ramin shamshiri abe6981 hw_03

7/30/2019 Ramin Shamshiri ABE6981 HW_03

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Ramin Shamshiri ABE 6986, HW #3 Due 09/17/08

Homework #3

Due 09/17/08

ABE 6986

Ramin Shamshiri

UFID # 90213353

Incidence of coronary heart disease with age

Reference: Hosmer, D.W and S. Lemeshow. 1989. Applied logistic regressionJohn Wiley & Sons. New York, NY

Table 1: Dependence of coronary heart disease with age

Age Group

yr

Age

yr

F Z

20-29 25.0 0.10 -2.079

30-34 32.5 0.13 -1.779

35-39 37.5 0.25 -0.95640-44 42.5 0.33 -0.547

45-49 47.5 0.46 0.044

50-54 52.5 0.63 0.847

55-59 57.5 0.76 1.692

60-69 65.0 0.80 2.079

Data are given in Table 1 for incidence of coronary heart disease with age group among 100 subjects,where F is the fraction of the 100 with significant symptoms. Assume the result can be described by the

logistic model given by:

= 1 + exp ( .)

Where A is the maximum value ofF at high age; b is the intercept parameter; and c is the response

coefficient, yr-1

. Now equation above can be rearranged to the linearized form

= ln 1 = .


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1- Plot F vs. Age on linear-linear graph paper.Answer:

Age

yr

F

25.0 0.10

32.5 0.13

37.5 0.25

42.5 0.33

47.5 0.46

52.5 0.63

57.5 0.76

65.0 0.80


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2- Calculate values of Z for each age for values of A= 0.90, 0.91, 0.92, 0.93, 0.94 and 0.95.Answer:

= ln 1 = . A=0.90

Age=25 => F=0.1

= ln 0.90.1

1 = 2.079.

.

.

A=0.95

Age=65 => F=0.8

= ln 0.950.8

1 = 1.6739

Table 2: Values of Z for each age for the given values of AAge

yr

Z1 Z2 Z3 Z4 Z5 Z6

25.0 -2.0794 -2.0919 -2.1041 -2.1163 -2.1282 -2.1401

32.5 -1.7789 -1.7918 -1.8045 -1.8171 -1.8295 -1.8418

37.5 -0.95551 -0.97078 -0.9858 -1.0006 -1.0152 -1.0296

42.5 -0.54654 -0.56394 -0.5810 -0.59784 -0.61437 -0.63063

47.5 0.044452 0.021979 0 -0.02151 -0.04256 -0.06318

52.5 0.8473 0.81093 0.7758 0.74194 0.70915 0.6774

57.5 1.6917 1.6227 1.5581 1.4975 1.4404 1.3863

65.0 2.0794 1.9841 1.8971 1.8171 1.743 1.674

A= 0.90 0.91 0.92 0.93 0.94 0.95


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3- Estimate values of parameters b and c corresponding to each values of A by linearregression of Z vs. Age. Include the correlation coefficient (r) to 5 decimal places.

Answer:

Linear model: = ln 1 = . = 0.1144 5.235Coefficients (with 95% confidence bounds):

c = 0.1144 (0.09751, 0.1313)

b = -5.235 (-6.022, -4.447)

Goodness of fit:

SSE: 0.3531 R-square: 0.9787 r=0.98929

Adjusted R-square: 0.9751 RMSE: 0.2426

Age

yr

Z

25.0 -2.07932.5 -1.779

37.5 -0.956

42.5 -0.547

47.5 0.044

52.5 0.847

57.5 1.692

65.0 2.079


5/13


Linear model: = ln 1 = . 1 = 0.1144 5.235Coefficients (with 95% confidence bounds):

c = 0.1144 (0.09754, 0.1312)b = -5.235 (-6.022, -4.448)

Goodness of fit:

SSE: 0.3523 R-square: 0.9787 r=0.98929Adjusted R-square: 0.9752 RMSE: 0.2423

Age (yr) Z1

25.0 -2.0794

32.5 -1.7789

37.5 -0.95551

42.5 -0.54654

47.5 0.044452

52.5 0.8473

57.5 1.6917

65.0 2.0794

A= 0.90


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c = 0.1123 (0.09589, 0.1288)

b = -5.178 (-5.946, -4.41)

