Problem No.1
  1. Derive the expected mean squares of a three-factor factorial experiment where factors A and B are random and C is fixed.
R R F R
a b c r
i j k l
\[ \alpha_i \] 1 b c r
\[\beta_j \] a 1 c r
\[\gamma_k \] a b 0 r
\[(\alpha\beta)_{ij} \] 1 1 c r
\[(\alpha\gamma)_{ik} \] 1 b 0 r
\[(\beta\gamma)_{jk} \] a 1 0 r
\[(\alpha\beta\gamma)_{ijk}\] 1 1 0 r
\[\epsilon_{(ijk)l} \] 1 1 1 1

\(E[MSA]=bcr\sigma^2_{\alpha}+cr\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\)
\(E[MSB]=acr\sigma^2_{\beta}+cr\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\) \(E[MSC]=abr\frac{\sum\gamma^2_k}{c-1}+br\frac{\sum\sum(\alpha\gamma)^2_{ik}}{(a-1)(c-1)}+ar\frac{\sum\sum(\beta\gamma)^2_{jk}}{(b-1)(c-1)}+r\frac{\sum\sum\sum(\alpha\beta\gamma)^2_{ijk}}{(a-1)(b-1)(c-1)}+\sigma^2_\epsilon\) \(E[MSAB]=cr\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\) \(E[MSAC]=br\frac{\sum\sum(\alpha\gamma)^2_{ik}}{(a-1)(c-1)}+r\frac{\sum\sum\sum(\alpha\beta\gamma)^2_{ijk}}{(a-1)(b-1)(c-1)}+\sigma^2_\epsilon\)
\(E[MSBC]=ar\frac{\sum\sum(\beta\gamma)^2_{jk}}{(b-1)(c-1)}+r\frac{\sum\sum\sum(\alpha\beta\gamma)^2_{ijk}}{(a-1)(b-1)(c-1)}+\sigma^2_\epsilon\)
\(E[MSABC]=r\frac{\sum\sum\sum(\alpha\beta\gamma)^2_{ijk}}{(a-1)(b-1)(c-1)}+\sigma^2_\epsilon\)
\(E[MSE]=\sigma^2_\epsilon\)

  1. Provide the exact and approximate F tests, if any, for testing the first- and second-order interactions and the main effects.

AB Interaction:
\(H_0:\sigma^2_{\alpha\beta}=0\)
\(H_1:\sigma_{\alpha\beta}>0\)
\(F=\frac{MSAB}{MSE}\)

AC Interaction:
\(H_0:(\alpha\gamma)_{ik}=0,\ \forall ik\)
\(H_1:(\alpha\gamma)_{ik}\not=0,\ \exists ik\)
\(F=\frac{MSAC}{MSABC}\)

BC Interaction:
\(H_0:(\beta\gamma)_{jk}=0,\ \forall jk\)
\(H_0:(\beta\gamma)_{jk}\not=0,\ \exists jk\)
\(F=\frac{MSBC}{MSABC}\)

ABC Interaction:
\(H_0:(\alpha\beta\gamma)_{ijk}=0,\ \forall ijk\)
\(H_0:(\alpha\beta\gamma)_{ijk}\not=0,\ \exists ijk\)
\(F=\frac{MSABC}{MSE}\)

Factor A:
\(H_0:\sigma^2_\alpha=0\)
\(H_0:\sigma^2_\alpha>0\)
\(F=\frac{MSA}{MSAB}\)

Factor B:
\(H_0:\sigma^2_\beta=0\)
\(H_0:\sigma^2_\beta>0\)
\(F=\frac{MSB}{MSAB}\)

Factor C:
\(H_0:\gamma_k=0,\ \forall k\)
\(H_0:\gamma_k\not=0,\ \exists k\)
\(F=\frac{MSC+MSABC}{MSAC+MSBC}\)

Problem No. 2

Suppose a botanist wants to understand the effects of sunlight (low vs. medium vs. high) and watering frequency (daily vs. weekly) on the growth (plant height measured seven weeks after planting) of a certain species of plant.

  1. Calculate the sums of squares by hand using the formulas presented in class. Summarize your computations in the ANOVA Table.

