Question

A corporation owns several companies. The strategic planner for the corporation believes dollars spent on advertising can to some extent be a predictor of total sales dollars. As an aid in long-term planning, she gathers the following sales and advertising information from several of the companies for 2017 (in millons)
advertising($12.5 , 3.7 ,21.0 ,60.0 ,37.6 ,6.1 ,17.9 ,41.2 )
sales($141, 55 ,338 ,994 ,541 ,89 ,126 ,379)

Construct a 98% confidence interval for the average value of y for the given data; use x = $20, s_ε = $111.9M, and the equation of the regression line, y^ = -48.54 + 15.26 x.

Do not round the intermediate values. Round your answers to 2 decimal places.

Enter a value.  ≤ E(y₂₀) ≤ Enter a value. 

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Construct a 98% prediction interval for a single value of y for the given data; use x = $20, s_ε = $111.9M, and the equation of the regression line, y^ = -48.54 + 15.26 x.

Do not round the intermediate values. Round your answers to 2 decimal places.

Enter a value.  ≤ y₂₀ ≤ Enter a value.  

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Which is wider? Why?

The prediction interval for the Select an option.                                      

                                  average value of ysingle value of y is wider than the confidence interval for the Select an option.                                                                        single value of yaverage value of y because the average is more towards the Select an option.                                                                        rightmiddleleft and individual values of y Select an option.                                                                       can vary more thancan't vary more than values of the average.

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Answer 1

I used R software to solve this question.

R codes and output:

> ad=c(12.5,3.7,21,60,37.6,6.1,17.9,41.2)
> sales=c(141,55,338,994,541,89,126,379)
> fit=lm(sales~ad)
> fit

Call:
lm(formula = sales ~ ad)

Coefficients:
(Intercept) ad
-48.54 15.26

> data=data.frame(ad=20)
> predict(fit,newdata=data,interval='confidence',level=0.98)
fit lwr upr
1 256.5926 127.7146 385.4707
> predict(fit,newdata=data,interval='prediction',level=0.98)
fit lwr upr
1 256.5926 -117.9173 631.1025

98% confidence interval for the average value of y for the given data:

127.71 < E(Y20) < 385.47

98% prediction interval for the average value of y for the given data:

-117.92 < (Y20) < 631.10

The prediction interval for the single value of y is wider than the confidence interval for the average value of y because the average is more towards the middle and individual values of y can vary more than values of the average.