Introduction to TV show Numb3rsHere are the important people:– Don (FBI) and Charlie (Prof.) Eppes– Their dad– Charlie’s colleagues

– Don’s agents

Determining a card’s color

00 11..55

Increasing Likelihood of Occurrence

Probability:

The eventis veryunlikelyto occur.

The eventis veryunlikelyto occur.

The occurrenceof the event isjust as likely asit is unlikely.

The occurrenceof the event isjust as likely asit is unlikely.

The eventis almostcertainto occur.

The eventis almostcertainto occur.

If you guess a card’s color (black or red), you have a 50% probability of being right

What if you guessed 25 times and got it wrong all the times?

Child Kidnappings56% of the children are found aliveIn 90% the parents are responsibleIt is far more likely to find the child alive

with help from the FBI– Let’s assume twice as likely

We want to know– a) the probability the parents will ask for help– b) the probability the parents are responsible

given they do not ask for help

Use conditional probability theory and Bayes’ Theorem

Child Kidnappings continued

We first want to know:

Let B denote the event of bringing the child back alive. Past history with these FBI cases indicate that with help the child was found alive in 95%:

Thus (using conditional probabilities):

P(B) = P(H)P(B|H) + P(HC)P(B|HC)

0.56 = P(H).95 + (1 – P(H)).475

P(H) = 0.18

H = parents ask for help from the FBI HC = parents do not ask for help from the FBI

H = parents ask for help from the FBI HC = parents do not ask for help from the FBI

a) P(H) = ?a) P(H) = ?

P(B|H) = .95P(B|H) = .95 P(B|HC) = .475P(B|HC) = .475

Child Kidnappings continuedNow we want to know:We apply Bayes’ theorem:

From past history we know that if one of the parents was responsible, they did not ask for help in 90% of the cases.

We revise the prior probabilities as follows:

b) P(R|HC) = ?b) P(R|HC) = ?

P(HC|R) = .90P(HC|R) = .90 P(H|R) = .10P(H|R) = .10

Shooting chainsGang related shootings result in shooting chains

of, on average, 2.8 shootings (i.e., the initial shooting, plus 1.8 additional shootings on average), with a standard deviation of 1.1.

Four shootings stand out, as they resulted in shooting chains of 4, 5, 6, and 7 shootings respectively.

Amita says: “Statistically that wouldn’t happen if he had chosen the victims at random”

Test this Hypothesis at the 1% level.

Shooting chains continued

Determine the hypotheses.Determine the hypotheses.

Specify the level of significance.Specify the level of significance. = .01

Compute the test statistic.Compute the test statistic.

Compute the Compute the pp –value. –value. P(z > 4.91) = .0000

Or: Determine the critical valueOr: Determine the critical value = .01, z.01 = 2.33

Determine whether to reject Determine whether to reject HH00..p-value: .0000 < 0.01Critical: 4.91 > 2.33

If the confidence interval contains the hypothesized value 0, do not reject H0. Otherwise, reject H0.

The 98% confidence interval for is

Because the hypothesized value for the population mean, 0 = 2.8, is not in this interval, the hypothesis-testing conclusion is that the null hypothesis (H0: = 2.8), can be rejected.

Shooting chains continued

Shooting chains continued

What if we do not actually know the standard deviation of the population of shooting chains?

We would have to use the standard deviation of the sample (4, 5, 6, and 7, leads to s = 1.3)

We would also have to use a t distribution:

For = .01 and d.f. = 3, t.01 = 4.54For t = 4.18, using n – 1 = 3 d.f., the p–value equals

0.012Now we cannot reject the null hypothesis of H0: = 2.8

at the 1% level!

Traffic accidentsThe FBI found that 5 of 13 traffic accident victims

were related to previous serious traffic accidents. If the population average is 40%, can we speak of a coincidence or not? Test with = 5%.

Note: in population proportions, we need to make sure that both np and n(1 – p) are greater than 5:13 x 0.4 = 5.2 > 5 & 13 x (1 – 0.4) = 7.8 > 5

Determine the hypotheses.Determine the hypotheses.

Specify the level of significance.Specify the level of significance. = .05

Traffic accidents continued

Compute the test statistic.Compute the test statistic.

Compute the Compute the pp –value. –value. P(z > -0.11) > 0.50

Or: Determine the critical valueOr: Determine the critical value = .05 z.95 = 1.64

Determine whether to reject Determine whether to reject HH00..p-value: .545 > 0.05Critical: -0.11 < 1.64

a commonerror is usingp in this formula

a commonerror is usingp in this formula

Traffic accidents continued

Compute the test statistic.Compute the test statistic.

Compute the Compute the pp –value. –value. P(z > 2.71) < 0.01

Or: Determine the critical valueOr: Determine the critical value = .05 z.95 = 1.64

Determine whether to reject Determine whether to reject HH00..p-value: .0033 < 0.05Critical: 2.71 > 1.64

Some more investigation leads to a further 5 victims who are related to serious traffic accidents. What can we say now?

Some other interesting video clips

Always go back to the DataGame showOn dating…Type I and Type II errorsBenford’s LawLogistic Regression

Some video clips on randomness

Randomness – random spacing is not equal

Randomness – raindrops on a sidewalk– Notice the conditional probability math

on the glass behind Charlie!

Randomness – shuffling cards