\(SST=\sum^a_{i=1}\sum^b_{j=1}\sum^c_{k=1} Y^2_{ijk}-\frac{Y^2...}{n}=(5^2+5.2^2+...+5.5^2)-\frac{165.4^2}{30}\approx10.935\)
\(SSTr=\frac{1}{r}\sum^a_{i=1}\sum^b_{j=1} Y^2_{ij}-\frac{Y^2...}{n}=(\frac{24.9^2}{5}+\frac{28.6^2}{5}+...+\frac{26.6^2}{5})-\frac{165.4^2}{30}\approx4.103\)
\(SSA=\frac{1}{br}\sum^a_{i=1} Y^2_{i..}-\frac{Y^2...}{n}=(\frac{51^2}{10}+\frac{58.9^2}{10}+\frac{55.5^2}{10})-\frac{165.4^2}{30}\approx3.141\)
\(SSB=\frac{1}{ar}\sum^a_{i=1} Y.^2_{j.}-\frac{Y^2...}{n}=(\frac{82.4^2}{15}+\frac{83^2}{15})-\frac{165.4^2}{30}\approx0.012\)
\(SSAB=SSTr-SSA-SSB=4.103-3.145-0.012\approx0.95\)
\(SSE=SST-SSTr=10.935-4.103\approx6.932\)

ANOVA Table (Fixed Model)

Variation df SS MS F p
A 2 3.141 1.571 5.512 0.011
B 1 0.012 0.012 0.042 0.839
AB 2 0.95 0.475 1.667 0.21
Error 24 6.832 0.285
Total 29 10.935

ANOVA Table (Random Model)

Variation df SS MS F p
A 2 3.141 1.571 3.307 0.09
B 1 0.012 0.012 0.025 0.878
AB 2 0.95 0.475 1.667 0.21
Error 24 6.832 0.285
Total 29 10.935
  1. Test the appropriate hypotheses if sunlight intensity (A) is treated as a random factor and watering frequency as a fixed factor.

Test of Significance of AB Interaction effect \(H_0:(\alpha\beta)_{ij}=0,\ \forall(ij)\)
\(H_0:(\alpha\beta)_{ij}\not=0,\ \exists(ij)\)
Test Statistic:\(F=1.66;p=0.21\)
Decision: We cannot reject the null hypothesis. Conclusion: We can then say that at the 5% level of significance, the effect of the intensity of sunlight and the watering frequency on the growth of a certain species of plant did not vary. Moreover, there is a non significant interaction effect.

Test of significance of the main effect of sunlight intensity (A)
\(H_0:\sigma^2_\alpha=0\)
\(H_0:\sigma^2_\alpha>0\)
\(Test\ Statistic:\ F=3.307;p=0.09\)
Decision: We cannot reject the null hypothesis. Conclusion: There is a non significant variation on the growth of a certain species of plant with the varying intensity of sunlight at the 5% level of significance.

Test of significance of the main effect of watering frequency (B)
\(H_0:\beta_j=0,\ \forall j\)
\(H_0:\beta_j\not=0,\ \exists j\)
\(Test\ Statistic:\ F=0.042;p=0.839\)
Decision: Do not reject the null hypothesis. Conclusion: At the 5% level of significance, it shows that there is a non significant variation on the growth of a certain species of plant with the different watering frequency.

  1. Estimate the variance in growth which can be attributed to sunlight intensity.

\(E[MSA]=br\sigma^2_\alpha\ +\sigma^2_\epsilon\)
\(E[MSB]=ar\frac{\sum\beta^2_j}{b-1}+r\frac{\sum\sum(\alpha\beta)^2_{ij}}{(a-1)(b-1)}+\sigma^2_\epsilon\)
\(E[MSAB]=r\frac{\sum\sum(\alpha\beta)^2_{ij}}{(a-1)(b-1)}+\sigma^2_\epsilon\)
\(E[MSE]=\sigma^2_\epsilon\)

\(a=3;\ b=2;\ r=5\)
\(\hat\sigma^2_\epsilon=MSE=0.285\)

\(E[MSAB]=r\frac{\sum\sum(\alpha\beta)^2_{ij}}{(a-1)(b-1)}+\sigma^2_\epsilon\)
\(0.475=5\frac{\sum\sum(\alpha\beta)^2_{ij}}{(a-1)(b-1)}+0.285\)
\(\frac{\sum\sum(\alpha\beta)^2_{ij}}{(a-1)(b-1)}=\frac{0.475-0.285}{5}\)
\(\frac{\sum\sum(\alpha\beta)^2_{ij}}{(a-1)(b-1)}=0.038\)

\(E[MSB]=ar\frac{\sum\beta^2_j}{b-1}+r\frac{\sum\sum(\alpha\beta)^2_{ij}}{(a-1)(b-1)}+\sigma^2_\epsilon\)
\(MSB=ar\frac{\sum\beta^2_j}{b-1}+MSAB\)
\(0.012=3(5)\frac{\sum\beta^2_j}{b-1}+0.475\)
\(\frac{\sum\beta^2_j}{b-1}=\frac{0.012-0.475}{15}\)
\(\frac{\sum\beta^2_j}{b-1}=-0.031\)

