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An article reports on an experiment in which each of five groups consisting of six rats was put on a diet with a different carbohydrate. At the conclusion of the experiment, the DNA content of the liver of each rat was determined (mg/g liver), with the following results:

Carbohydrate xi.
Starch 2.58
Sucrose 2.63
Fructose 2.13
Glucose 2.41
Maltose 2.49

Assuming also that Σ Σ xij^2 = 185.4, does the data indicate that true average DNA content is affected by the type of carbohydrate in the diet? Construct an ANOVA table and use a 0.05 level of significance.

Respuesta :

akiran007

Answer:

H0 is rejected, Hence the true average DNA content is affected by the type of carbohydrate in the diet between samples

Step-by-step explanation:

A              B               C          D              E

2.58 2.63 2.13          2.41    2.49

∑A=2.58 ∑B=2.63 ∑C=2.13  ∑D=2.41   ∑E=2.49

A²                           B²                C²               D²             E²

6.6564               6.9169 4.5369             5.8081 6.2001

∑A²=6.6564 ∑B²=6.9169 ∑C²=4.5369 ∑D²=5.8081 ∑E²=6.2001

Calculations

Group A            B           C           D             E       Total

N      n1=1         n2=1 n3=1 n4=1 n5=1 n=5

∑xi T1=2.58    T2=2.63 T3=2.13 T4=2.41 T5=2.49 ∑x=12.24

∑x²i  6.6564      6.9169 4.5369 5.8081 6.2001 ∑x²=30.1184

Mean xi 2.58 2.63 2.13          2.41         2.49 Total2.448

Σ Σ xij^2 = 185.4 given

Let k = the number of different samples = 5

n=n1+n2+n3+n4+n5=1+1+1+1+1=5

Overall ˉx=12.24/5  =2.448

∑x=T1+T2+T3+T4+T5=2.58+2.63+2.13+2.41+2.49=12.24----1

Correction Factor=(∑x)²/n=12.242/5=29.9635--------(2)

∑T²i/ni=(2.582/1+2.632/1+2.132/1+2.412/1+2.492/1)=30.1184------(3)

∑x²=∑x²1+∑x²2+∑x²3+∑x²4+∑x²5

=6.6564+6.9169+4.5369+5.8081+6.2001=30.1184-------(4)

ANOVA:

Step-1 : sum of squares between samples

SSB=(∑T2i/ni)-(∑x)2/n

=(3)-(2)

=30.1184-29.9635

=0.1549

Step-2 : Total sum of squares

SST=SSB-CF

= 185.4-2.488

=182.952

Step-3 : Sum of Squares Within Samples

SSW=∑x²-(∑T²i/ni)=SST- SSB=

        = 182.952-0.1549=182.7971

Step-4 : variance between samples

MSB=SSB/k-1

=0.1549/4

=0.0387

Step-5 : variance within samples

MSW=SSW/n-k

=182.7971/5-5

=36.55942

Step-6 : test statistic F for one way ANOVA test

F=MSB/MSW

=0.0387/36.55942 = 0 .001058

ANOVA table

Source               Sums                  Degrees        Mean

of Variation    of Squares       of freedom        Squares

                                SS                        DF                 MS              F

B/w samples         SSB = 0.1549        k-1 = 4 MSB = 0.0387 0.001

Within samples  SSW = 0               n-k = 0 MSW = 36.55942  

Total                  SST = 0.1549       n-1 = 4  

H0 : The true average DNA content is not affected by the type of carbohydrate in the diet between samples

Ha : The true average DNA content is affected by the type of carbohydrate in the diet between samples

F(4,0) at 0.05 level of significance =0.5

As calculated F=0.001 >0.5

So, H0 is rejected, Hence the true average DNA content is affected by the type of carbohydrate in the diet between samples

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