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# Regression: ANOVA

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An experiment was conducted to investigate the effect of four treatments, A, B, C and D on the yield of penicillin in a manufacturing process. It was necessary to use a different blend for each application if the four treatments. The results of the yields for this randomised block experiment are given in the table below.

Construct an ANOVA table to analysis these data to determine whether the treatment affect the yield.
Blend A B C D
1 89 88 97 94
2 84 77 92 79
3 81 87 87 85
4 87 92 89 84
5 79 81 80 88

? Construct an ANOVA table to analysis these data to determine whether the treatment affect the yield.

? Also, give a 95% confidence interval for the standard deviation of the 'errors'.

See attached for full question.

https://brainmass.com/statistics/regression-analysis/regression-anova-37766

## SOLUTION This solution is FREE courtesy of BrainMass!

An experiment was conducted to investigate the effect of four treatments, A, B, C and D on the yield of penicillin in a manufacturing process. It was necessary to use a different blend for each application if the four treatments. The results of the yields for this randomised block experiment are given in the table below.

Construct an ANOVA table to analysis these data to determine whether the treatment affect the yield.

Hence, fill in the missing values, denoted (**) in the GLIM report below (I have been given the solutions that I have included in red, please show me manually, showing all working, that those answers are correct).

Also, give a 95% confidence interval for the standard deviation of the 'errors'.

Blend A B C D
1 89 88 97 94
2 84 77 92 79
3 81 87 87 85
4 87 92 89 84
5 79 81 80 88

Blend A B C D
1 89 88 97 94
2 84 77 92 79
3 81 87 87 85
4 87 92 89 84
5 79 81 80 88 Grand total
Total=T i= 420 425 445 430 1720
ni= 5 5 5 5 20

Total number of treatments=k= 4

Null Hypothesis: H0: u1=u2=u3=u4 (All population means are equal)
Research (Alternative) Hypothesis: H1: At least two of the population means are different

1) Calculation of CM
CM= correction for mean=( total of all observations )^2 / n = ( sigma yi )^2 / n

total of all observations = 1720 = 420+425+445+430
n=total number of observations = 20

Therefore, CM = 147920 = 1720^2/ 20

2) Calculation of Total SS
Total SS= ( Sum of squares of each observation) - CM =( sigma yi^2 ) / n

Blend A yi 2 B yi 2 C yi 2 D yi 2
1 89 7921 = 89^2 88 7744 97 9409 94 8836
2 84 7056 = 84^2 77 5929 92 8464 79 6241
3 81 6561 = 81^2 87 7569 87 7569 85 7225
4 87 7569 = 87^2 92 8464 89 7921 84 7056
5 79 6241 = 79^2 81 6561 80 6400 88 7744
35348 36267 39763 37102

( sigma yi^2 ) = 148480 =35348+36267+39763+37102

Total SS= 560 =148480-147920

3) Claculation of SST = Sum of squares for treatment

SST=sum of squares for treatment
= (Sum of squares of treatment totals with each square divided by the number of obseravtions for that treatment) - CM

= T1^2 / n1 + T2^2 / n2 +T3^2 / n3 +T4^2 / n4 - CM

70 =420^2/5+425^2/5+445^2/5+430^2/5-147920

df=k-1= 3 (df=degrees of freedom)

4) Calculation of SSE = Sum of squares for error
SSE= (total sum of squares) - ( sum of squares for treatments)
= Total SS - SST
490 =560-70

df=(sigma*nj)- k=nT - k= 16 = 5 + 5 + 5 + 5 - 4 df=degrees of freedom

5) MST=mean square for treatments = sum of squares for treatments / ( number of treatments -1)

MST= 23.3333 =70/(4-1)

6) MSE= Mean Square error=sum of squares for error / (total number of observations - number of treatments)

MSE= 30.625 =490/(20-4)

7) F= test statistic for resrarch hypothesis that at least two population means differ
= MST/MSE

F= 0.7619 =23.3333/30.625

8) Critical Value of F

significance level of the test alpha(alpha)= 0.05 (Confidence interval 95%)
degrees of freedom associated with MST=k-1= 3
degrees of freedom associated with MSE=n-k= 16
For significance level= 0.05 and degrees of freedom for numerator and denominator = and respectively the value of F from the tables is
F= 3.2389
Therefore the critical value is= 3.2389
The acceptance region for the Null Hypothesis is less than 3.2389

9) Conclusion:
Since the value of F (test statistic)= 0.7619
is within the acceptance region of F= 3.2389
we will accept the null hypothesis that there is no difference between the means

There are no differences in the means at 0.05 level of significance

The probability associated with F statistic= 53.18%
This is the probability of getting an F value as large or larger than 0.7619 if Null hypothesis is true
Since the probability value is larger than the significance level of 5.00%
We accept the Null hypothesis

Construct an ANOVA table to analysis these data to determine whether the treatment affect the yield.
ANOVA table:

Source Degree of freedom Sum of squares Mean Square

Model 3 70 23.3333
Error 16 490 30.625
Total 19 560
Model F= 0.7619 Pr > F= 0.5318

Since the value of F (test statistic)= 0.7619
is within the acceptance region of F= 3.2389
we will accept the null hypothesis that there is no difference between the means

There are no differences in the means at 0.05 level of significance

Also, give a 95% confidence interval for the standard deviation of the 'errors'.

Variance =s 2 = 30.625

We use the chi square distribution for the confidence interval for the variance

degrees of freedom=n-k= 16
Confidence interval of 95%
Corresponding to this confidence interval and degrees of freedom the chi square values are
ΧL^2 = 6.9077
ΧU^2 = 28.8453

sigma*L^2 =(n-k) s^2 / ΧU^2 = 16.9872 =16*30.625 / 28.8453
or σL = 4.12 = sqrt16.9872

sigma*U^2 =(n-k) s^2 / ΧL^2 = 70.9353 =16*30.625 / 6.9077
or σU = 8.42 = sqrt70.9353