STAT 350

Assignment 3: Solutions

The data and assignment comments are here

1. Vehicle mileage vs emissions.
   1. Plot the data.
   2. Consider the following 4 models for the data:
      1. Two straight lines, one for each vehicle, with different slopes and intercepts,
      2. Two parallel straight lines.
      3. Two lines with the same intercept but different slopes.
      4. One straight line.

      Write out model equations for data points number 2 and 22 for the first 3 models.

      FIRST MODEL

      

      and

      

      Here and are the two intercepts while and are the slopes.

      SECOND MODEL

      

      and

      

      Here and are the two intercepts while is the common slope.

      THIRD MODEL

      

      and

      

      Here and are the two slopes while is the common intercept.

   3. Fit all 4 models. Hand in: estimates of the slopes and intercepts and of . Do NOT just hand in output from SAS or Minitab.

      I create a data file which contains the design matrix for the first model.

        50 1     0        0     0 
        56 1  1000        0     0 
        58 1  2000        0     0
        60 1  3000        0     0
        58 1  4200        0     0
        63 1  5000        0     0
        73 1  6000        0     0
        71 1  6900        0     0
        76 1  8000        0     0
        73 1  9200        0     0
        80 1 10000        0     0
        40 0     0        1     0 
        49 0     0        1  1100  
        58 0     0        1  2200 
        65 0     0        1  3000 
        75 0     0        1  4000
        77 0     0        1  5300
        86 0     0        1  6000
        93 0     0        1  7000
        98 0     0        1  8100
       103 0     0        1  9000
       109 0     0        1 10000

      I used the following SAS code to fit the models.

      options pagesize=60 linesize=80;
      data mileage;
        infile 'mile1.dat' ;
        input emiss car1 mile1 car2 mile2 ;
        mile = mile1+mile2;
      proc glm  data=mileage;
         model emiss =  car1 mile1 car2 mile2 / NOINT ;
         estimate 'sloped' mile1 1 mile2 -1 /E ;
         estimate 'intd' car1 1 car2 -1 /E ;
      run ;
      proc glm  data=mileage; 
         model emiss =  car1 car2 mile /NOINT  ; 
      run ;
      proc glm  data=mileage; 
         model emiss =  mile1 mile2  ; 
      run ;
      proc glm  data=mileage; 
         model emiss =  mile ; 
      run ;
      proc glm  data=mileage;
         model emiss =  car1 mile1 car2 mile2 / NOINT ;
         estimate 'veh1em' car1 10000 mile1 50000000 /E ;
         estimate 'veh2em' car2 10000 mile2 50000000 /E ;
         estimate 'diff'   car1 10000 mile1 50000000 car2 -10000 mile2 -50000000 /E;
      run ;

      Here are the estimates;

      1 51.28 0.00278 42.93 0.00684 2.79

      2 41.20 0.00479 53.29 -- 7.41

      3 47.16 0.00337 -- 0.00623 3.62

      4 47.19 0.00480 -- -- 9.61

   4. Using formal hypothesis tests and plots select the best of these models. Again, I do not want computer output but discussion. You may attach computer output in order to say things like: ``The Sum of Squares for ... is on page ..." and to hand in plots on which you comment in the discussion but I will not be looking through the output.

      I begin by testing the hypothesis that . You can do this either using the extra sum of squares F-test or a suitable t-test. When I assigned the question, however, you only really knew how to do the t-tests. The line estimate 'sloped' gets standard errors and a t-statistic. The output lines corresponding to the estimate lines is

                                              T for H0:    Pr > |T|   Std Error of
      Parameter                  Estimate    Parameter=0                Estimate
      
      sloped                  -0.00405367         -10.75     0.0001     0.00037706
      intd                     8.35473800           3.72     0.0016     2.24468995

      Each of these tests is quite significant so that you can't get by with either a common slope or a common intercept. That is, the first model is preferred. You can also do extra sum of squares tests. The needed information is in the Error SS from the various runs of glm:

      MODEL     Error DF     Error SS          MSE
      1           18        140.47791         7.80433
      2           19       1042.46491        54.86657
      3           19        248.59354        13.0838705
      4           20       1847.50378        92.3751888

