Question

Use the data "Orange" in R. You should include the r code as well as the output in your file, with appropriate answer to questions. Answer the following questions in your document 1. Fit a simple linear regression using "circumterence" as response and "age" as predictor. Is there a significant linear relationship between the two variables? State the null and alternative hypothesis, test statistic and p-value 2. Find which observation has the largest residual in absolute value) Give the value of this residual, the response value of the observation associated with this residual, and the fitted value for this observation 3, Construct an 90% CI for the true mean circumference at age equal to 600 4. Construct a 90% PI for a new observation at age equal to 800.
Do 2.3.4

Answer 1

2) First we fit a linear regression as circumference as response and age as predictor. Then we calculate the residuals from the fitted model and find the largest residual and corresponding response value and the fitted value.The R code is given as below:

d=Orange
y=d$circumference #response   
x=d$age #predictor

l=lm(y~x) #fitted model
summary(l)
y_hat=predict(l) #predicted values of y
res=abs(y-y_hat) # absolute residuals
maxres=max(res) #maximum residual
y_maxres=y[which(res==maxres)] #corresponding response
y_hat_maxres=y_hat[which(res==maxres)] #corresponding fitted response

3) Here we have to find 90% CI. So here the level alpha=0.1. Let the linear regression model is denoted by $y=\beta_0+\beta_1x+e$ where $e\sim N(0,\sigma^2)$ . Let the true mean circumference is denoted by $\mu_{y|x_0}=E(y|x_0)=\beta_0+\beta_1x_0$ . So fitted mean circumference is given by $\hat\mu_{y|x_0}=\hat\beta_0+\hat\beta_1x_0$

So   $E(\hat\mu_{y|x_0})=\mu_{y|x_0}$ . Now , $V(\hat\mu_{y|x_0})=V(\hat\beta_0+\hat\beta_1x_0)=V(\hat\beta_0)+x_0^2V(\hat\beta_1)+2x_0Cov(\hat\beta_0,\hat\beta_1)$

$V(\hat\beta_0)=\sigma^2[(\frac{1}{n}+\frac{\bar x^2}{s_{xx}})]$ ,   $V(\hat\beta_1)=\sigma^2(\frac{1}{s_{xx}})$ , $Cov(\hat\beta_0,\hat\beta_1)=-\sigma^2(\frac{\bar x}{s_{xx}})$ \

putting these values we get -   $V(\hat\mu_{y|x_0})=\sigma^2[\frac{1}{n}+\frac{(x_0-\bar x)^2}{s_{xx}}]$

Therefore $\frac{\hat\mu_{y|x_0}-\mu_{y|x_0}}{\sqrt{\sigma^2[\frac{1}{n}+\frac{(x_0-\bar x)^2}{s_{xx}}]}}\sim N(0,1)$ and $\frac{(n-2)\hat\sigma^2}{\sigma^2}\sim \chi^2_{n-2}$

So $\frac{\hat\mu_{y|x_0}-\mu_{y|x_0}}{\sqrt{\hat\sigma^2[\frac{1}{n}+\frac{(x_0-\bar x)^2}{s_{xx}}]}}\sim t_{n-2}$

The 90% CI can be obtained from the following equation

$P(-t_{0.05;n-2}<=\frac{\hat\mu_{y|x_0}-\mu_{y|x_0}}{\sqrt{\hat\sigma^2[\frac{1}{n}&plus;\frac{(x_0-\bar x)^2}{s_{xx}}]}}<=t_{0.05;n-2})=0.90$

which will give the 90%CI as

$(\hat\mu_{y|x_0}-t_{0.05;n-2}{\sqrt{\hat\sigma^2[\frac{1}{n}&plus;\frac{(x_0-\bar x)^2}{s_{xx}}]}},\hat\mu_{y|x_0}&plus;t_{0.05;n-2}{\sqrt{\hat\sigma^2[\frac{1}{n}&plus;\frac{(x_0-\bar x)^2}{s_{xx}}]}} \,\,\,)$

Here   $s_{xx}=\sum_{i=1}^{n}(x_i-\bar x)^2$ =   8225644 , n = 35, x₀ = 600 , t_0.05;33 = 1.69236, xbar=922.1429, $\hat \sigma^2=SSRes/33$ = 18594.74 / 33 = 563.477 , $\hat\mu_{y|x_0}=\hat\beta_0&plus;\hat\beta_1x_0=$ 17.3997 +  0.1068*600 = 81.4797

Putting these values we will get the 90% CI for true mean circumference which is equal to (73.3267,89.6327)

4) Here we need to find 90% PI for a new obsevation corresponding to age=800.

Here $y_0=\beta_0&plus;\beta_1x_0&plus;e_0$ and $\hat y_0=\hat \beta_0&plus;\hat \beta_1x_0$ . Take  $z=\hat y_0-y_0$ So E(z)=0 and $V(z)=\sigma^2[1&plus;\frac{1}{n}&plus;\frac{(x_0-\bar x)^2}{s_{xx}}]$ .The variance of z is derived similarly as before only extra term here is V(y₀) which is $\sigma^2$. Similarly we will get that

$\frac{\hat y_0-\-y_0}{\sqrt{\hat\sigma^2[1&plus;\frac{1}{n}&plus;\frac{(x_0-\bar x)^2}{s_{xx}}]}}\sim t_{n-2}$

The 90% CI can be obtained from the following equation

$P(-t_{0.05;n-2}<=\frac{\hat y_0-y_0}{\sqrt{\hat\sigma^2[1&plus;\frac{1}{n}&plus;\frac{(x_0-\bar x)^2}{s_{xx}}]}}<=t_{0.05;n-2})=0.90$

which will give the 90% PI as

$(\hat y_0-t_{0.05;n-2}{\sqrt{\hat\sigma^2[1&plus;\frac{1}{n}&plus;\frac{(x_0-\bar x)^2}{s_{xx}}]}},\hat y_0&plus;t_{0.05;n-2}{\sqrt{\hat\sigma^2[1&plus;\frac{1}{n}&plus;\frac{(x_0-\bar x)^2}{s_{xx}}]}} \,\,\,)$

Where the values of all quantities will remain same except x₀ which is equal to 800 here. $\hat y_0=\hat \beta_0&plus;\hat \beta_1x_0$ =  17.3997 +  0.1068*800=102.8397

So putting all other values we will get the desired 90% PI for a new observation which is equal to ( 62.0597,143.6197)

All the calculations which are done in R is given below:

x_bar=mean(x)
sxx=sum((x-x_bar)**2)
n=length(y)
tab_t=qt(0.95,33)
res1=(y-y_hat)
ssres=sum(res1**2)