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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), at-bats (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 at-bats (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%