row

Alex Towell (atowell@siue.edu)

Refer to the data from Exercise 1.20

Data has been collected on [\(45\)]{.math .inline} calls for routine maintenance. The goal is to explore the relationship between the number of copiers serviced [\((x)\)]{.math .inline} and the time in minutes spent to complete the service [\((y)\)]{.math .inline}. Let [\(x_h = 3\)]{.math .inline} copiers.

Part (1)

[\(\mu_h\)]{.math .inline} is the mean response [\(\operatorname{E}(Y_h|x_h=3)\)]{.math .inline}. A confidence interval for [\(\mu_h\)]{.math .inline} is given by [\[ \hat{Y}_h \pm t_{\alpha/2,n-2} \operatorname{se}(\hat{Y}_h). \]]{.math .display}

alpha = .05
x.h = 3
data = read.table('CH01PR20.txt')
colnames(data)=c("time","copiers")
mod = lm(time~copiers, data=data)
predict(mod,level=1-alpha,data.frame(copiers=x.h),interval="confidence")

##        fit     lwr      upr
## 1 44.52559 41.1476 47.90357

We see that a CI for [\(\mu_h\)]{.math .inline} is given by [\([44.526,47.904]\)]{.math .inline}.

Part (2)

We denote [\(Y|X=x_h\)]{.math .inline} by [\(Y_h\)]{.math .inline}. Assume [\[ Y_h \sim \mathcal{N}(\beta_0 + \beta_1 x_h, \sigma^2). \]]{.math .display} Then, if we wish to predict [\(Y_h\)]{.math .inline} with [\((\beta_0,\beta_1,\sigma)\)]{.math .inline} known, the interval [\[ \beta_0 + \beta_1 x \pm z_{\alpha/2} \sigma. \]]{.math .display} includes the observed value of [\(Y_h\)]{.math .inline} with probability [\(1-\alpha\)]{.math .inline}. For predicting a single outcome [\(Y_h\)]{.math .inline}, the uncertainty of the prediction is quantified by [\(\operatorname{V}(Y_h) = \sigma^2\)]{.math .inline}. For a given [\(\alpha\)]{.math .inline}, the smaller the variance the smaller the interval.

However, if [\((\beta_0,\beta_1,\sigma^2)'\)]{.math .inline} is unknown, we have the additional source uncertainty due to a random sample [\(\{(Y_h,x_h)\}\)]{.math .inline} being used to estimate [\((\beta_0,\beta_1,\sigma^2)'\)]{.math .inline}.

Let pred = [\(P_h = Y_{h(\rm{new})} - \hat{Y}_h\)]{.math .inline} denote the random . Then, [\(\operatorname{E}(P_h) = 0\)]{.math .inline} and [\[\begin{align*} \operatorname{V}(P_h) &= \operatorname{V}(Y_{h(\rm{new})}) + \operatorname{V}(\hat{Y}_h)\\ &= \sigma^2\left[1 + \frac{1}{n} + \frac{(x_h - \bar{x})^2}{\rm{SS}_x}\right]. \end{align*}\]]{.math .display}

Since we do not know [\(\sigma^2\)]{.math .inline}, we estimate it with [\(\rm{MSE}\)]{.math .inline} and thus a prediction interval for [\(Y_{h(\rm{new})}\)]{.math .inline} is given by [\[ \hat{Y}_h \pm t_{\alpha/2,n-2} \hat{\operatorname{sd}}(P_h) \]]{.math .display} where [\[ \hat{\operatorname{sd}}(P_h) = \sqrt{\rm{MSE}\left(1 + \frac{1}{n} + \frac{(x_h-\bar{x})^2}{\rm{SS}_x}\right)}. \]]{.math .display}

predict(mod,data.frame(copiers=3),level=1-alpha,interval="predict")

##        fit      lwr      upr
## 1 44.52559 26.23515 62.81603

We see that the PI for [\(Y_{h(\rm{new})}\)]{.math .inline} is given by [\([26.235,62.816]\)]{.math .inline}.