Goodness of fit:

SSE: 0.3358 R-square: 0.979 r=0.98944


Age (yr) Z2

25.0 -2.0919

32.5 -1.7918

37.5 -0.97078

42.5 -0.56394

47.5 0.021979

52.5 0.81093

57.5 1.622765.0 1.9841

A= 0.91


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c = 0.1105 (0.09435, 0.1266)

b = -5.127 (-5.88, -4.374)

Goodness of fit:

SSE: 0.3226 R-square: 0.9791 r=0.989797


Age (yr) Z3

25.0 -2.1041

32.5 -1.8045

37.5 -0.9858

42.5 -0.5810

47.5 0

52.5 0.7758

57.5 1.5581

65.0 1.8971

A= 0.92


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c = 0.1088 (0.09291, 0.1246)

b = -5.082 (-5.823, -4.341)

Goodness of fit:

SSE: 0.312 R-square: 0.9791 r=0.989494


Age (yr) Z4

25.0 -2.1163

32.5 -1.8171

37.5 -1.0006

42.5 -0.59784

47.5 -0.02151

52.5 0.74194

57.5 1.497565.0 1.8171

A= 0.93


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c = 0.1072 (0.09156, 0.1228)

b = -5.041 (-5.772, -4.311)

Goodness of fit:

SSE: 0.3035 R-square: 0.9791 r=0.989494


Age (yr) Z5

25.0 -2.1282

32.5 -1.8295

37.5 -1.0152

42.5 -0.61437

47.5 -0.04256

52.5 0.70915

57.5 1.4404

65.0 1.743

A= 0.94


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c = 0.1057 (0.09028, 0.1212)

b = -5.004 (-5.726, -4.283)

Goodness of fit:

SSE: 0.2964 R-square: 0.979 r=0.989444


Age (yr) Z6

25.0 -2.1401

32.5 -1.8418

37.5 -1.0296

42.5 -0.63063

47.5 -0.06318

52.5 0.6774

57.5 1.3863

65.0 1.674

A= 0.95


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4- Select the values of A, b, and c for the optimum r.Answer:Based on the results of problem 3, we have the below table for the r values:

A r

0.90 0.989290.91 0.98944

0.92 0.98979

0.93 0.98949

0.94 0.98949

0.95 0.98944

The correlation coefficient value is desired to be closer to 1.0. According to the above table, r=0.98979

has the larger value and closet to one, thus we consider it as the optimum r. This value corresponds to:

3 = 0.1105 5.127A=0.92,b = -5.127 (-5.88, -4.374)

c = 0.1105 (0.09435, 0.1266)

5- Plot Z vs. Age for this case on linear-linear graph paper. Plot the regression line as well.Answer:


12/13


6- Plot the estimation equation on part (1)Answer:First, using the F and Age data set, I have plotted F vs. Age on linear-linear paper and a regression line

(Figure 9). The using MATLAB, I have symbolically plotted (Figure 10) Eq.1 with:

A=0.92

b = -5.127

c = 0.1105

= 0.921 + exp (5.127 0.1105.)

Linear model to fit on F vs. Age:

= 0.02024 0.4784R-square: 0.9605 r=0.98005

Age

yr

F

25.0 0.10

32.5 0.13

37.5 0.25

42.5 0.33

47.5 0.46

52.5 0.63

57.5 0.76

65.0 0.80


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7- Discuss your results.Answer:I tried to fit Eq.1 to Age and F data set using MATLAB to find out the constants, b & c. The fit

equation was then calculated as below:

General model:

F(Age) = 0.92/(1+exp(b-c*Age))Coefficients (with 95% confidence bounds):

b = 0.402

c = -9.724

Goodness of fit:SSE: 2.024

R-square: -2.834

The results show that this model is not capable of fitting this data set and cannot be considered as a

prediction model for this data. Converting the F values to linear values which has resulted the Z values

seems to have solved this problem. We can see from the results that a simple linear regression model witha very good correlation coefficient is capable to fit the data.

Using the b and c constant from the linear model, I plotted Eq.1 which appears to better fit the F data,

however it is still not a good prediction because for instance, at Age=65 the given value for F is 0.8 whilethe value from the prediction model is around 0.92!

ramin shamshiri abe6981 hw_03

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