08092BHW8Solution

08092BHW8Solution - 2009 Math 2b Homework 8 Solutions...

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2009 Math 2b Homework 8 Solutions Problem 1. The stopping distance ( y ) of an automobile on a certain road was studied s a function of velocity (Brownlee 1960). The data are listed in the following table. Fit y and y as linear functions of velocity and examine the residuals in each case. Which ±t is better? Can you suggest a physical reason that explains why? Solution. We use the models y i = β 0 + β 1 x i + ± i and y i = β ± 0 + β ± 1 x i + ± ± i . To calculate our esimators for the β i we use the formulae given on page 565 in Rice ˆ β =( X T X ) - 1 X T Y. Plugging in our data and doing the calculation gives ˆ β = ± - 62 . 0454 3 . 4930 ² , ˆ β ± = ± - 0 . 8776 0 . 2277 ² . To compare the quality of the ±ts, we will compare the residuals, but ±rst we must rescale the data because the y -±t will have smaller residuals simply because it takes smaller values than the other ±t. To do this, we look at the y -±t, take the square of the values and then look at the residuals, i.e. ( y ) 2 - ( ˆ β ± 0 - ˆ β ± 1 v ) 2 where we use the values for β ± i found earlier. Calculating the errors in this way gives: 1 . 03 , - 1 . 07 , - 2 . 92 , 3 . 46 , 8 . 24 , - 8 . 32 whose squares add up to 159 . 8238. On the other hand, the y -±t errors are: 5 . 84 , 3 . 74 , - 10 . 59 , - 6 . 32 , 4 . 59 , 2 . 75 whose squares add up to 228 . 81. This suggests that the y -±t is better than the y -±t. Physically, we know that under constant acceleration a , it is true that v 2 f = v 2 0 +2 ay . In the case where acceleration is negative and v f = 0, this gives y = v 2 0 2 | a | , so y v . Problem 2. The data in the problem were obtaind in a Stanford study of the survival time of heart transplant patients. The “mismatch score” is an indication of how well the transplanted heart should ±t the patient. Consider the regression model log(Survival Time) = β 1 + β 2 (Age) + β 3 (Mismatch Score) + error, where we assume the errors are independent normal with mean 0 and unknown variance σ 2 . ( a ) Fit the model i.e. calculate the least squares estimates of β 1 , β 2 and β 3 and calculate the sample standard deviation s . ( b ) Calculate R 2 , the coe²cient of determination. ( c ) Test the hypothesis β 2 = 0 at the α =0 . 1 level. ( d ) Give a 90% con±dence interval for β 3 . ( e ) Give a 90% con±dence interval for the expectation of the logarithm of the survival time of patients with a mismatch score of 1.00 and age 55. 1
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Solution. (a) We have Y = + ± where Y = log(44) log(550) . . . log(48) ,X = 1 36 0 . 00 1 48 0 . 12 1 45 0 . 16 . . . . . . . . . 1 53 3 . 05 , β = β 1 β 2 β 3 . One of the data points has a survival time of 0, so to prevent taking log(0), we replace this 0 with 0 . 5. Therefore ˆ β =( X T X ) - 1 X T Y = 6 . 39417 - 0 . 0746735 - 0 . 0154375 and s 2 = || Y - X ˆ β || 2 157 - 3 =3 . 24106 .
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This note was uploaded on 01/08/2011 for the course MA 2b taught by Professor Makarov,n during the Winter '08 term at Caltech.

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08092BHW8Solution - 2009 Math 2b Homework 8 Solutions...

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