Question

A local tire dealer wants to predict the number of tires sold each month. He believes that the number of tires sold is a linear function of the amount of money invested in advertising. He randomly selects 6 months of data consisting of tire sales (in thousands of tires) and advertising expenditures (in thousands of dollars). Based on the data set with 6 observations, the simple linear regression model yielded the following results.

?X =24
?X²=124
?Y  = 42
?Y² =338
?XY =196

Calculate the coefficient of determination and the coefficient of correlation between X and Y. Interpret the coefficient of Determination. also find the slope and intercept and write the estimated Regression equation. What would the predicted sales of tires be if he spends five thousand dollars in advertising? Perform the test of significance for the slope coefficient. Use 5% level of significance.

Answer 1

Σn	ΣX	ΣY	ΣXY	ΣX²	ΣY²
6	24.00	42.00	196.00	124.00	338.00

sample size ,   n =   6      
here, x̅ =   4.0000       ȳ =   7
              
SSxx =    Σx² - (Σx)²/n =   28.000      
SSxy=   Σxy - (Σx*Σy)/n =   28.000      
SSyy =    Σy²-(Σy)²/n =   44.000      

coefficient of determination=(Sxy)²/(Sx.Sy) =    0.6364

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correlation coefficient ,    r = Sxy/√(Sx.Sy) =   0.7977
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Interpret the coefficient of Determination

79.77 % of variation in observation of variable Y is explained by variable X

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slope ,    ß1 = SSxy/SSxx =   1          
                  
intercept,   ß0 = y̅-ß1* x̄ =   3          
                  
so, regression line is   Ŷ =   3   +   1   *x

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predicted sales of tires be if he spends five thousand dollars in advertising

Ŷ =   3   +   1   *x

Ŷ =   3   +   1   *5 = 8

so, predicted sales of tires = $8000

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SSE=   (Sx*Sy - S²xy)/Sx =    16
      
std error ,Se =    √(SSE/(n-2)) =    2

slope hypothesis test                  
Ho:   ß1=   0          
H1:   ß1╪   0          
                  
alpha=   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    0.3780
                  
t stat =    ß1 /Se(ß1) =        2.646
                  
df=n-2 = 4
p-value =    0.0572 [excel function: t.dist.2t(2.646,4) ]
decision :    p-value>α , do not reject Ho      

so, there is enough evidence to conclude that slope is not significant