Question

Refer to Copier maintenance Problem 1.20. a. Obtain a 90 percent confidence interval for the mean gervice time on calls in which six copiers are serviced. Interpret your confidence interval. b. Obtain a 90 percent prediction interval for the service time on the next call in which six copiers are serviced. Is your prediction interval wider than the corresponding confidence interval in part (a)? Should it be?

Answer 1

Answer:

Step-by-step explanation:

Hello!

You have the variables:

Yi: Service time on call.

Xi: Number of copiers that were serviced.

The data set attached.

This data belongs to linear regression analysis, the objective is to check the relationship between the service time and the number of copies that were serviced.

a. You need to calculate a 90% CI for the average service time in which six copiers are serviced.

Symbolically: E(Y/X=6)

The point estimate of the average value of Y is the estimated regression line:

E(Y)= a + bX

And to estimate it using a confidence interval you have to use the following formula:

a + bx₀ ± $t_{n-2;1-\alpha /2}$ * $\sqrt{Se^2*(\frac{1}{n}+\frac{(x_0-X[bar])^2}{sumX^2-\frac{(sumX)^2}{n} } )}$

a= -0.58

b= 15.04

n= 45

∑X= 230

∑X²= 1516

X[bar]= 5.11

Se^2= 79.45

$t_{n-2;1-\alpha /2} = t_{43;0.95}= 1.681$

[(-0.58+15.04*6) ± 1.681*$\sqrt{79.45*(\frac{1}{45}+\frac{(6-5.11)^2}{1516-\frac{(230)^2}{45} } )}$]

[87.312;92.007]

With a 90% confidence level, you'd expect that the interval [87.30;92.02] includes the average service time in call in which 6 copiers where serviced.

b. In this item instead of calculating a confidence interval for the average time of Y given a certain value of X, you will predict a value that the dependent variable Y will take given a known value of the response variable X

Symbolically $Y/X_{n-1}=6$

Another difference between the confidence interval and the prediction interval is that the amplitude of the prediction interval is weider because the standard error of this interval is greater as you can see in the formula:

$(a + bx_{n-1})$ ± $t_{n-2;1-\alpha /2}$ * $\sqrt{Se^2*(1+\frac{1}{n}+\frac{(x_{n-1}-X[bar])^2}{sumX^2-\frac{(sumX)^2}{n} } )}$

[(-0.58+15.04*6) ± (1.681 * $\sqrt{79.45*(1+\frac{1}{45}+\frac{(6-5.11)^2}{1516-\frac{(230)^2}{45} } )}$)]

[74.494;104.826]

Using a 90% level you'd expect the interval [74.494;104.826] will contain the value of the service time call in wich six copiers are serviced.

The amplitude of the interval in item a is: 92.007 - 87.312= 4.695

The amplitude of the interval in itemb is: 104.826 - 74.494= 30.332

As expected, the amplitude of the prediction interval is greater than the amplitude of the confidence interval.

I hope it helps!