##########
# SOlutions to hw 1
##########
# 1
####

The estimate of parameters are 6.7935(intercept) and -0.4133(slope).
The standard errors are 1.2365 and 0.2863 respectively. 
The t-values are 5.494 and -1.444 respectively.
The p-values are 0.00000175 and 0.156 respectively.
The covariance matrix of the parameter estimates is

            (Intercept)        temp
(Intercept)   1.5289709 -0.35317581
temp         -0.3531758  0.08194334

####
# 2
####

See output

####
# 3
####

The percentages are 

0.9785 0.9639 0.9521 0.9492 0.9507.

Note, the student's answer should be close but not exactly equal to my answer.

########
# R output
########
# 1
#####
> star <- read.table("u:\\stat526\\dataset\\ascdata\\star.txt",h=T)
> star
> g <- lm(light~temp,data=star)
> summary(g)

Call:
lm(formula = light ~ temp, data = star)

Residuals:
    Min      1Q  Median      3Q     Max 
-1.1052 -0.5067  0.1327  0.4423  0.9390 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   6.7935     1.2365   5.494 1.75e-06 ***
temp         -0.4133     0.2863  -1.444    0.156    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

Residual standard error: 0.5646 on 45 degrees of freedom
Multiple R-Squared: 0.04427,    Adjusted R-squared: 0.02304 
F-statistic: 2.085 on 1 and 45 DF,  p-value: 0.1557 

> (summary(g)$sigma^2)*(summary(g)$cov.unscaled)
            (Intercept)        temp
(Intercept)   1.5289709 -0.35317581
temp         -0.3531758  0.08194334

######
# 2
######

> A <- cbind(c(5,-2,0),c(4,3,-1),c(0,7,1))
> A
     [,1] [,2] [,3]
[1,]    5    4    0
[2,]   -2    3    7
[3,]    0   -1    1
> 
> b <- c(1,2,-1)
> b
[1]  1  2 -1

> A.inverse <- solve(A)
> A.inverse
           [,1]        [,2]       [,3]
[1,] 0.17241379 -0.06896552  0.4827586
[2,] 0.03448276  0.08620690 -0.6034483
[3,] 0.03448276  0.08620690  0.3965517
> A.inverse%*%b
           [,1]
[1,] -0.4482759
[2,]  0.8103448
[3,] -0.1896552
> t(A)%*%b
     [,1]
[1,]    1
[2,]   11
[3,]   13
> A.inverse%*%b
           [,1]
[1,] -0.4482759
[2,]  0.8103448
[3,] -0.1896552

#######
# 3
#######
> run <- 10000
> b <- matrix(0,run,5)
> 
> for(i in 1:run){
+ 
+ x1 <- rt(10,1)
+ x2 <- rt(10,2)
+ x3 <- rt(10,5)
+ x4 <- rt(10,20)
+ x5 <- rt(10,50)
+ 
+ lower1 <- mean(x1)-qt(0.975,9)*sqrt(var(x1))/sqrt(10)
+ upper1 <- mean(x1)+qt(0.975,9)*sqrt(var(x1))/sqrt(10)
+ 
+ lower2 <- mean(x2)-qt(0.975,9)*sqrt(var(x2))/sqrt(10)
+ upper2 <- mean(x2)+qt(0.975,9)*sqrt(var(x2))/sqrt(10)
+ 
+ lower3 <- mean(x3)-qt(0.975,9)*sqrt(var(x3))/sqrt(10)
+ upper3 <- mean(x3)+qt(0.975,9)*sqrt(var(x3))/sqrt(10)
+ 
+ lower4 <- mean(x4)-qt(0.975,9)*sqrt(var(x4))/sqrt(10)
+ upper4 <- mean(x4)+qt(0.975,9)*sqrt(var(x4))/sqrt(10)
+ 
+ lower5 <- mean(x5)-qt(0.975,9)*sqrt(var(x5))/sqrt(10)
+ upper5 <- mean(x5)+qt(0.975,9)*sqrt(var(x5))/sqrt(10)
+ 
+ 
+ b[i,1] <- (lower1<=0)*(upper1>=0)
+ b[i,2] <- (lower2<=0)*(upper2>=0)
+ b[i,3] <- (lower3<=0)*(upper3>=0)
+ b[i,4] <- (lower4<=0)*(upper4>=0)
+ b[i,5] <- (lower5<=0)*(upper5>=0)
+ 
+ }
> 
> 
> 
> apply(b,2,mean)
[1] 0.9785 0.9639 0.9521 0.9492 0.9507