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Answer:
Step-by-step explanation:
The null and the alternative hypothesis are:
H₀=μ₀=μ₁=μ₂=μ₃=μ₄
H₁=two or more μ are different X
Let denote the jth observation of the ith sample. The mean and the variance of the ith sample is given by
=
where J=1 to
and
is ith sample size
The sample sizes are J₁ =12 J₂=10 J₃=18 J₄=9
the total number in all samples combined is 49
finding Xbar₁ and s₁
Xbar₁= 1÷12(17.2+....+13.4)
Xbar₁= 17.0500
s₁²= 1÷(12-1) [(17.0500-17.2)+...+(17.0500-13.4)]
s₁²=5.1336
Similarly find the means and variances of other samples
the sample grand mean denoted by Xbar is the average of all sampled items taken together:
Xbar= (17.2+.....+16.7)
Xbar=16.2255
Compute the treatment sum of squares SSTr, the sum of squares SSE and the total sum of squares SST
SSTr = ∑ from i=1 to 49
SSTr= 20.9910
SSE=(12-1)(5.1336)+...(9-1)(7.4428)
SSE=353.1796
SST=SSTr+SSE
SST=374.1706
Find the treatment mean square MSTr and the error mean square MSE:
MSTr= SSTr/(I-1)
MSTr=6.9970
MSE=SSE/(N-I)
MSE=7.8484
F=0.89
The degrees of freedom are I-1=3 for the numerator and N-I=45 for the denominator. Under H₀, F has an distribution. To find the P value we consult the F table.
P>0.100
The complete ANOVA table is below
(b) The P-value is large. A follower of 5% rule would not reject H₀. Therefore, we cannot conclude the concentration ratios differ among the age groups