Extra Brownie Points! Lottery To Win: choose the 5 winnings numbers from 1 to 49 AND Choose the "Powerball" number from 1 to 42 What is the probability you will win? Use combinations to answer this question

p of winning jackpot Total number of ways to win / total number of possible outcomes

Total Number of Outcomes Different 5 number combinations Different Powerball outcomes Thus, there are 1,906,884 * 42 = 80,089,128 ways the drawing can occur

Total Number of Ways to Win Only one way to win –

Remember Playing perfect black jack – the probability of winning a hand is .498 What is the probability that you will win 8 of the next 10 games of blackjack?

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events p = .0429

Binomial Distribution What if you are interested in the probability of winning at least 8 games of black jack? To do this you need to know the distribution of these probabilities

Probability of Winning Blackjack Number of Wins p 1 2 3 4 5 6 7 8 9 10 p = .498, N = 10

Probability of Winning Blackjack Number of Wins p .001 1 2 3 4 5 6 7 8 9 10 p = .498, N = 10

Probability of Winning Blackjack Number of Wins p .001 1 .010 2 .045 3 .119 4 .207 5 .246 6 .203 7 .115 8 .044 9 .009 10 p = .498, N = 10

Probability of Winning Blackjack Number of Wins p .001 1 .010 2 .045 3 .119 4 .207 5 .246 6 .203 7 .115 8 .044 9 .009 10 1.00 p = .498, N = 10

Binomial Distribution p Games Won

Hypothesis Testing You wonder if winning at least 7 games of blackjack is significantly (.05) better than what would be expected due to chance. H1= Games won > 6 H0= Games won < or equal to 6 What is the probability of winning 7 or more games?

Binomial Distribution p Games Won

Binomial Distribution p Games Won

Probability of Winning Blackjack Number of Wins p .001 1 .010 2 .045 3 .119 4 .207 5 .246 6 .203 7 .115 8 .044 9 .009 10 1.00 p = .498, N = 10

Probability of Winning Blackjack Number of Wins p .001 1 .010 2 .045 3 .119 4 .207 5 .246 6 .203 7 .115 8 .044 9 .009 10 1.00 p = .498, N = 10 p of winning 7 or more games .115+.044+.009+.001 = .169 p > .05 Not better than chance

Practice The probability at winning the “Statistical Slot Machine” is .08. Create a distribution of probabilities when N = 10 Determine if winning at least 4 games of slots is significantly (.05) better than what would be expected due to chance.

Probability of Winning Slot Number of Wins p .434 1 .378 2 .148 3 .034 4 .005 5 .001 6 .000 7 8 9 10 1.00

Binomial Distribution p Games Won

Probability of Winning Slot Number of Wins p .434 1 .378 2 .148 3 .034 4 .005 5 .001 6 .000 7 8 9 10 1.00 p of winning at least 4 games .005+.001+.000 . . . .000 = .006 p< .05 Winning at least 4 games is significantly better than chance

Binomial Distribution These distributions can be described with means and SD. Mean = Np SD =

Binomial Distribution Black Jack; p = .498, N =10 M = 4.98 SD = 1.59

Binomial Distribution p Games Won

Binomial Distribution Statistical Slot Machine; p = .08, N = 10 M = .8 SD = .86

Binomial Distribution Note: as N gets bigger, distributions will approach normal p Games Won

Next Step You think someone is cheating at BLINGOO! p = .30 of winning You watch a person play 89 games of blingoo and wins 39 times (i.e., 44%). Is this significantly bigger than .30 to assume that he is cheating?

Hypothesis H1= .44 > .30 H0= .44 < or equal to .30 Or H1= 39 wins > 26.7 wins H0= 39 wins < or equal to 26.7 wins

Distribution Mean = 26.7 SD = 4.32 X = 39

Z-score

Results (39 – 26.7) / 4.32 = 2.85 p = .0021 p < .05 .44 is significantly bigger than .30. There is reason to believe the person is cheating! Or – 39 wins is significantly more than 26.7 wins (which are what is expected due to chance)

BLINGOO Competition You and your friend enter at competition with 2,642 other players p = .30 You win 57 of the 150 games and your friend won 39. Afterward you wonder how many people A) did better than you? B) did worse than you? C) won between 39 and 57 games You also wonder how many games you needed to win in order to be in the top 10%

Blingoo M = 45 SD = 5.61 A) did better than you? (57 – 45) / 5.61 = 2.14 p = .0162 2,642 * .0162 = 42.8 or 43 people

Blingoo M = 45 SD = 5.61 A) did worse than you? (57 – 45) / 5.61 = 2.14 p = .9838 2,642 * .9838 = 2,599.2 or 2,599 people

