roblem ap eview Practlce Problems, Problem 20 In Problem redictionj interval for

ID: 3305833 • Letter: R

Question

roblem ap eview Practlce Problems, Problem 20 In Problem redictionj interval for the future observation and [a 99% confidence interval for its expected value E(YK-x), at X-5.50 and X-8.20. Data on the effect of force applied (X) on the boiler plate elongation (Y) are reproduced below: Test Piece # 27 2.6850 67 83 5.35101 6.24 117 7.14 134 8.93154 9.82188 10.70 206 1.33 3.57 4.46 3 4 6 8 10 Fit a regression line to these data, and fill in the equation coefficients below: Equation: Elongation Force Expected Value E(Y| X = 5.50) 99% prediction interval for the observation Y when X = 5.50: ( 99% confidence interval for the expected value E(Y| X= 5.50): ( Expected Value EYI X 8.2) 99% prediction interval for the observation Y when X = 8.20: ( 99% confidence interval for the expected value E(Y|X = 8.20): (

Explanation / Answer

Solution

Back-up Theory

Let X = force applied and Y = elongation. Then, the linear regression is given by the model: Y = + X + , where is the error term, which is assumed to be Normally distributed with mean 0 and variance 2. Then,

Let (xi, yi) be a pair of sample observation on (X, Y), i= 1, 2, …., n,where n = sample size.

Then, Mean X = Xbar = (1/n)sum of xi over I = 1, 2, …., n; ……………….(1)

Variance of X, V(X) = (1/n)Sxx where Sxx = sum of (xi – Xbar)2 over i = 1, 2, …., n …..(2)

Standard Deviation of X = SDX = sq.rt of V(X). ……………………………(3)

Similarly, Mean Y = Ybar =(1/n)sum of yi over i= 1, 2, …., n;…………….(4)

Variance of Y, V(Y) = (1/n)Syy where Syy = sum of (yi – Ybar)2 over i = 1, 2, …., n …(5)

Standard Deviation of Y = SDY = sq.rt of V(Y). ………………………….……………….(6)

Covariance of X and Y, Cov(X, Y)

= (1/n)Sxy where Sxy = sum of {(xi – Xbar)(yi – Ybar)} over i = 1, 2, …., n………(7)

Estimated Y is given by: Ycap = a + bX, where

b = Cov(X, Y)/V(X) = Sxy/Sxx and a = Ybar – b.Xbar..…………………….(8)

Estimate of 2 is given by s2 = (Syy – b2Sxx)/(n - 2)……………………………(9)

100(1 - )% Confidence Interval (CI) for ycap at x = x0 is E(Y/X = x0) = (a + bx0) ± tn – 2,/2xs[(1/n) + {(x0 – Xbar)2/Sxx}] ………………….(10)

100(1 - )% Prediction Interval (PI) for ycap at x = x0 is (a + bx0) ± tn – 2,/2xs[1 + (1/n) + {(x0 – Xbar)2/Sxx}]

[Note: The only difference in the above formulae is that the PI formula has an additional 1 within the radical sign.

Now, the calculations:

1.33

2.68

3.57

4.46

5.35

101

6.24

117

7.14

134

8.93

154

9.82

188

10.7

206

xbar

ybar

114.7

Sxx

88.14996

Syy

29468.1

Sxy

1596.216

18.1079606

5.65386144

s^2

70.4854512

8.3955614

0.01

n-2

tn-2,/2

3.35538733

Given x = 5.5

E(Y/x = 5.5)

105.6873

CIYcapLB

96.202759

CIYcapUB

114.29253

PI LB

75.6608374

PIUB

134.834452

Given x = 8.2

E(Y/X = 8.2) =

154.1386

CIYcapLB

143.090976

CIYcapUB

165.1873

PI LB

123.879741