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Exercise 12 p.179 ... Jeter and Justice
1995  1996  

Jeter  12 H, 48 AB,

183 H, 582 AB,

Justice  104 H, 411 AB,

45 H, 140 AB,

Notations: hits (H), atbats (AB), batting average (AVG=H/AB). Which player had higher AVG in 1995, 1996, and over the period of two years?
Clearly, Justice had a higher AVG both in 1995 (253 vs 250) and in 1996 (321 vs 314).
Over the period of two years:
Jeter: 12+183=195 H, 48+582=630 AB, 195/630=0.309 ...
309 AVG
Justice: 104+45=149 H, 411+140=551 AB, 149/551=0.270 ...
270 AVG
Jeter has a higher AVG (309 vs 270)
Explanation: Jeter had a very small amount of atbats (48) in 1995. That almost does not influence his high AVG (314) from 1996.
Math SAT scores of high school students in 1988 and 1998
% Students  SAT Score  

Grade Average  1988  1998  1988  1998  Change 
A+  4  7  632  629 

A  11  15  586  582 

A−  13  16  556  554 

B  53  48  490  487 

C  19  14  431  428 

Overall average  504  514  +10 
While within every grade category the average dropped, the overall average has increased from 504 to 514 points.
That is because the fraction of higher grades is bigger in 1998 than in 1988.
Time Improvement with Weight Training  Time Improvement without Weight Training  Team Average Time Improvement  

Gazelles  10 s  2 s  6.0 s 
Cheetahs  9 s  1 s  6.2 s 
Explanation: more Cheetahs improved by 9s than Gazelles improved by 10s.
Here is a specific example which yields this outcome.
Out of 20 Gazelles, 10 improved by 10s, and 10 only by 2s
with an average improvement of \( \frac{10 \times 10+10 \times 2}{20} = \frac{120}{20}=6\) s
Out of 20 Cheetahs, 13 improved by 9s, and 7 only by 1s
with an average improvement of \( \frac{13 \times 9+7 \times 1}{20} = \frac{124}{20}=6.2\) s
Women  Men  

Drug A  5 of 100 cured  400 of 800 cured 
Drug B  101 of 900 cured  196 of 200 cured 
While A cured 5 out of 100, which is 5% of women, and 400 out of 800, which is 50% of men,
B cured 101 out of 900, which is 11.2% of women, and 196 out of 200, which is 98%
Thus B is more effective for both men and women.
Women  Men  

Drug A  5 of 100 cured  400 of 800 cured 
Drug B  101 of 900 cured  196 of 200 cured 
While A cured 5+400=405 out of 900 people, which is 45%,
B cured 101+196=297 out of 1100, which is 27% of the patients
Thus A is more effective for the patients.
Women  Men  

Drug A  5 of 100 cured  400 of 800 cured 
Drug B  101 of 900 cured  196 of 200 cured 
Drug B is better.
Drug B is more effective for both men and women.
Note that both drugs are much more effective for men than for women.
It is because drug B was tested to mostly on women, its overall performance looks poorer than A.
Tumor Is Malignant  Tumor Is Benign  Total  

Positive Mammogram  90  990  1080 
true positives  false positives  
Negative Mammogram  10  8910  8920 
false negatives  true negatives  
Total  100  9900  10,000 
Suppose that a patient has a positive mammogram. What is the chance that she really has cancer?
Solution. There are 1080 women with positive mammograms. Out of them, there are 90 with cancer. That is \( \frac{90}{1080}=0.083 \), that is 8.3%Tumor Is Malignant  Tumor Is Benign  Total  

Positive Mammogram  90  990  1080 
true positives  false positives  
Negative Mammogram  10  8910  8920 
false negatives  true negatives  
Total  100  9900  10,000 
c) What is the chance of positive mammogram given a patient has cancer?
Out of 100 women with cancer, 90 got a positive mammogram.
The chance in question is 90%
In fact, it was just assumed that the mammogram has an accuracy of 90%
Tumor Is Malignant  Tumor Is Benign  Total  

Positive Mammogram  90  990  1080 
true positives  false positives  
Negative Mammogram  10  8910  8920 
false negatives  true negatives  
Total  100  9900  10,000 
d) What is the chance for a patient with negative mammogram to have cancer?
Out of 8920 women with negative mammogram, 10 actually have cancer.
The chance in question is \( \frac{10}{8920}=0.0011 \), which is 0.11%