The fourth assignment of Linear Regression. The assignment is written in Rmarkdown, a smart syntax supported by RStudio helping with formula, plot visualization and plugin codes running.

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# 3.6
```{r}
hardness<- read.table('CH01PR22_947709365.txt')
names(hardness) <- c("y","x")
is.data.frame(hardness)
hardness.fit <- lm(y~x,data=hardness)
sumtable<-summary(hardness.fit)
sumtable

a

1	boxplot(resid(hardness.fit),main="Box Plot of Hardness Data", ylab="Residuals")

The mean of residuals is zero

b

1
2
3

yhat <- fitted(hardness.fit); resid <- resid(hardness.fit)
plot(yhat,resid,main="Plot of residuals against the fitted values Yhat", 
     ylab="Residuals",xlab='yhat')

There is one residual equals to 5.575 a little higher than others.

c

hardness.stdres = rstandard(hardness.fit)
qqnorm(resid, 
       ylab="Residuals", 
       xlab="Expected", 
       main=" Normal Probability Plot of Residuals") 
qqline(resid)

StdErr = summary(hardness.fit)$sigma
n = 16
ExpVals = sapply(1:n, function(k) StdErr * qnorm((k-.375)/(n+.25)))
cor(ExpVals,sort(resid))

With n=16, from Table B.6, the critical value for the coefficient of correlation between the ordered residuals and the expected values under normality when the distribution of error terms is normal using a 0.05 significance level is 0.941. Since 0.9916733 > 0.941, the assumption of normality appeared reasonable.

4.5

a

interval <- function(dat){
  names(dat) <- c("y","x")
  is.data.frame(dat)
  data.fit <- lm(y~x,data=dat)
  sumtable<-summary(data.fit)
  coef <- sumtable$coefficients
  signif<-1-(1-0.9)/4.0
  tvalue <-qt(signif, 14)
  beta_0 = list(coef[1]-tvalue*coef[3],coef[1]+tvalue*coef[3])
  beta_1 = list(coef[2]-tvalue*coef[4],coef[2]+tvalue*coef[4])
  return(list(beta_0,beta_1))
}
interval(hardness)

From R result, $b_0 = 168.6, s(b_0)=2.65702, b_1 = 2.03438, s(b_1)=0.09039.$ Since
t(0.975, 14) = 2.145, Bonferroni joint confidence intervals for $β_0$ and $β_1$, using a 90% percent family confidence coefficient, are 168.6±2.145(2.65702) = [162.901, 174.299] for $β_0$ and 2.03438±2.145(0.09039) = [1.840, 2.228] for $β_1$. At least 90% of the time, both coefficients will be within the limits stated.

c

The 90% joint confidence interval means that both will be in the interval at least 90% of the time.
Restated, at least one of them will be out of the interval no more than 10% of the time. We cannot get more specific than this.

4.9

a

For Bonferroni, use $b_0+b_1X_j±t(1−0.1/6, 14)s{\hat Y_h}$, with t(1−.10/6, 14) = 2.35982.

meaninterval <-function(X_h){
mse <- mean(sumtable$residuals^2)
sYh <- (mse * ( 1/16.0 + (X_h -ave(hardness$x)[1])**2/(sum((hardness$x - ave(hardness$x))**2)/16.0)) )**0.5
beta_0 = coef[1]
beta_1 = coef[3]
list(beta_0 +beta_1*X_h - qt(1-.10/6, 14) * sYh,beta_0 +beta_1*X_h + qt(1-.10/6, 14) * sYh)
}

Using this function we have CI of 20,30,40 are [215.1106，228.3704], [245.9163，250.7052], [265.1384，284.6236] respectively.

The 90% joint confidence interval means that all three mean hardness will be in their respective interval at least 90% of the time.Restated, at least one of them will be out of the interval no more than 10% of the time. We cannot get more specific than this.

4.12

1 2	galleys.x <- c(7, 12, 10, 10, 14, 25, 30, 25, 18, 10, 4, 6) cost.y <- c(128, 213, 191, 178, 250 , 446, 540, 457, 324, 177, 75, 107)

a

1 2	typos.lm <- lm(cost.y~galleys.x-1) typos.lm

$\hat Y_h=18.03X$

b

1 2	plot(galleys.x,cost.y,xlab= "Galleys", ylab="Cost") abline(typos.lm)

It appears that the model fits good

c

historical.norm <- data.frame(galleys.x=1)
alpha <- 0.02
typos.int <- predict.lm(typos.lm, newdata = historical.norm , interval = "confidence", level = 1-alpha)
typos.int

Alternatives: $H0:E[Y]=β{10}=17.50$

$H0:E[Y]≠β{10}=17.50$

CI: $17.81226≤E[Y_h]≤18.24435$

Decision rule:

If $β_{10}$ falls within the confidence interval for $E[Y_h]$, conclude $H_0$;

If $β_{10}$ does not fall within the confidence interval for $E[Y_h]$, conclude $H_A$

Conclusion:

Since 17.50<17.81226; therefore, accept $H_A$.

d

1
2
3

newdata.galleys <- data.frame(galleys.x=10)
typos.pred <- predict(typos.lm, newdata.galleys, level=0.98, interval = "predict", se.fit = TRUE)  
typos.pred

$\hat Y_h=180.283$

s[pred]=4.506806

180.283±2.738769(4.506806)

$167.8441≤Y_{h(new)}≤192.722$