Course Notes - Department of Mathematics and Statistics

2.( nP r (X = x) = πk)x (1 − π) n−x( ) 20P r (X = 15) = 0.75 15 (1 − 0.75) 20−151520!=15! (20 − 15)! 0.7515 (1 − 0.75) 5= 0.2023Or just use RcmdrDistributions > Discrete Distributions > BinomialDistribution > Binomial Probabilities3. P r (X ≤ 11) = P r (X = 0, 1, 2, . . . , 11) can be replaced by P r (Y ≥ 9) =P r (Y = 20, 19, 18, . . . , 9) . Just use RcmdrDistributions > Discrete Distributions > BinomialDistribution > Binomial tail Probabilities = 0.04104. If π = 0.75 is assumed for Caucasian families, Pr(X ≤ 11) is verysmall (0.0410 is less than 0.05). This gives us evidence that theprobability is less than 75% for NZ Caucasian children. We rejectthe claim that π = 0.75 for Caucasian families can conclude fewerlive with both parents. (because 11 is in the direction of fewerrather than more)Note that instead if 12 out of 20 of the NZ Caucasian childrenwere living with both parents there would now be no evidence fromour data to suppose the situation is any different among Caucasianfamilies. Pr(X ≤ 12) = 0.1010 which is not small.Wait, What?• Where do the conclusions come from?• The p-value, the probability that an event will occur given a setn and π.66