Part (3)

A confidence interval is for the mean service time for all service calls with [\(3\)]{.math .inline} copiers to service.

A prediction interval is for the service time of a single service call with [\(3\)]{.math .inline} copiers to service.

Part (4)

Let pred.mean = [\(\bar{P}_h = \bar{Y}_{h(\rm{new})} - \hat{Y}_h\)]{.math .inline} denote the random . Then, [\(\operatorname{E}(\bar{P}_h) = 0\)]{.math .inline} and [\[\begin{align*} \operatorname{V}(\bar{P}_h) &= \operatorname{V}(\bar{Y}_{h(\rm{new})}) + \operatorname{V}(\hat{Y}_h)\\ &= \sigma^2 \left[\frac{1}{m} + \frac{1}{n} + \frac{(x_h - \bar{x})^2}{\rm{SS}_x}\right]. \end{align*}\]]{.math .display}

Since we do not know [\(\sigma^2\)]{.math .inline}, we estimate it with [\(\rm{MSE}\)]{.math .inline} and thus a prediction interval for [\(\bar{Y}_{h(\rm{new})}\)]{.math .inline} is given by [\[ \hat{Y}_h \pm t_{\alpha/2,n-2} \hat{\operatorname{sd}}(\bar{P}_h) \]]{.math .display} where [\[ \hat{\operatorname{sd}}(\bar{P}_h) = \sqrt{\rm{MSE}\left(\frac{1}{m} + \frac{1}{n} + \frac{(x_h-\bar{x})^2}{\rm{SS}_x}\right)} \]]{.math .display}

b0 = mod$coefficients[1]; names(b0) = NULL
b1 = mod$coefficients[2]; names(b1) = NULL

residual = mod$residuals
n = length(residual)
sse = sum(residual^2)
mse = sse / (n-2)
x = data$copiers

x.bar = mean(x)
ssx = sum((x-x.bar)^2)

y.hat = b0 + b1*x.h 
m = 10
var.pred.mean = mse*(1/m + 1/n + (x.h-x.bar)^2/ssx)

lower.ynew = y.hat - qt(alpha/2,n-2,lower.tail=F)*sqrt(var.pred.mean)
upper.ynew = y.hat + qt(alpha/2,n-2,lower.tail=F)*sqrt(var.pred.mean)

c(lower.ynew,upper.ynew)

## [1] 37.91320 51.13798

We see that the PI for [\(\bar{Y}_{h(\rm{new})}\)]{.math .inline} is given by [\[ [37.913, 51.138]. \]]{.math .display}

Part (5)

We predict that the mean service time for [\(10\)]{.math .inline} service calls with [\(3\)]{.math .inline} copiers each will be between [\(37.913\)]{.math .inline} and [\(51.138\)]{.math .inline}.

Part (6)

The variance of pred.mean = [\(\bar{P}_h\)]{.math .inline} is given by [\[ \operatorname{V}(\bar{P}_h) = \sigma^2 \left(\frac{1}{m} + \frac{1}{n} + \frac{(x_h-\bar{x})^2}{\rm{SS}_x}\right) \]]{.math .display} which we may rewrite as [\[\begin{align*} \operatorname{V}(\bar{P}_h) &= \frac{\sigma^2}{m} + \sigma^2\left(\frac{1}{n} + \frac{(x_h-\bar{x})^2}{\rm{SS}_x}\right)\\ &= \operatorname{V}(\bar{Y}_{h(\rm{new})}) + \operatorname{V}(\hat{Y}_h). \end{align*}\]]{.math .display}

As [\(m \to \infty\)]{.math .inline}, [\(\operatorname{V}(\bar{Y}_{h(\rm{new})}) \to 0\)]{.math .inline}, and thus [\[ \lim_{m \to \infty} \operatorname{V}(\bar{P}_h) = \operatorname{V}(\hat{Y}_h). \]]{.math .display}

A CI estimate of the mean service time [\(\mu_h\)]{.math .inline} can be interpreted as a prediction of the sample mean [\(\bar{Y}_{h(\rm{new})}\)]{.math .inline} for a large number of service time responses.