      The extra SS F-statistic for testing model 2 against model 1 is [(1042.46-140.48)/1]/[140.48/18] and this is compared to F tables with 1 and 18 degrees of freedom. The statistic value is 115.6 which is very significant. Similarly model 3 is rejected in favour of model 1. Model 4, requiring both models 2 and 3 to be correct is untenable. It can be tested directly against model 1 using [(1847.50378-140.477910/2]/7.80433 as an F-test.
   5. For the final selected model estimate the total emissions of CO in grams for each vehicle over the first 10000 miles. (This is the area under the fitted straight line from 0 to 10000 and is a linear combination of the parameter estimates.) Attach a standard error.

      In terms of the coefficients in the model the emissions for vehicle1 are while those for vehicle 2 are . These are estimated by plugging in least squares estimates. These two estimates and their difference are all linear combinations of the form for which the standard error is . You can calculate these standard errors using estimate statements as in the last run of proc glm. The corresponding output is

                                              T for H0:    Pr > |T|   Std Error of
      Parameter                  Estimate    Parameter=0                Estimate
      
      veh1em                   651968.435          77.40     0.0001      8423.4004
      veh2em                   771104.319          91.53     0.0001      8424.8153
      diff                    -119135.884         -10.00     0.0001     11913.4876

      The last line shows that the two cars have different emissions in total over the first 10000 miles (answering the next part) while the previous 2 permit confidence intervals of the form .
   6. Are the emissions of the two vehicles different over the first 10000 miles?

      The answer is yes, the second vehicle clearly has higher emissions. See the previous question for the test.

   7. Suppose the two cars are of the same make but that one vehicle was equipped with a special pollution control device. In 4 or 5 sentences comment on the experimental design as a method of determining whether or not the device reduces emissions and on what else you would want to find out from the experimenter to help interpret the results.

      The key problem is replication. In any case we need to be sure that the two cars are similar in make, model, driver, road conditions, nature and extent of maintenance and so on. The trouble is that with only two cars the variation from car to car under identical conditions cannot be allowed for. How do you know that if you took 2 cars identically equipped with no different pollution control devices you wouldn't see just as big a difference?

2. Data below are from a nitrogen balance experiment on Kangaroo Island Wallabies, taken from Barker,S. (1968). ``Nitrogen balance and Water Intake in the Kangaroo Island Wallaby'' Austral. J. Experimental Biology and Medical Science, 46, 17-32.

   Nitrogen Body Dry Water Nitrogen

   Excreted Weight Intake Intake Intake

   162 3.386 16.6 41.7 54

   174 3.033 18.1 40.9 99

   119 3.477 13.4 25.0 46

   205 3.278 22.6 39.2 188

   312 3.368 26.5 47.4 345

   157 2.932 21.4 51.6 66

   184 3.128 30.3 71.6 171

   155 3.251 17.6 27.1 81

   192 3.396 21.3 37.7 175

   331 3.497 29.9 50.5 399

   114 3.182 12.8 28.4 38

   159 3.234 19.6 34.3 106

   260 3.139 36.2 77.6 228

   265 3.434 35.0 58.9 291

   387 2.970 32.9 55.3 449

   146 3.230 22.9 46.2 72

   233 3.470 32.9 67.4 176

   261 3.000 35.7 77.1 235

   287 3.224 34.4 74.9 288

   412 3.366 36.2 60.7 485

   174 3.264 29.9 65.4 92

   171 3.292 21.7 51.2 126

   259 3.525 35.0 66.8 224

   298 3.036 29.7 65.8 276

   407 3.356 29.2 48.1 386

   Fit the model

   

   by least squares. Get estimates and standard errors for all the parameters and an estimate of . Suggest a simpler model for the data, and fit it. Check the fit of the model, graphically and, if the model seems poor, modify it appropriately. Hand in a discussion of your findings bolstered by output used only as an appendix. I will be marking the discussion, not sorting through the output.

   The final fitted model has only. An extra sum of squares F-test comparing this test to the full model accepts the null hypothesis that . The plots look quite alright though observation 25 has a surprisingly large residual. Deletion of this observation changes the conclusions, however; variable is retained.

   Here is SAS code and output.

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