Question

The relationship between the amount of remaining carbon monoxide (CO) in an individual’s lungs and the time since that person last smoked a cigarette can be summarized using a linear regression approach.

(a) Write down a simple linear regression model and the underlying assumptions. The following summary data was collected from 12 different smokers.

x = time since last smoked a cigarette (hours) y = amount of CO in ppm

n=12, x=1.88 sxx =25.8, syy =1805 SSE=877.4 Fitted regression line: ?̂ = ??. ?? − ?. ??? ?

2. (b) State the values of ? and ? from the fitted regression equation. Not that ? and ? are the estimates of intercept parameter ? and slope parameter ? respectively.

3. (c) What is the estimated change in the remaining CO in the lungs if the time since last smoked is increased by 1 hour?

4. (d) State the hypotheses to test the significance of the linear regression.

5. (e) Compute MSE.

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6. (f) Calculate the test statistic, state the degrees of freedom, and find the p-value to test the hypotheses in (c).

7. (g) Would you reject H0 or fail to reject H0 at 5% level of significance?

8. (h) What do you conclude about the regression of CO in the lungs on time since last

   smoked?

9. (i) Compute the coefficient of determination, r2 using 1-SSE/Syy. Interpret the value.

10. (j) Predict the amount of remaining CO in the lungs for an elapsed time of 2.25 hours

    after smoking.

11. (k) What is the residual at ?0 = 2.25 if the corresponding observed amount CO is

    28ppm?

12. (l) Estimate the mean amount of CO in the lungs for an elapsed time of 2.25 hours.

(m) Construct a 95% confidence interval for the true mean amount of CO in the lungs

for an elapsed time of 2.25 hours.

Answer 1

Answer:

a)  Ŷ =   45.1700   +   -5.9960   *x

b)

estimated slope , b = -5.99600
                  
intercept, a = 45.17000          

c) decrease by 5.996

d)

Ho:   ß1=   0
H1:   ß1╪   0

e) MSE=SSE/df = 877.4/10 = 87.74

f) test stat=10.5721

p value=0.0087

g) reject ho

i) R²= SSR/SST=0.5139

about 51.39% of variation in observation of variably y, is xplained by variable x

j)

Predicted Y at X=   2.25   is          
Ŷ =   45.1700   +   -5.9960   *2.25=   31.679

k) residual = 28-31.679 =  -3.6790

l)  Ŷ =   45.1700   +   -5.9960   *2.25=   31.679

m)

X Value=   2.25              
Confidence Level=   95%              
                  
                  
Sample Size , n=   12              
Degrees of Freedom,df=n-2 =   10              
critical t Value=tα/2 =   2.228   [excel function: =t.inv.2t(α/2,df) ]          
                  
X̅ =    1.88              
Σ(x-x̅)² =Sxx   25.80              
Standard Error of the Estimate,Se=   9.3670              
                  
Predicted Y at X=   2.25   is          
Ŷ =   45.1700   +   -5.9960   *2.25=   31.679
                  
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    2.789              
margin of error,E=t*Std error=t* S(ŷ) =   2.2281   *   2.789   =   6.2138
                  
Confidence Lower Limit=Ŷ +E =    31.679   -   6.214   =   25.465
Confidence Upper Limit=Ŷ +E =   31.679   +   6.214   =   37.893

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