Question

At a local department store, let X be an independent variable for the number of salespeople on the floor and y the dependent variable for daily sales in thousands of dollars. The next screen will present the data.

X={5, 6, 7, 8, 9, 10}                  Y = {7, 8, 9, 12, 15, 20}

1.   Write down the prediction equation.

2. Write down SSE, S2, SSyy and S(std. dev).

3. Predict the retail sales when there are 10 salespeople on the floor and then calculate the prediction error.

4. Construct a 95% confidence interval for b1.

5. Test if the number of salespeople on the floor is significant to the prediction of y.

6. Find the coefficient determination and interpret.

7. Find the 95% confidence interval for E(Y) when x = 10 salespeople.

8. Find a 95% prediction interval for Y when x = 10 salespeople.            

9.   Predict Y if x = 4 salespeople and find the confidence interval for E(y) and the prediction interval for Y. Is there a residual? If not, why not? If yes then what is it?

Answer 1

1)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	45.00	71.00	17.50	122.83	44.50
mean	7.50	11.83	SSxx	SSyy	SSxy

`   n =   6      
here, x̅ = Σx / n=   7.500          
ȳ = Σy/n =   11.833          
SSxx =    Σ(x-x̅)² =    17.5000      
SSxy=   Σ(x-x̅)(y-ȳ) =   44.5      
              
estimated slope , ß1 = SSxy/SSxx =   44.5/17.5=   2.5429      
intercept,ß0 = y̅-ß1* x̄ =   11.8333- (2.5429 )*7.5=   -7.2381      
              
Regression line is, Ŷ=   -7.2381   + (   2.5429   )*x

...............

2)

SSE=   (SSxx * SSyy - SS²xy)/SSxx =    9.6762
std error ,Se =    √(SSE/(n-2)) =    1.5553
SSyy =122.83
..............

3)

Predicted Y at X=   10   is          
Ŷ=   -7.23810   +   2.54286   *10=   18.1905

error =1.8095

.............

4)

confidence interval for slope                  
α=   0.05              
t critical value=   t α/2 =    2.776   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    1.5553/√17.5=   0.372          
                  
margin of error ,E= t*std error =    2.776   *   0.372   =   1.032268
estimated slope , ß^ =    2.5429              
                  
lower confidence limit = estimated slope - margin of error =   2.5429   -   1.032   =   1.5106
upper confidence limit=estimated slope + margin of error =   2.5429   +   1.032   =   3.5751

.............

5)

Ho:   β1=   0
H1:   β1╪   0
n=   6  
alpha =   0.05  
estimated std error of slope =Se(ß1) = Se/√Sxx =    1.5553/√17.5=   0.3718
t stat = estimated slope/std error =ß1 /Se(ß1) =    (2.5429-0)/0.3718=   6.84
Degree of freedom ,df = n-2=   4  
      
p-value =    0.0024  
decison :    p-value<α , reject Ho  
Conclusion:   Reject Ho and conclude that slope is significantly different from zero  
................

6)

R² =    (SSxy)²/(SSx.SSy) =    0.9212
Approximately    92.12%   of variation in observations of variable Y, is explained by variable x
.....

7)

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    1.126              
margin of error,E=t*Std error=t* S(ŷ) =   2.7764   *   1.126   =   3.1253
                  
Confidence Lower Limit=Ŷ +E =    18.190   -   3.125   =   15.0651
Confidence Upper Limit=Ŷ +E =   18.190   +   3.125   =   21.3158

.............

8)

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   1.9199              
margin of error,E=t*std error=t*S(ŷ)=    2.776   *   1.920   =   5.3306
                  
Prediction Interval Lower Limit=Ŷ -E =   18.190   -   5.331   =   12.8599
Prediction Interval Upper Limit=Ŷ +E =   18.190   +   5.331   =   23.5211

............

9)

Predicted Y at X=   4   is          
Ŷ=   -7.23810   +   2.54286   *4=   2.9333
                  
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    1.448              
margin of error,E=t*Std error=t* S(ŷ) =   2.7764   *   1.448   =   4.0201
                  
Confidence Lower Limit=Ŷ +E =    2.933   -   4.020   =   -1.0868
Confidence Upper Limit=Ŷ +E =   2.933   +   4.020   =   6.9534
                  
For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   2.1250              
margin of error,E=t*std error=t*S(ŷ)=    2.776   *   2.125   =   5.8999
                  
Prediction Interval Lower Limit=Ŷ -E =   2.933   -   5.900   =   -2.9666
Prediction Interval Upper Limit=Ŷ +E =   2.933   +   5.900   =   8.8332

.........

observed value of y at x=4 is not given, so, we cannot calalulate residual for y at x = 4

...................

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