Question

A random sample of 24 companies from the Forbes 500 list was selected, and the relationship between their annual sales (, in hundreds of thousands of dollars) and annual profits (, in hundreds of thousands of dollars) was investigated by the researchers. [Note: if x = 200, that means the company had annual sales of $20 million; similarly for y values]. The following simple linear regression model was used: = + + .

Answer the following questions:

(a) In order for the researchers to validate their choice for simple linear regression, list 3 assumptions that the error terms, , must satisfy? List these assumptions one by one, followed by a short description of how to check that particular assumption.

(b) You are given the following summary data: ∑x= 6,600 ∑x²= 1,818,000 ∑y= 4,008 ∑y²= 673,868 ∑xy= 1,104,920 SSE = 1,905.075

(i) Calculate the sample regression equation. Provide a clear interpretation of the slope coefficient in the context of the question.

(ii) Calculate an interval (confidence or prediction) for the profit of one randomly selected Forbes 500 company whose sales are 270.5 hundred thousand dollars. Use a level of 95% and provide an interpretation of your interval in the context of the question.

Answer 1

a)

1)No or little multicollinearity
2)No auto-correlation
3)Homoscedasticity

b)

sample size ,   n =   24          
1)

here, x̅ =Σx/n =   275.0000   ,   ȳ = Σy/n =   167  
                  
SSxx =    Σx² - (Σx)²/n =   3000.000          
SSxy=   Σxy - (Σx*Σy)/n =   2720.000          
SSyy =    Σy²-(Σy)²/n =   4532.000          
estimated slope , ß1 = SSxy/SSxx =   2720.000   /   3000.000   =   0.9067
                  
intercept,   ß0 = y̅-ß1* x̄ =   -82.3333          
                  
so, regression line is   Ŷ =   -82.33   +   0.91   *x

It means if we incraese the sales by 1 million then profit will increase by .91 milliuon.

2)

X Value=   270.5
Confidence Level=   95%
  
  
Sample Size , n=   24
Degrees of Freedom,df=n-2 =   22
critical t Value=tα/2 =   2.074

Predicted Y at X=   270.5   is                  
Ŷ =   -82.333   +   0.907   *   270.5   =   162.920
                          
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    2.132                      
margin of error,E=t*Std error=t* S(ŷ) =   2.0739   *   2.1322   =   4.4220      
                          
Confidence Lower Limit=Ŷ +E =    162.920   -   4.4220   =   158.498      
Confidence Upper Limit=Ŷ +E =   162.920   +   4.4220   =   167.342     

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