Mar. 3 Statistic for the day: Average number of pieces of

mail that end up in the dead letter box each year: 57,100,000

Assignment:Assignment:

Read Chapter 17Read Chapter 17

Exercises p. 309-311: 1, 4, 7, 10, 15Exercises p. 309-311: 1, 4, 7, 10, 15

These slides were created by Tom Hettmansperger and in some cases modified by David Hunter

Shuffle two decks of cards.

Stack the two decks side-by-side, face down Stack the two decks side-by-side, face down next to each other.next to each other.

One by one, flip over one card from each One by one, flip over one card from each deck.deck.

I bet I see at least one match. Do you want I bet I see at least one match. Do you want to bet against me?to bet against me?

Probability of no match: Probability of no match on 1Probability of no match on 1stst flip: 51/52 flip: 51/52 Probability of no match on 2Probability of no match on 2ndnd flip: 51/52 flip: 51/52 …… Probability of no match on 52Probability of no match on 52ndnd flip: 51/52 flip: 51/52

These events are NOT independent; however, they are APPROXIMATELY independent

because, say, whether a match occurs on the 36th flip doesn’t influence whether a match occurs on the 47th flip very strongly.

52Thus, Pr(no match) (51/52) .364

Efron Dice

333333D

662222C

555111B

444400A

Side value

4

4

4

4

0

0

555111

Die B

Die A

4

4

4

4

0

0

555111

Die A

Die B

Pr(B beats A) = (12 + 12)/36 = 24/36 = 2/3

5

5

5

1

1

1

662222

Die B

Die C

Pr(C beats B) = (18 + 6)/36 = 24/36 = 2/3

6

6

2

2

2

2

333333

Die D

Die C

Pr(D beats C) = 24/36 = 2/3

3

3

3

3

3

3

444400

Die A

Die D

Pr( A beats D ) = 24/36 = 2/3

333333D

662222C

555111B

444400A

Side value

Pr( B beats A ) = 2/3

Pr( C beats B ) = 2/3

Pr( D beats C ) = 2/3

Pr (A beats D ) = 2/3

Hence, there is NO best die! You can always pick a winner if you pick second.

Percent tables and count tables

A stratified population is one that is divided intomutually exclusive subgroups and the subgroupsexhaust all members of the population.

Cancer testing: confusion of the inverseSuppose we have a cancer test for a certain type of cancer.

Sensitivity of the test:If you have cancer then the probability of a positive testis .98. Pr(+ given you have C) = .98

Specificity of the test:If you do not have cancer then the probability of a negativetest is .95. Pr(- given you do not have C) = .95

Base rate:The percent of the population who has the cancer. This is the probability that someone has C.Suppose for our example it is 1%. Hence, Pr(C) = .01.

++

PositivePositive

--

NegativeNegative

CC

(Cancer)(Cancer)

.98.98 .02.02 .01.01

no Cno C

(no Cancer)(no Cancer)

.05.05 .95.95 .99.99

Sensitivity

Specificity

BaseRate

Percent table

Suppose you go in for a test and it comes back positive.What is the probability that you have cancer?

false positive false negative

Count table from a percent table

++ --

CC .98.98 .02.02 .01.01

no Cno C .05.05 .95.95 .99.99

++ --

CC 9898 22 100100

no Cno C 495495 94059405 9,9009,900

593593 94079407 10,00010,000

Pr(C given a + test) = 98/593 = .165

Do you have a tattoo?

What is the probability that a randomly chosenperson from the class will say yes?

Rows: Sex Columns: Tattoo No Yes All Female 105 31 136 Male 85 15 100 All 190 46 236

Need a count table to estimate the probabilities:

Rows: Sex Columns: Tattoo No Yes All Female 77.21 22.79 100.00 Male 85.00 15.00 100.00 All 80.51 19.49 100.00

Percent table:

Pr(yes) = 46/236 = .1949

Pr(yes given the person is a female) = .2279Pr(yes given the person is a male) = .1500

Are the events ‘yes’ and ‘female’ independent?

Pr(no given the person is female) = .7721

Pr(no given the person is a male) = .8500

Suppose I tell you that a stat100 student came into office hours and they said that they did not have a tattoo.

Which is more likely:

•The student was female.•The student was male.

Rows: Sex Columns: Tattoo No Yes All Female 105 31 136 Male 85 15 100 All 190 46 236

Pr(female given the student said no) = 105/190 = .553

Pr(male given the student said no) = 85/190 = .447

More likely that the student is a female!

Rows: Sex Columns: Tattoo No Yes All Female 105 31 136 Male 85 15 100 All 190 46 236

Pr(yes) = 46/236 = .195 Pr(no) = 190/236 = .805Pr(female) = 136/236 = .576Pr(male) = 100/236 = .424Pr(yes given the student is a female) = 31/136 = .228Pr(yes given the student is a male) = 15/100 = .150Pr(no given the student is a female) = 105/136 = .772Pr(no given the student is a male) = 85/100 = .850Pr(female given the student said yes) = 31/46 = .674Pr(male given the student said yes) = 15/46 = .326Pr(female given the student said no) = 105/190 = .553Pr(male given the student said no) = 85/190 = .447

The count table gives the ability to calculate everything. If you have a percent table, you should create a count table.

Rows: Sex Columns: Tattoo No Yes All Female 77.21 22.79 100.00 Male 85.00 15.00 100.00 All 80.51 19.49 100.00

Note: It’s not always possible to reconstruct a representative count table. In the above, you can’t do it unless you also know the percentage of females.

The count table gives the ability to calculate everything. If you have a percent table, you should create a count table.

Rows: Sex Columns: Tattoo No Yes All Female 77.21 22.79 100.00 Male 85.00 15.00 100.00 All 80.51 19.49 100.00

Also, 57.63% are females

NN YY

FF 44504450 13131313 57635763

MM 636636 36013601 42374237

10000 (arbitrary)10000 (arbitrary)

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