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Archaeologists can determine the diets of ancient civilizations by measuring the ratio of carbon-13 to carbon-12 in bones found at burial sites. Large amounts of carbon-13 suggest a diet rich in grasses such as maize, while small amounts suggest a diet based on herbaceous plants. The article a Climate and Diet in Fremont Prehistory: Economic Variability and Abandonment of Maize Agriculture in the Great Salt Lake Basina (J. Coltrain and S. Leavitt, American Antiquity, 2002:453-485) reports ratios, as a difference from a standard in units of parts per thousand, for bones from individuals in several age groups. The data are presented in the following table.
Age Group (years) Ratio
0–11 17.2 18.4 17.9 16.6 19.0 18.3 13.6 13.5 18.5 19.1 19.1 13.4
2–24 14.8 17.6 18.3 17.2 10.0 11.3 10.2 17.0 18.9 19.2
25–45 18.4 13.0 14.8 18.4 12.8 17.6 18.8 17.9 18.5 17.5 18.3 15.2 10.8 19.8 17.3 19.2 15.4 13.2
46+ 15.5 18.2 12.7 15.1 18.2 18.0 14.4 10.2 16.7
(a) Construct a complete ANOVA table.
(b) Can you conclude that the concentration ratios differ among the age groups?

Respuesta :

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 [tex]X_{ij}[/tex] denote the jth observation of the ith sample. The mean and the variance of the ith sample is given by

[tex]Xbar_{i}[/tex] =[tex]\frac{1}{J_{i}} summation(X_{ij})[/tex] where J=1 to [tex]J_{i}[/tex] and  [tex]J_{i}[/tex]  is ith sample size

[tex]s_{i} ^{2} =\frac{1}{J_{i}-1 }summation (X_{ij} - Xbar_{i})^{2}[/tex]

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

[tex]\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][/tex]

the sample grand mean denoted by Xbar is the average of all sampled items taken together:

Xbar=[tex]\frac{1}{49}[/tex] (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 = ∑ [tex]J_{i} (Xbar_{i} -Xbar)^{2}[/tex]  from i=1 to 49

SSTr= 20.9910

[tex]SSE= \sum\limits^I_{i=1}\sum\limits^{J_i}_{j=1}(Xbar_{ij}-Xbar_i)^{2}[/tex]

[tex]SSE=\sum\limits^I_{i=1}(J_i-1)s_i^{2}[/tex]

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

[tex]F=\frac{MSTr}{MSE}[/tex]

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 [tex]F_{3,45}[/tex] distribution. To find the P value we consult the F table.

P>0.100

The complete ANOVA table is below

[tex]\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][/tex]

(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

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