Use R to re-draw Figs. 8-11, 8-15 and 9-4 from the text. [Hint: See the example below.]
Analyze the following data from a study to determine the effect of air voids on percentage retained strength of asphalt. Air voids were controlled at three levels: low (2-4%), medium (4-6%) and high (6-8%). Give an appropriate graph. Give a 95% confidence interval for the residual variance. State any assumptions you make and do what you can to test the assumptions. State your conclusions.
Air Voids Retained Strength (%) Low 106 90 103 90 79 88 Medium 80 69 94 91 70 83 High 78 80 62 69 76 85
A chemical reaction was run 9 times at different temperatures. The efficiency of the reaction was observed each time.
Temperature (°C) 10 30 20 50 40 10 20 10 40 Efficiency (%) 50 65 55 70 50 55 60 45 60(a) Fit a straight line to the data by least squares, with efficiency as the dependent variable. Plot the data and the fitted line on a graph. Can efficiency be predicted as a linear function of temperature? Present your analysis in an ANOVA table with F-Tests for non-linearity and for the slope of the regression line. Give a 95% confidence interval for the residual variance. State your assumptions and your conclusions.
(b) Predict the efficiency to be obtained at 30°C, 60°C and 100°C. How reliable do you think your predictions are?
Analyze the following data from a study of ion-beam-assisted etching of aluminum with chlorine. The independent variable x is chlorine flow and the dependent variable y is the etch rate. Give an appropriate graph. State any assumptions you make and do what you can to test the assumptions. State your conclusions.
x 1.5 1.5 2.0 2.5 2.5 3.0 3.5 3.5 4.0 y 23.0 24.5 25.0 30.0 33.5 40.0 40.5 47.0 49.0
13-4 (p. 639). [Hint: see the example below.]
The following examples will show you how to set up these problems in R.
> xgr <- seq(0,12,length=80) > plot(xgr,dexp(xgr,2),type="l",lty=1,xlab="x",ylab="f(x)") > lines(xgr,dexp(xgr,0.5),lty=2) > lines(xgr,dexp(xgr,0.1),lty=3) > legend(6,1.8,c("lambda = 2","lambda = 0.5","lambda = 0.1"),lty=1:3)Note that I plotted the highest curve first, to set the limits on the y-axis; I could also have done this by specifying ylim = c(0, 2)
> tvtube <- data.frame(bright=c(280,290,285,230,235,240,300,310,295,260,240,235,290,285,290,220,225,230),
glass=as.factor(rep(rep(1:2,rep(3,2)),3)), phosphor=as.factor(rep(1:3,rep(6,3))))
> tvtube
bright glass phosphor
1 280 1 1
2 290 1 1
3 285 1 1
4 230 2 1
5 235 2 1
6 240 2 1
7 300 1 2
8 310 1 2
9 295 1 2
10 260 2 2
11 240 2 2
12 235 2 2
13 290 1 3
14 285 1 3
15 290 1 3
16 220 2 3
17 225 2 3
18 230 2 3
> anova(lm(bright~glass*phosphor, data=tvtube))
Analysis of Variance Table
Response: bright
Df Sum Sq Mean Sq F value Pr(>F)
glass 1 14450.0 14450.0 273.7895 1.259e-09 ***
phosphor 2 933.3 466.7 8.8421 0.004364 **
glass:phosphor 2 133.3 66.7 1.2632 0.317801
Residuals 12 633.3 52.8
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
> exten9 <- data.frame(x=c(1,1,2,3.3,3.3,4,4,4,5,5.6,5.6,5.6,6,6,6.5,6.9),
y=c(2.3,1.8,2.8,1.8,3.7,2.6,2.6,2.2,2,3.5,2.8,2.1,3.4,3.2,3.4,5))
> exten9
x y
1 1.0 2.3
2 1.0 1.8
3 2.0 2.8
4 3.3 1.8
5 3.3 3.7
6 4.0 2.6
7 4.0 2.6
8 4.0 2.2
9 5.0 2.0
10 5.6 3.5
11 5.6 2.8
12 5.6 2.1
13 6.0 3.4
14 6.0 3.2
15 6.5 3.4
16 6.9 5.0
> exten9$xf <- as.factor(exten9$x)
> exten9
x y xf
1 1.0 2.3 1
2 1.0 1.8 1
3 2.0 2.8 2
4 3.3 1.8 3.3
5 3.3 3.7 3.3
6 4.0 2.6 4
7 4.0 2.6 4
8 4.0 2.2 4
9 5.0 2.0 5
10 5.6 3.5 5.6
11 5.6 2.8 5.6
12 5.6 2.1 5.6
13 6.0 3.4 6
14 6.0 3.2 6
15 6.5 3.4 6.5
16 6.9 5.0 6.9Analysis as a simple linear regression:
> anova(lm(y~x,data=exten9))
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x 1 3.4928 3.4928 6.6645 0.02174 *
Residuals 14 7.3372 0.5241
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
Analysis as a one-factor design:
> anova(lm(y~xf,data=exten9))
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
xf 8 7.7933 0.9742 2.2456 0.1515
Residuals 7 3.0367 0.4338 Analysis as a simple linear regression with a test for lack of fit:
> anova(lm(y~x+xf,data=exten9))
Analysis of Variance Table
Response: y
Df Sum Sq Mean Sq F value Pr(>F)
x 1 3.4928 3.4928 8.0514 0.02513 *
xf 7 4.3005 0.6144 1.4162 0.32882
Residuals 7 3.0367 0.4338
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
Since the lack of fit is not significant, we look further at the simple linear regression model:
> fitexten9 <- lm(y~x,data=exten9)
> summary(fitexten9)
Call:
lm(formula = y ~ x, data = exten9)
Residuals:
Min 1Q Median 3Q Max
-1.04505 -0.39160 -0.08988 0.34722 1.51873
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.6967 0.4730 3.587 0.00298 **
x 0.2586 0.1002 2.582 0.02174 *
---
Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
Residual standard error: 0.7239 on 14 degrees of freedom
Multiple R-Squared: 0.3225, Adjusted R-squared: 0.2741
F-statistic: 6.665 on 1 and 14 DF, p-value: 0.02174
> plot(exten9$x,exten9$y,pch=19)
> abline(fitexten9)
Of course, we should have looked at the graph before starting any analysis!