with yij
so the null hypothesis H0:u1=u2=u3 is now equvalent in testing
H0: a1=a2=a3 vs H1: not H0
let TSS be the total sum of squares, SSE be the error sum of squares and SSF be the sum of squres due to factors
so TSS=SSF+SSE
now there are total n=18+18+16=52 observations
hence total df=n-1=51
there are 3 factors. so df of factors=3-1=2
hence df of error=51-2=49
hence MSE=SSE/49 and MSF=SSF/2 with MSF and MSE being independent
hence the test statistic for testing H0 is F=MSF/MSE which under H0 follows an F distribution with df 2 and 49
using MINITAB we get the anova table as
Source DF SS MS F P
Factor 2 28.93 14.46 2.26 0.115
Error 49 313.31 6.39
Total 51 342.23
so here p value is p=0.115
and level of significance=alpha=0.1
so p>alpha
hence H0 is accepted
hence the conclusion is: The test fails to reject the null hypothesis that the three means are equal. [answer]
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