Question

31. Short Answer Question We have a dataset with n = 10 pairs of observations (Ci,y), and n 2 = 683, 813, =1 i1 ** = 47,405, «iyi = 56,089, y² = 66,731. What is an approximate 95% prediction interval for the response yo at :20 = 60?

Answer 1

Given

n = 10

Σ = 683

Σy = 813

x2 = 47405

ту 56089

$\sum y^2=66731$

$x_{0}=60$

So,

$\overline{x}=\frac{\sum x}{n}=\frac{683}{10}=68.3$

$\overline{y}=\frac{\sum y}{n}=\frac{813}{10}=81.3$

Στy – (Στ)(Σ9) Slope = b = Σ2 – (Στ)2 56089 – 6 (683) (813) 47405 – 6 (683)? 0.7421

$Intercept=b_{0}=\overline{y}-b_{1}\overline{x}=81.3-0.7421*68.3=30.6146$

The equation of regression line is,

$\widehat{y}=30.6146+0.7421x$

When x₀ = 60

$\widehat{y}_{0}=30.6146+0.7421*60=75.1406$

$SS_{Total}=\sum y^2-\frac{1}{n}(\sum y)^2=66731-\frac{1}{10}(813)^2=634.1$

$SS_{Residuals}=b_{1}\left ( \sum xy-\frac{1}{n}(\sum x)(\sum y) \right )$

  $=0.7421\left ( 56089-\frac{1}{10}(683)(813) \right )$

$=416.3923$

$SS_{Error}=SS_{Total}-SS_{Residuals}=634.1-416.3923=217.7077$

$S_{\epsilon }=\sqrt{\frac{SS_{Error}}{n-2}}=\sqrt{\frac{217.7077}{10-2}}\approx 5.2167$

Here

Confidence level = 95%

100(1-$\alpha$ ) = 95

$\alpha =1-0.95=0.05$

and

$\alpha/2 =0.05/2=0.025$

Degrees of freedom = n-2 = 10 -2 =8

From t distribution table,

$t_{\alpha /2,n-2}=t_{0.025,8}=2.306$

Now, the 95% prediction interval is

$\widehat{y}_{0}\pm t_{\alpha /2,n-2}*S_{\epsilon }\sqrt{1+\frac{1}{n}+\frac{(x_{0}-\overline{x})^2}{\sum x^2-\frac{1}{n}(\sum x)^2}}$

$75.1406\pm 2.306*5.2167\sqrt{1+\frac{1}{10}+\frac{(60-68.3)^2}{47405-\frac{1}{10}(683)^2}}$

$75.1406\pm 13.1290$

( 62.0116 , 88.2696)

( 62.0 , 88.3) ( for round to one decimal place )

An approximate 95% prediction interval for the response y₀ at x₀ = 60 is ( 62.0 , 88.3)