**NOTE: For your homework download and use the template** (https://math.dartmouth.edu/~m50f17/HW2.Rmd)

**Read the green comments in the rmd file to see where your answers should go.**

Given a fixed confidence interval percentage (say 95%) at what value of x does CI on the mean response achieve its minimum width?

Write an R-chunk using the propellant data which computes the following.

- Fit a simple linear regression model relating shear strength to age.
- Plot scatter diagram.
- Plot two curves (in blue color) that traces upper and lower limits of 95% confidence interval on \(E(y|x_0)\)
- Plot two curves (in red color) that traces upper and lower limits of 95% prediction interval for \(y\)
- Print the 95% quantile of the corresponding t distribution

The width of the interval is \[ 2 t_{\alpha/2,n-2} \sqrt{MS_{Res} ((1/n) + (x_0 - \bar{x})^2 / S_{xx} } \] and all terms inside the square root are positive. Therefore it is minimized when \(x_0=\bar{x}\).

```
# Computation part of the answer :
prop<-read.table("https://math.dartmouth.edu/~m50f17/propellant.csv", header=T, sep=",")
shearS<-prop$ShearS
age<-prop$Age
plot(age, shearS, xlab = "Propellant Age (weeks)", ylab = "Shear S. (psi)", main = "Rocket Propellant")
fitted <- lm(shearS ~ age)
ageList <- seq(0,25,0.5)
cList <- predict(fitted, list(age = ageList), int = "c", level = 0.95)
pList <- predict(fitted, list(age = ageList), int = "p", level = 0.95)
matlines(ageList, pList, lty='solid' , col = "red")
matlines(ageList, cList, lty = 'solid', col = "blue")
```

```
# since n=20 we look at the t_18 distribution
wantedQuantile <- qt( 0.95, 18) ;
cat("95% quantile is of t_18 is : ", wantedQuantile ) ;
```

`## 95% quantile is of t_18 is : 1.734064`

Plot the same graph as in Question-1 without using R function predict, but instead directly calculating the interval limits we discussed in class. In particular, what are the limits of 95% confidence interval on \(E(y|x_0)\)?

```
# Computation part of the answer :
Sxx<-sum((age-mean(age))^2)
Sxy<-sum((age-mean(age))*shearS)
beta1Hat<-Sxy/Sxx
beta0Hat<-mean(shearS)-Sxy/Sxx*mean(age)
plot(age,shearS)
abline(c(beta0Hat,beta1Hat), lwd=2, col='blue')
N <- length(age) ;
dof <- N - 2 ;
muHat <- beta0Hat + beta1Hat * ageList ;
yHat <- fitted$fitted.values ;
SSres <- sum((shearS - yHat)^2);
MSres <- SSres / dof ;
xBar <- mean (age)
tQuantile_975 <- qt( 1 - 0.05/2 , 18) ;
upperCI <- muHat + tQuantile_975 * sqrt ( MSres * ( (1/N) + (ageList-xBar)^2 / Sxx ) )
lowerCI <- muHat - tQuantile_975 * sqrt ( MSres * ( (1/N) + (ageList-xBar)^2 / Sxx ) )
matlines(ageList, upperCI, lty = 'solid', col = "blue")
matlines(ageList, lowerCI, lty = 'solid', col = "blue")
yHatList <- beta0Hat + beta1Hat * ageList ;
upperPI <- yHatList + tQuantile_975 * sqrt ( MSres * ( 1 + (1/N) + (ageList-xBar)^2 / Sxx ) )
lowerPI <- yHatList - tQuantile_975 * sqrt ( MSres * ( 1 + (1/N) + (ageList-xBar)^2 / Sxx ) )
matlines(ageList, upperPI, lty = 'solid', col = "red")
matlines(ageList, lowerPI, lty = 'solid', col = "red")
```

Load the propellant data and fit a simple linear regression model relating shear strength to age.

- Test the hypothesis \(\beta_1 = -30\) using confidence level 97.5%.
- Calculate the limits of 97.5% confidence interval for \(\beta_0\) and \(\beta_1\)

- Is there any relation between the answers you find in part (a) and (b) ?
- Calculate \(R^2\)

```
# Computation part of the answer :
seBeta1Hat = sqrt ( MSres / Sxx ) ;
guess = -30 ;
testStat = ( beta1Hat - guess ) / seBeta1Hat ;
tQuantile_9875 <- qt( 1 - 0.025/2 , 18) ;
cat ("Part (a)
Absolute value of test statistic is " , abs(testStat) , " Since it is greater than t_quantile=" , tQuantile_9875 , "
we REJECT the hypothesis \n\n")
```

```
## Part (a)
## Absolute value of test statistic is 2.476056 Since it is greater than t_quantile= 2.445006
## we REJECT the hypothesis
```

```
seBeta0Hat = sqrt ( MSres * ( (1/N) + (xBar)^2 / Sxx ) ) ;
cat ("Part (b)
Lower and upper bounds of 97.5% CI for beta0 are : " , beta0Hat - tQuantile_9875 * seBeta0Hat ," and ", beta0Hat + tQuantile_9875 * seBeta0Hat , "
Lower and upper bounds of 97.5% CI for beta1 are : " , beta1Hat - tQuantile_9875 * seBeta1Hat ," and ", beta1Hat + tQuantile_9875 * seBeta1Hat, "\n\n")
```

```
## Part (b)
## Lower and upper bounds of 97.5% CI for beta0 are : 2519.792 and 2735.852
## Lower and upper bounds of 97.5% CI for beta1 are : -44.21747 and -30.08971
```

```
cat ("Part (c)
Yes, t-test is equivalent to testing whether -30 is in the CI for beta1. We observe that -30 is out of the above confidence interval for beta1, thus hypothesis is rejected in part b \n\n")
```

```
## Part (c)
## Yes, t-test is equivalent to testing whether -30 is in the CI for beta1. We observe that -30 is out of the above confidence interval for beta1, thus hypothesis is rejected in part b
```

```
yBar <- mean (shearS)
SSt <- sum((shearS - yBar)^2);
SSr <- sum((yHat - yBar)^2);
cat ("Part (d)
R-square is : " , SSr / SSt )
```

```
## Part (d)
## R-square is : 0.9018414
```

Load the propellant data. This time let us consider a relation between square of shear strength and propellant age.

- Fit a simple linear regression model relating
**square**of shear strength to age. Plot scatter diagram and fitted line. - Using analysis-of-variance test for significance of regression (using the formulas we discussed in class)
- Use t-test and check significance of regression (using the formulas we discussed in class)
- Does the regression analysis predict a linear relationship between square of shear strength and propellant age ?

```
# Computation part of the answer :
prop<-read.table("https://math.dartmouth.edu/~m50f17/propellant.csv", header=T, sep=",")
shearS<-prop$ShearS
age<-prop$Age
shear_SQ<- (prop$ShearS)^2
plot(age, shear_SQ, xlab = "Propellant Age (weeks)", ylab = "Shear S. Squared", main = "Rocket Propellant")
fitted_SQ <- lm(shear_SQ ~ age)
abline(fitted_SQ$coef, lwd=2, col='blue')
```