   # The data from exercise 1 follow. x i 1 2 3 4 5 y i 3 7 5 11 14 a. Use equation (14.23) to estimate the standard deviation of ŷ * when x = 4. b. Use expression (14.24) to develop a 95% confidence interval for the expected value of y when x = 4. c. Use equation (14.26) to estimate the standard deviation of an individual value of y when x = 4. d. Use expression (14.27) to develop a 95% prediction interval for y when x = 4. ### Essentials Of Statistics For Busin...

9th Edition
David R. Anderson + 4 others
Publisher: South-Western College Pub
ISBN: 9780357045435

#### Solutions

Chapter
Section ### Essentials Of Statistics For Busin...

9th Edition
David R. Anderson + 4 others
Publisher: South-Western College Pub
ISBN: 9780357045435
Chapter 14.6, Problem 32E
Textbook Problem
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## The data from exercise 1 follow. xi 1 2 3 4 5 yi 3 7 5 11 14 a. Use equation (14.23) to estimate the standard deviation of ŷ* when x = 4. b. Use expression (14.24) to develop a 95% confidence interval for the expected value of y when x = 4. c. Use equation (14.26) to estimate the standard deviation of an individual value of y when x = 4. d. Use expression (14.27) to develop a 95% prediction interval for y when x = 4.

a.

To determine

Estimate the standard deviation of y^* for x=4.

### Explanation of Solution

Calculation:

The estimate of standard deviation of y^* is given as,

sy^*=s1n+(x*x¯)2(xix¯)2

Where x* be the given value of the independent variable x, y^* be the possible values of the dependent variable y when x=x* and x¯ be the sample of independent variable x for sample of size n.

Simple linear regression model can be expressed as y=β0+β1x+ and the estimated simple linear regression equation is y^=b0+b1x.

Least squares criterion:

The least square criterion can be obtained as by minimizing sum of squares of difference between observed and predictor variable, that is, min(yiy^i)2.

Where yi be the observed value of the dependent variable for the i–th observation and y^i be the predicted value of the dependent variable for the i–th observation.

Slope and y–intercept of the estimated regression equation:

The slope can be obtained as,

b1=(xix¯)(yiy¯)(xix¯)2

And the y–intercept is,

b0=y¯b1x¯

where xi is the value of the independent variable of the i-th observation.

yi is the value of the dependent variable of the i-th observations.

x¯ is the mean value for the independent variable.

y¯ is the mean value for the dependent variable.

n is the total number of observations.

It is known for a sample size n that mean of a random variable x can be obtained as,

x¯=1nxin.

Thus, mean of the random variable x is,

x¯=(1+2+3+4+5)5=155=3

Thus, mean of the random variable y is,

y¯=(3+7+5+11+14)5=405=8

The value of (xix¯)(yiy¯) and (xix¯)2 is calculated as follows:

 xi (xi−x¯) yi (yi−y¯) (xi−x¯)(yi−y¯) (xi−x¯)2 1 –2 3 –3.67 7.34 4 2 –1 7 0.33 –0.33 1 3 0 5 –1.67 0 0 4 1 11 4.33 4.33 1 5 2 14 7.33 14.66 4 1 –2 3 –3.67 7.34 4 Total 26 10

Thus, using the table the slope of the estimated regression equation is,

b1=2610=2.6

Thus, using the value of slope estimate the y–intercept of the estimated regression equation is,

b0=8(2.6)(3)=87.8=0.20

Hence, the estimated regression equation is y^=0.20+2.6x.

SSE:

The Sum of squares due to error (SSE) is obtained as SSE=i(yiy^i)2 where yi be the observed value of the dependent variable for the i-th observation and y^i be the predicted value of the dependent variable for the i-th observation

b.

To determine

Obtain a 95% confidence interval for the expected value of y when x=4.

c.

To determine

Estimate the standard deviation of an individual value of y for x=4.

d.

To determine

Obtain a 95% prediction interval for y when x=4.

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