Slide 1Survey TerminologySlide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 411 - 1Population Parameter and Sample StatisticsSampleSampleSample...Populationp = ?ˆpˆpˆpIf certain conditions hold, the distribution of sample proportions is1 - 2Conditions for Central Limit TheoremqThe sample is randomly selected and individuals within the sample are independent.qThe sample size n is such that np ≥ 10 and nq = n(1-p) ≥10.qThe population size N is such that N ≥ 10n. Equivalently, n ≤ (10%)*N.1 - 3Central Limit Theorem (General Version)Reading: Chapter 9, pg 388 – pg 395qIn general, Central Limit Theorem states that for a numerical variable X with mean µ and s.d. σ, the sampling distribution of X is a normal distribution with mean µ and standard error σ/√n, when sample size is large enough.qAs n → ∞, X ~ N(µ, σ/√n).qThis normal model specifies the distribution of all sample means.qIf σ is not known, use s, sample s.d.1 - 41 - 5Sampling Distribution Vs. Distribution of A SampleqDistribution of streak data from one sample (single simulation of 134 shots)qDistribution of sample means from 1000 samples.5101520253035stre ak0 2 4 6 8streak mean = 1.14286Measures from Sample of Hot HandHistogram0.050.100.150.200.250.30m ean_streak0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8mean_streak mean = 0.828029Measures from Measures from Sample of Hot HandHistogram1 - 6CLICKER QUESTION:1 - 7Sampling Distribution of Sample Means (When Population Distribution Is Normal)http://onlinestatbook.com/stat_sim/sampling_dist/index.html1 - 8Sampling Distribution of Sample Means (When Population Is Not Normal)http://onlinestatbook.com/stat_sim/sampling_dist/index.html1 - 9Central Limit TheoremWhen samples are drawn from a population that is (A) not normally distributed, the shape of the sampling distribution of sample means is:nearly normal when sample size n is large enough. (n>30)(B) nearly normal, the shape of the sampling distribution of sample means is:nearly normal.1 - 10Example: Evaluating Normal Probabilities Using Normal Model (CLT)1 - 111 - 12Example: Evaluating Normal Probabilities Using Normal Model (CLT)1 - 131 - 14Additional Exercises1 - 15Additional Exercises1 - 16Additional Exercises1 - 17Additional Exercises1 - 181441 - 19Connection Between CLT for Sample Mean and CLT for Sample ProportionWhen investigating a categorical variable, we can code its values into discrete numbers. In the case of success/failure scenario, we may assume that every subject in the population is holding a numerical value that is either 0 or 1. The population mean, i.e. µ = (sum of all values in population)/(population size) = (sum of 1's)/N, in this case, is exactly µ = population proportion = p.1 - 20CLT for Sample Proportion is a Special Case When All Values are BernoulliLikewise, a sample with replacement from such population extracts at random an independent collection of n 0's and 1's.The sample mean, i.e. X = (sum of all values in sample)/(population size) = (sum of 1's) / n, in this case, is exactly X = sample proportion = .So we agree on the expected value analogy, but how about standard error, σ/√n vs. √p(1-p)/√n?ˆp1 - 21Sampling From Population of 0's and 1's with replacement = BinomialqWhen a random sample of size n is chosen from a population containing Bernoulli-type values, we can model the result, X= # of 1's, using Binomial(n, p).qTherefore, the sampling distribution of many such sample proportions can be summarized to haveqE[X] = np, Var[X] = np(1-p), sd[X] = √np(1-p).qSample proportion = X/n = X, is .qE[X] = p, Var[X] = p(1-p)/n, sd[X] = √p(1-p)/n.qHere, n – sample size; p – population proportion.ˆp1 - 22Verifying Conditions for CLTqRestating the CLT that X ~ N(µ, σ/√n) yields the result ~ N(p, √p(1-p)/n)qThe sampling distribution of sample proportions is a normal model. qIt is often safe to assume that a sample is random and independent. But how about the normality of the underlying distribution?ˆp1 - 23Normal Approximation to BinomialqWhen a random sample of size n is chosen, with replacement, from a population containing Bernoulli-type values, we can model the result,qLet X = # of 1's, X ~ Binomial(n, p).qThe Success/Failure condition provides a sufficient sample size so that a normal model can be assumed even when p is not close to 0.5qSince = X/n ~ N(p, √p(1-p)/n) = N(µ, σ/√n),we have X ~ N(n*p, √np(1-p)) = N(E[X],SD[X]).ˆp1 - 24Example: Normal ApproximationqIn other words, when the sample size is large enough (so that np and n(1-p) are at least 10), we may use the normal model to approximate binomial probabilities.qExample: A manufacturer of LCD screens has a defective rate of 1.4%. During a routine inspection, a sample of 1500 was taken with replacement. Find the probability that the number of defects is no more than 15.qLet X = # of defective items, X ~ Bin(1500, 0.014).qCheck: n*p = 1500*0.014 = 21.1 - 25Example: Normal ApproximationqNow, instead of using the binomial pdf formula to calculate qP(X<=15) = P(X=0) + P(X=1) + … + P(X=15),qWe calculate instead, as if X ~ N(E[X],SD[X]).qE[X] = np = 21,qSD[X] = √np(1-p) = 4.55qThus, the corresponding Z-score is (15-21)/4.55=-1.32qHence, P(X<=15) can be approximated by P(Z<-1.32), which is ...1 - 261 - 27Example: Two Models for the Same Problem7.41 (pg330) A true/false test has 40 questions. A passing grade is 60% or more correct answers. What is the probability that a person who is just randomly guessing will pass the test?Solution 1: Use CLT for sample proportion.We know that ~ N(p, √p(1-p)/n), where p = 0.5.CLT applies since np = 20 = n(1-p) >= 10. It is reasonable to assume that the answers are independent if one is truly guessing at random.Hence, P( >= .60) = P(Z >= 1.26) = .1038ˆpˆpˆp1 - 28Example: Two Models for the Same ProblemSolution 2: Use binomial model and normal approximation.Let X = # of correctly guessed items.Therefore, X ~ Bin(n = 40, p = 0.5). It makes sense to assume that each answer is independent.Passing the test in this case means that X is 24 or larger.Hence, P(X >= 24) = P(X = 24) + P(X = 25) + … + P(X = 40)where P(X = k) = 40Ck(.5)k(0.5)40-k.CLT applies