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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.

    © BrainMass Inc. brainmass.com December 24, 2021, 5:15 pm ad1c9bdddf
    https://brainmass.com/statistics/regression-analysis/regression-anova-37766

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    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

    Answer Upper confidence limit= 8.42
    Lower confidence limit= 4.12

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:15 pm ad1c9bdddf>
    https://brainmass.com/statistics/regression-analysis/regression-anova-37766

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