STAT 587 Homework Assignment No



STAT 587 Homework Assignment No.1

Problem 2

A. Brand preference. In a small-scale experimental study of the relation between degree of brand liking [pic] and moisture content [pic]and sweetness [pic]of the product, the following results were obtained from the experiment based on a completely randomized design (data are coded):

i: |1 |2 |3 |4 |5 |6 |7 |8 |9 |10 |11 |12 |13 |14 |15 |16 | |[pic] |4 |4 |4 |4 |6 |6 |6 |6 |8 |8 |8 |8 |10 |10 |10 |10 | |[pic] |2 |4 |2 |4 |2 |4 |2 |4 |2 |4 |2 |4 |2 |4 |2 |4 | |[pic] |64 |73 |61 |76 |72 |80 |71 |83 |83 |89 |86 |93 |88 |95 |94 |100 | |

a. Fit regression model to the data. State the estimated regression function. How is [pic]interpreted here?

The estimated regression function is Y=37.6500+4.4250*X1+4.3750*X2.

The second coefficient provides the dependency of brand likelihood of moisture content. It indicates that given fixed amount of sweetness and as moisture content increases on unit, the degree of brand liking increases about 4.425 on average.

b. Obtain the residuals and prepare a box plot of the residuals. What information does this plot provide?

[pic]

The boxplot for residuals shows that they are symmetrically distributed around zero and also have the zero mean.

c. Plot the residuals against [pic]and [pic] on separate graphs. Also prepare a normal probability plot. Analyze the plots and summarize your findings.

Normal probability plot for residuals shows more or less normality of them:

[pic]

The plot of residuals vs all the fitted and X1 and X2 and their product shows uniform spread, so regression function is linear in any of these:

[pic]

d. Conduct a formal test for lack of fit of the first-order regression function; use a = .01. State the alternatives, decision rule, and conclusion.

[pic]

|J=1 |J=2 |J=3 |J=4 |J=5 |J=6 |J=7 |J=8 | |Replicate |X1=4, X2=2 |X1=4, X2=4 |X1=6, X2=2 |X1=6, X2=4 |X1=8, X2=2 |X1=8, X2=4 |X1=10, X2=2 |X1=10, X2=4 | |I=1 |64 |73 |72 |80 |83 |89 |88 |95 | |I=2 |61 |76 |71 |83 |86 |93 |94 |100 | |Mean j |62.5 |74.5 |71.5 |81.5 |84.5 |91 |91 |97.5 | |Sum([pic])^2 |4.5 |4.5 |0.5 |4.5 |4.5 |8 |18 |12.5 | |c=8

SSPE(from above)=57 df=n-c=16-8=8

SSE=94.3

Then SSLF=SSE-SSPE=37.3 df=c-p=8-3=5

F*=(SSLF/5)/(SSPE/8)=1.0470175 < F(1-0.01,5,8)= 6.631825

So F* ................
................

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