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
![\left[\begin{array}{ccc}i&Mean&variance\\1&17.0500&5.1336\\2&15.4500&13.2317\\3&16.4972&6.9460\\4&15.4444&7.4428\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26Mean%26variance%5C%5C1%2617.0500%265.1336%5C%5C2%2615.4500%2613.2317%5C%5C3%2616.4972%266.9460%5C%5C4%2615.4444%267.4428%5Cend%7Barray%7D%5Cright%5D)
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
![\left[\begin{array}{cccccc}source&DF&SS&MS&F&P\\Agegroup&3&20.9910&6.9970&0.89&0.453\\Error&45&353.1796&7.8484\\Total&48&374.1706\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccccc%7Dsource%26DF%26SS%26MS%26F%26P%5C%5CAgegroup%263%2620.9910%266.9970%260.89%260.453%5C%5CError%2645%26353.1796%267.8484%5C%5CTotal%2648%26374.1706%5Cend%7Barray%7D%5Cright%5D)
(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