lab06, homework
 

Homework

Polynomials * Homework 7, Due /*------------------------------------------------------------------ | Draper and Smith report radial growth of ice crystals in a room | | kept at -5 degrees C. The exposure time X and radial diameter of | | ice crystals Y are recorded for several replications of the | | experiment. | ------------------------------------------------------------------- Questions ------------------------------------------------------------------- | 1. An attempt is made to fit a degree 21 polynomial below. What | | happened? Why do you think this happened? Why was a degree | | 21 polynomial investigated? | | 2. Theoretically, it should be possible to fit a degree 21 | | polynomial. Put TIME in a class statement in GLM and from | | the output, find the (corrected) regression sum of squares | | that WOULD HAVE resulted had our attempt to fit a degree 21 | | polynomial been successful. | | 3. Plot the data using the code shown below. | | Why is the plot instruction repeated 3 times | | with an overlay option? What does the i=R option in the | | symbol statement control? (use the HELP menu --> GRAPHICS | | to find out). What is your guess as to the need for a | | higher than linear model? | | 4. Run regressions to determine the equations for the three | | curves in your plot. (this should match the log window | | message from gplot) | | 5. Using question 2 and 4 results, test for lack of fit of your | | linear model (versus a degree 21 model). Also look at the | | cubic you fit in problem 4 and test to see if its quadratic | | and cubic terms can be omitted (this is a pseudo lack of fit | | test assuming no higher than degree 3 is needed). | | 6. Put in a variable LOF which is just a copy of the TIME | | variable. Now submit this code: | | PROC GLM; CLASS LOF; MODEL Length = Time LOF/SS1; | | Explain how the output relates to some things we've done | | above. | | IF the linear model lack of fit had been significant, we might| | have tried a quadratic. Show how to modify the above code to | | check lack of fit of a quadratic model (even though it might | | not make sense to fit a quadratic to this particular data) | | Suppose someone fits the quadratic, notes that it has a | | maximum and proceeds to state that beyond that point the | | variable Y= length will decrease. Criticize his comments. | | 7. (optional - not graded) Sometimes even roughly centering the | | data can help. Replace time by (time-115)/50 and note that | | we are in better shape but still have problems. Also you | | might run the cubic regression with the centered data to | | make sure you understand the algebraic relationship among the | | coefficients. | | You may have noticed in part 6 that the output is incorrect.| | The use of time*time has caused numerical inaccuracies in the | | algorithm that decides the rank of our X matrix thereby | | producing an inaccurate number of df for LOF. Rerun the | | quadratic from part 6 with the CENTERED time variable and | | note that the problem disappears. | | 8. (optional - not graded) For the simple linear regression, | | perform White's (1980) specification test. Do the usual | | assumptions on the errors seem to hold? Compute a standard | | error for the slope that would be valid in large samples | | even in the presence of unequal error variances. | | | | Note: Questions of rank in computer programs depend on values | | of "pivotol elements" Tuning parameters determine how close | | these can get to 0 before being interpreted as 0s and thus | | the matrix being deemed rank deficient. The critical values | | of these tuning parameters are sometimes reset in updates of | | SAS. These homework questions are accurate for version 6.12 | | If you are using a later version and something changes, | | simply note this in your answers. | -----------------------------------------------------------------*/ data ice; input Time n @; * time=(time-115)/50; do i=1 to n; input length @; output; end; cards; 20 1 . <--- trick to peg X axis left end 50 1 19 (why isn't all this text messing 60 2 20 21 up my data step ???) 70 2 17 22 80 2 25 28 90 3 21 25 31 95 1 25 100 3 30 29 33 105 2 35 32 110 3 30 28 30 115 3 31 36 30 120 3 36 25 28 125 1 28 130 2 31 32 135 2 34 35 140 2 26 33 145 1 31 150 2 36 33 155 2 41 33 160 3 40 30 37 165 1 32 170 1 35 180 1 38 210 1 . <-- trick to peg X axis right end. ; proc print; proc gplot; plot length*time length*time length*time/overlay; symbol1 v=diamond h=3 i=RL c=red; symbol2 v=none i=join i=RQ c=green; symbol3 v=none i=join i=RC c=blue; proc glm; model length = time|time|time|time|time|time|time|time|time|time| time|time|time|time|time|time|time|time|time|time| time; proc means; run;