Simple Regression Analysis - Exam Questions with Solutions | STAT 202, Exams of Business Statistics

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STAT 202

Recitation Problem: Simple Regression Analysis

Betty’s Bodacious Burgers sells burgers (duh), and Betty has recorded advertising expenses and

gross sales data for n = 12 randomly selected months. Expense and sales values are measured in $1k

units. Here’s Betty’s data and some other useful results. For accurate calculation I recommend that

you work to five (or more) significant digits!

x = advertising

expense

($1k units)

y = gross

sales

($1k units)

2.4 22.

3.1 31.

4.2 34.

1.8 12.

3.5 30.

3.7 39.

4.1 45.

3.6 28.

3.4 25.

4.1 38.

3.7 42.

2.9 33.

0 5 10 15 20 25 30 35 40 45 50 1 2 3 4 5 ad expenses gross sales  xi ^40.^50  xi^2 ^142.^43 yi^385.^2  y i^2 ^13 ,^328.^38  xiyi ^1 ,^361.^89

(a) Find the equation of the sample regression line.

(b) Next month, Betty plans to spend $2,500 on advertising. Predict (point predictor) next month’s

gross sales.

(c) Calculate s y.x, the ‘standard error of estimate.’

(d) At the 5% level of significance, test to see if gross sales is related to advertising.

(e) Refer to Part (b). Find a 95% prediction interval for next month’s gross sales.

‘Take-home’ problem:

Use Excel or Minitab to generate a regression analysis printout for this data. Verify that your results

for (a), (c), and (d) match those given on the printout, give or take rounding error. Also calculate the

correlation coefficient and verify that the square of your correlation coefficient matches the R -

squared value on the printout.

Possibly Useful Formulas:

 

 

( )( ) x n x xy nx y b i i i   





b 0 (^)  y  b 1 x    

( )

( ˆ ) 0 1

2 2 .



 









   

n

y b y b x y

n

y y

s y x i i i i i i

CI:

 2  2

2 . 1 (x value ) ˆ ( ) x n x x n y t s i y x 



 



PI:

 2  2

2 . 1 (x value ) ˆ ( ) 1 x n x x n y t s i y x 



  



    

       

           i i i i i i i i n x x n y y n x y x y r    2 2 . 1 x n x s s i y x b 1

s b

b

t





Solution:

(a) First note that X ^40.^5 /^12 ^3.^3750 ; Y ^385.^2 /^12 ^32.^10

 

 

10. 7688
11. 7425
12. 840
13. 43 12 ( 3. 3750 ) ( )( ) 1361. 89 12 ( 3. 3750 )( 32. 10 ) (^1 )        





x n x xy n x y b i i i b 0  y  b 1 x  32. 10  10. 7688 ( 3. 3750 ) 4. 2447

So the sample regression line is y ˆ^^ ^4.^2447 ^10.^7688 x

(b) y ˆ^^ ^4.^2447 ^10.^7688 (^2.^5 )^22.^6773 but that’s in $1k units so predicted gross sales is

$22,677.30.

(c)

   

5. 4545 10

13328. 38 ( 4. 2447 )( 385. 2 ) 10. 7688 ( 1361. 89 )

( 2 ) 0 1

. 

  





 



   n y b y b x y s (^) y x i i i i

(d) Let’s use a t -test to test H 0 :  1 = 0 (no relation) vs. H 1 :  1 ≠ 0 (is a relation).

Test stat is

1

s b

b

t



 where

2. 27616
3. 7425
4. 4545 2 2 . 1    

 x nx

s s i y x

b . So

1

^1  ^  

s b

b

t . Degrees of freedom = n – 2 = 10, the decision rule is to

reject H 0 if t < -2.2281 or t > +2.2281. Since 4.7277 > 2.2281 we reject H 0 and conclude that

there is a relation (the slope is different from zero).

(e)  

22. 6773 2. 2281 ( 5. 4545 ) 1. 216659 22. 6773 13. 4052 ( 9. 2721 , 36. 0825 )

( 2. 5 3. 3750 )

22. 6773 2. 2281 ( 5. 4545 ) 1

1 (x value ) ˆ ( ) 1 2 2 2 2 .     



   





  

 x nx

x n y t s i y x

So there is a 0.95 probability that next month’s gross sales will fall between $9,272.10 and

$36,082.50.

Here’s most of the regression printout:

SUMMARY OUTPUT Regression Statistics Multiple R 0. R Square 0. Adjusted R Square 0. Standard Error 5. Observations 12 ANOVA df SS MS F Significanc e F Regression 1 665.9443796 665.9444 22.38351 0. Residual 10 297.5156204 29. Total 11 963. Coefficient s Standard Error t Stat P-value Lower 95% Upper 95% Intercept -4.2448 7.841777188 -0.54131 0.600148 -21.7174 13. x 10.76883 2.276168371 4.731121 0.000803 5.69721 15.