\(E[MSA]=br\sigma^2_\alpha\ +\sigma^2_\epsilon\)
\(1.571=2(5)\sigma^2_\alpha+0.285\)
\(\sigma^2_\alpha=\frac{1.571-0.475}{15}\)
\(\sigma^2_\alpha=0.1285\)

Hence,
\(V(Y_{ijk})=\sigma^2_\alpha+\frac{\sum\beta^2_j}{b-1}+\frac{\sum\sum(\alpha\beta)^2_{ij}}{(a-1)(b-1)}+\sigma^2_\epsilon\)
\(V(Y_{ijk})=0.1285+(-0.031)+0.038+0.285\)
\(V(Y_{ijk})=0.4205\)

  1. Draw the appropriate overall conclusion from the experiment.
Problem No. 3

An experiment was set up to compare the effect of different soil pH and calcium additives on the increase in trunk diameters for orange trees. Annual applications of elemental sulfur, gypsum, soda ash, and other ingredients were applied to provide pH value levels of 4, 5, 6, and 7. Three levels of a calcium supplement (100, 200, and 300 pounds per acre) were also applied. All factor–level combinations of these two variables were used in the experiment. At the end of a 2-year period, trunk diameters of three orange trees were determined at each factor–level combination.

  1. Calculate the sums of squares by hand using the formulas presented in class. Summarize your computations in the ANOVA Table.

\(SST=(5.2^2+5.9^2+...+6.4^2)-\frac{253.9^2}{36}\approx10.810\)
\(SSTr=(\frac{17.4^2}{3}+\frac{22^2}{3}+...+\frac{19.8^2}{3})-\frac{253.9^2}{36}\approx9.183\)
\(SSA=(\frac{58.5^2}{9}+\frac{65.8^2}{9}+...+\frac{63^2}{9})-\frac{253.9^2}{36}\approx4.461\)
\(SSB=(\frac{83.5^2}{12}+\frac{88^2}{12}+\frac{82.4^2}{12})-\frac{253.9^2}{36}\approx1.467\)
\(SSAB=SSTr-SSA-SSB=9.183-4.461-1.467\approx3.255\)
\(SSE=SST-SSTr=10.810-9.183\approx1.627\)

ANOVA Table (Fixed Model)

Variation df SS MS F p
A 3 4.461 1.487 21.868 5.955
B 2 1.467 0.734 10.794 0.0005
AB 6 3.255 0.543 7.985 17.647
Error 24 1.627 0.068
Total 35 10.81

ANOVA Table (Random Model)

Variation df SS MS F p
A 3 4.461 1.487 2.738 0.066
B 2 1.467 0.734 1.352 0.278
AB 6 3.255 0.543 7.985 17.647
Error 24 1.627 0.068
Total 35 10.81
  1. Suppose the pH and calcium levels used in the experiment are random samples from respective populations of pH and calcium levels. Estimate the variance components.

\(E[MSA]=br\sigma^2_\alpha+r\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\)
\(E[MSB]=ar\sigma^2_\beta+r\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\)
\(E[MSAB]=r\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\)
\(E[MSE]=\sigma^2_\epsilon\)

\(a=4;\ b=3;\ r=3\)
\(\hat\sigma^2_\epsilon=MSE=0.068\)

\(E[MSAB]=r\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\)
\(0.543=3\sigma^2_{\alpha\beta}+0.068\)
\(\sigma^2_{\alpha\beta}=\frac{0.543-0.068}{3}\)
\(\sigma^2_{\alpha\beta}=0.158\)

\(E[MSB]=ar\sigma^2_\beta+r\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\)
\(E[MSB]=ar\sigma^2_\beta+E[MSAB]\)
\(0.743=4(3)\sigma^2_\beta+0.543\)
\(\sigma^2_\beta=\frac{0.743-0.543}{12}\)
\(\sigma^2_\beta=0.0167\)

\(E[MSA]=br\sigma^2_\alpha+r\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\)
\(E[MSA]=br\sigma^2_\alpha+E[MSAB]\)
\(1.487=3(3)\sigma^2_\alpha+0.543\)
\(\sigma^2_\alpha=\frac{1.487-0.543}{9}\)
\(\sigma^2_\alpha=0.105\)

Hence, \(V(Y_{ijk})=\sigma^2_\alpha+\sigma^2_\beta+\sigma^2_{\alpha\beta}+\sigma^2_\epsilon\)
\(V(Y_{ijk})=0.105+0.0167+0.158+0.068\)
\(V(Y_{ijk})\approx0.348\)

\(E[MSE]=\sigma^2_\epsilon\)

  1. What proportion of the total variance in trunk diameter can be attributed to pH? Calcium?