Presentation on theme: "Introduction to TV show Numb3rs Here are the important people: – Don (FBI) and Charlie (Prof.) Eppes – Their dad – Charlies colleagues – Dons agents."— Presentation transcript:
Introduction to TV show Numb3rs Here are the important people: – Don (FBI) and Charlie (Prof.) Eppes – Their dad – Charlies colleagues – Dons agents
Determining a cards color 0 1.5 Increasing Likelihood of Occurrence Probability: The event is very unlikely to occur. The event is very unlikely to occur. The occurrence of the event is just as likely as it is unlikely. The occurrence of the event is just as likely as it is unlikely. The event is almost certain to occur. The event is almost certain to occur. If you guess a cards 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 Kidnappings 56% of the children are found alive In 90% the parents are responsible It is far more likely to find the child alive with help from the FBI – Lets 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 ( H C ) P ( B|H C ) 0.56 = P ( H ).95 + (1 – P ( H )).475 P ( H ) = 0.18 H = parents ask for help from the FBI H C = parents do not ask for help from the FBI H = parents ask for help from the FBI H C = parents do not ask for help from the FBI a) P ( H ) = ? P ( B | H ) =.95 P ( B | H C ) =.475
Child Kidnappings continued Now 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 | H C ) = ? P ( H C | R ) =.90 P ( H | R ) =.10
Shooting chains Gang 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 wouldnt happen if he had chosen the victims at random Test this Hypothesis at the 1% level.
Shooting chains continued Determine the hypotheses. Specify the level of significance. =.01 Compute the test statistic. Compute the p –value. P (z > 4.91) =.0000 Or: Determine the critical value =.01, z.01 = 2.33 Determine whether to reject H 0. p-value :.0000 2.33
If the confidence interval contains the hypothesized value 0, do not reject H 0. Otherwise, reject H 0. 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 ( H 0 : = 2.8), can be rejected. 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 = 4.54 For t = 4.18, using n – 1 = 3 d.f., the p –value equals 0.012 Now we cannot reject the null hypothesis of H 0 : = 2.8 at the 1% level!
Traffic accidents The 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. Specify the level of significance. =.05
Traffic accidents continued Compute the test statistic. Compute the p –value. P ( z > -0.11) > 0.50 Or: Determine the critical value =.05 z.95 = 1.64 Determine whether to reject H 0. p-value :.545 > 0.05 Critical: -0.11 < 1.64 a common error is using p in this formula a common error is using p in this formula
Traffic accidents continued Compute the test statistic. Compute the p –value. P ( z > 2.71) < 0.01 Or: Determine the critical value =.05 z.95 = 1.64 Determine whether to reject H 0. p-value :.0033 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 Data Game show On dating… Type I and Type II errors Benfords Law Logistic 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