Blingoo M = 45 SD = 5.61 A) won between 39 and 57 games? (57 – 45) / 5.61 = 2.14 ; p = .4838 (39 – 45) / 5.61 = -1.07 ; p = .3577 .4838 + .3577 = .8415 2,642 * .8415 = 2,223.2 or 2, 223 people

Blingoo M = 45 SD = 5.61 You also wonder how many games you needed to win in order to be in the top 10% Z = 1.28 45 + 5.61 (1.28) = 52.18 games or 52 games

Is a persons’ size related to if they were bullied You gathered data from 209 children at Springfield Elementary School. Assessed: Height (short vs. not short) Bullied (yes vs. no)

Results Ever Bullied

Results Ever Bullied

Results Ever Bullied

Results Ever Bullied

Results Ever Bullied

Results Ever Bullied

Is this difference in proportion due to chance? To test this you use a Chi-Square (2) Notice you are using nominal data

Hypothesis H1: There is a relationship between the two variables i.e., a persons size is related to if they were bullied H0:The two variables are independent of each other i.e., there is no relationship between a persons size and if they were bullied

Logic 1) Calculate an observed Chi-square 2) Find a critical value 3) See if the the observed Chi-square falls in the critical area

Chi-Square O = observed frequency E = expected frequency

Results Ever Bullied

Observed Frequencies Ever Bullied

Expected frequencies Are how many observations you would expect in each cell if the null hypothesis was true i.e., there there was no relationship between a persons size and if they were bullied

Expected frequencies To calculate a cells expected frequency: For each cell you do this formula

Expected Frequencies Ever Bullied

Expected Frequencies Ever Bullied

Expected Frequencies Row total = 92 Ever Bullied

Expected Frequencies Row total = 92 Column total = 72 Ever Bullied

Expected Frequencies Ever Bullied Row total = 92 N = 209 Column total = 72 Ever Bullied

Expected Frequencies E = (92 * 72) /209 = 31.69 Ever Bullied

Expected Frequencies Ever Bullied

Expected Frequencies Ever Bullied

Expected Frequencies E = (92 * 137) /209 = 60.30 Ever Bullied

Expected Frequencies Ever Bullied E = (117 * 72) / 209 = 40.30

Expected Frequencies Ever Bullied The expected frequencies are what you would expect if there was no relationship between the two variables! Ever Bullied

How do the expected frequencies work? Looking only at: Ever Bullied

How do the expected frequencies work? If you randomly selected a person from these 209 people what is the probability you would select a person who is short? Ever Bullied

How do the expected frequencies work? If you randomly selected a person from these 209 people what is the probability you would select a person who is short? 92 / 209 = .44 Ever Bullied

How do the expected frequencies work? If you randomly selected a person from these 209 people what is the probability you would select a person who was bullied? Ever Bullied

How do the expected frequencies work? If you randomly selected a person from these 209 people what is the probability you would select a person who was bullied? 72 / 209 = .34 Ever Bullied

How do the expected frequencies work? If you randomly selected a person from these 209 people what is the probability you would select a person who was bullied and is short? Ever Bullied

How do the expected frequencies work? If you randomly selected a person from these 209 people what is the probability you would select a person who was bullied and is short? (.44) (.34) = .15 Ever Bullied

How do the expected frequencies work? How many people do you expect to have been bullied and short? Ever Bullied

How do the expected frequencies work? How many people would you expect to have been bullied and short? (.15 * 209) = 31.35 (difference due to rounding) Ever Bullied

Back to Chi-Square O = observed frequency E = expected frequency

2

2

2

2

2

2

2

Significance Is a 2 of 9.13 significant at the .05 level? To find out you need to know df

Degrees of Freedom To determine the degrees of freedom you use the number of rows (R) and the number of columns (C) DF = (R - 1)(C - 1)

Degrees of Freedom Rows = 2 Ever Bullied

Degrees of Freedom Rows = 2 Columns = 2 Ever Bullied

Degrees of Freedom To determine the degrees of freedom you use the number of rows (R) and the number of columns (C) df = (R - 1)(C - 1) df = (2 - 1)(2 - 1) = 1

Significance Look on page 736 df = 1  = .05 2critical = 3.84

Decision Thus, if 2 > than 2critical Reject H0, and accept H1 If 2 < or = to 2critical Fail to reject H0

Current Example 2 = 9.13 2critical = 3.84 Thus, reject H0, and accept H1

Current Example H1: There is a relationship between the the two variables A persons size is significantly (alpha = .05) related to if they were bullied