Question

Q1 (30 points) Consider Problem 11.45, Page 637. Please note that for this problem the data will be entered in R as follows: #Enter data on x = Dose Level of Drug, and y = Potency of Drug (Problem 11.45, page 637) x<-c(2, 2, 2, 4, 4, 8, 8, 16, 16, 16, 32, 32, 64, 64, 64) y<-c(5, 7, 3, 10, 14, 15, 17, 20, 21, 19, 23, 29, 28, 31, 30) For this problem, answer the following questions. In answering the questions, you should use R wherever you can. You must provide a print out of your R-Code, and the corresponding output. From the output, choose the numbers or plots as necessary and appropriate. For Parts i.-iv., you must provide the appropriate expression before you use any number from the output. Also, whenever you use a plot, you must write comments below the plot that reflect what you understand from the plot. i. Draw the scatter plot and comment whether simple linear regression of Drug Potency on Drug Dosage Level is plausible. ii. Write down the simple linear regression model for yon x. Then estimate the regression parameters, and error variance.Formulate the appropriate hypotheses to test whether the regression model is significant. Now carryout the test and draw your conclusions. iii. What proportion of total variation is explained by the simple liner regression? How strong is the linear association between y and x? iv. Now state the assumptions behind the simple linear regression model and perform model diagnostics (residual analysis) verify the validity of the assumptions made. Based on the model diagnostics, would you conclude that the model is appropriate or inappropriate. v. Consider a log transformation on the x variable by introducing a new variable x' = log(x). Now write down a regression model of y on x'. Repeat questions i.-iv. for the regression of y on x' (replace x by x' in questions i.-iv.). Here, you do not have to give expressions any more when you borrow any number from the Routput. However, you must provide the R-Code and the output. You should also insert comments below every plot. vi. Based on your work for questions i.-V., would you prefer regression model of y on x or regression model of y on x'? Explain why. vii. Based on the preferred model you have chosen in question vi, construct a plot that consists of scatter plot, fitted regression line, confidence band for true average of y at a given x, and prediction band for a single y at a given x.

Answer 1

1) we first use scatter plot to visualize their relationship, we saw a positive correlation among them

30 000 25 20 potency 15 10 5 0 10 20 30 40 50 60 dose

2)###model
###potency(estimated)=10.6188+0.3375dose

dose=c(2,2,2,4,4,8,8,16,16,16,32,32,64,64,64)
> dose
[1] 2 2 2 4 4 8 8 16 16 16 32 32 64 64 64
> potency=c(5,7,3,10,14,15,17,20,21,19,23,29,28,31,30)
> potency
[1] 5 7 3 10 14 15 17 20 21 19 23 29 28 31 30
> plot(dose,potency)
> fit=lm(potency~dose)
> fit

Call:
lm(formula = potency ~ dose)

Coefficients:
(Intercept) dose
10.6188 0.3375

> ###model
> ###potency(estimated)=10.6188+0.3375dose
> summary(fit)

Call:
lm(formula = potency ~ dose)

Residuals:
Min 1Q Median 3Q Max
-8.294 -3.217 1.582 3.331 7.582

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 10.61883 1.68749 6.293 2.78e-05 ***
dose 0.33748 0.05288 6.382 2.41e-05 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 4.682 on 13 degrees of freedom
Multiple R-squared: 0.7581, Adjusted R-squared: 0.7394
F-statistic: 40.73 on 1 and 13 DF, p-value: 2.41e-05

> residuals(fit)
1 2 3 4 5 6 7 8
-6.293781 -4.293781 -8.293781 -1.968737 2.031263 1.681351 3.681351 3.981528
9 10 11 12 13 14 15
4.981528 2.981528 1.581882 7.581882 -4.217411 -1.217411 -2.217411
> predict(fit)
1 2 3 4 5 6 7 8
11.29378 11.29378 11.29378 11.96874 11.96874 13.31865 13.31865 16.01847
9 10 11 12 13 14 15
16.01847 16.01847 21.41812 21.41812 32.21741 32.21741 32.21741
> confint(fit)
2.5 % 97.5 %
(Intercept) 6.9732274 14.2644240
dose 0.2232401 0.4517156
>
> ###############################################
> dose=c(2,2,2,4,4,8,8,16,16,16,32,32,64,64,64)
> dose
[1] 2 2 2 4 4 8 8 16 16 16 32 32 64 64 64
> dosenew=log(dose+1)
> dosenew
[1] 1.098612 1.098612 1.098612 1.609438 1.609438 2.197225 2.197225 2.833213
[9] 2.833213 2.833213 3.496508 3.496508 4.174387 4.174387 4.174387
> plot(dosenew,potency)
> fit1=lm(potency~dosenew)
> fit1

Call:
lm(formula = potency ~ dosenew)

Coefficients:
(Intercept) dosenew
-2.181 7.828

> ###model2
> ###potency(estimated)=-2.181+7.828dosenew
> summary(fit1)

Call:
lm(formula = potency ~ dosenew)

Residuals:
Min 1Q Median 3Q Max
-3.4194 -1.2087 -0.0195 0.7913 3.8095

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -2.1808 1.4061 -1.551 0.145
dosenew 7.8282 0.4987 15.698 7.87e-10 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.131 on 13 degrees of freedom
Multiple R-squared: 0.9499, Adjusted R-squared: 0.946
F-statistic: 246.4 on 1 and 13 DF, p-value: 7.867e-10

> residuals(fit1)
1 2 3 4 5 6
-1.419353837 0.580646163 -3.419353837 -0.418188154 3.581811846 -0.019487262
7 8 9 10 11 12
1.980512738 0.001878959 1.001878959 -0.998121041 -2.190506770 3.809493230
13 14 15
-2.497070331 0.502929669 -0.497070331
> predict(fit1)
1 2 3 4 5 6 7 8
6.419354 6.419354 6.419354 10.418188 10.418188 15.019487 15.019487 19.998121
9 10 11 12 13 14 15
19.998121 19.998121 25.190507 25.190507 30.497070 30.497070 30.497070
> confint(fit1)
2.5 % 97.5 %
(Intercept) -5.218565 0.8570054
dosenew 6.750860 8.9054980