Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Week 7 Sample Means & Proportions Variability of Summary Statistics Variability in shape of distn of sample Variability in summary statistics Mean, median, st devn, upper quartile, … Summary statistics have distributions Parameters and statistics Parameter describes underlying population Summary statistic Constant Greek letter (e.g. , , , …) Unknown value in practice Random Roman letter (e.g. m, s, p, …) We hope statistic will tell us about corresponding parameter Distn of sample vs Sampling distn of statistic Values in a single random sample have a distribution Single sample --> single value for statistic Sample-to-sample variability of statistic is its sampling distribution. Means Unknown population mean, Sample mean, X, has a distribution — its sampling distribution. Usually x ≠ A single sample mean, x, gives us information about Sampling distribution of mean If sample size, n, increases: Spread of distn of sample is (approx) same. Spread of sampling distn of mean gets smaller. x is likely to be closer to x becomes a better estimate of Sampling distribution of mean Population with mean , st devn Random sample (n independent values) Sample mean, X, has sampling distn with: Mean, X St devn, X n (We will deal later with the problem that and are unknown in practice.) Weight loss Estimate mean weight loss for those attending clinic for 10 weeks Random sample of n = 25 people Sample mean, x How accurate? Let’s see, if the population distn of weight loss is: X ~ normal 8lb, 5lb Some samples Four random samples of n = 25 people: 1. Mean = 8.32 pounds, st devn = 4.74 pounds 2. Mean = 8.32 pounds, st devn = 4.74 pounds 3. Mean = 8.48 pounds, st devn = 5.27 pounds 4. Mean = 7.16 pounds, st devn = 5.93 pounds N.B. In all samples, x ≠ Sampling distribution Means from simulation of 400 samples Theory: mean = = 8 lb, s.d.( x ) = n 5 1 lb 25 (How does this compare to simulation? To popn distn?) Errors in estimation Population X ~ normal 8lb, 5lb Sampling distribution of mean mean = = 8 lb, n 5 1 lb 25 From 70-95-100 rule s.d.( x ) = x will be almost certainly within 8 ± 3 lb x is unlikely to be more than 3 lb in error Even if we didn’t know x is unlikely to be more than 3 lb in error Increasing sample size, n If we sample n = 100 people instead of 25: s.d.( x ) = 5 0.5 lb. n 100 Larger samples more accurate estimates Central Limit Theorem If population is normal (, ) X ~ normal, n If popn is non-normal with (, ) but n is large X approx ~ normal, n Guideline: n > 30 even if very non-normal Other summary statistics E.g. Lower quartile, proportion, correlation Usually not normal distns Formula for standard devn of samling distn sometimes Sampling distn usually close to normal if n is large Lottery problem Pennsylvania Cash 5 lottery 5 numbers selected from 1-39 Pick birthdays of family members (none 32-39) P(highest selected is 32 or over)? Statistic: H = highest of 5 random numbers (without replacement) Lottery simulation Theory? Fairly hard. Simulation: Generated 5 numbers (without replacement) 1560 times Highest number > 31 in about 72% of repetitions Normal distributions Family of distributions (populations) Shape depends only on parameters (mean) & (st devn) All have same symmetric ‘bell shape’ = 65 inches, = 2.7 inches Importance of normal distn A reasonable model for many data sets Transformed data often approx normal Sample means (and many other statistics) are approx normal. Standard normal distribution Z ~ Normal ( = 0, = 1) -3 Prob ( Z < z* ) -2 -1 0 1 2 3 Probabilities for normal (0, 1) Check from tables: P(Z -3.00) P(Z −2.59) P(Z 1.31) P(Z 2.00) P(Z -4.75) = = = = = 0.0013 0 .0048 0 .9049 0 .9772 0 .000001 Probability Z > 1.31 P(Z > 1.31) = 1 – P(Z 1.31) = 1 – .9049 = .0951 Prob ( Z between –2.59 and 1.31) P(-2.59 Z 1.31) = P(Z 1.31) – P(Z -2.59) = .9049 – .0048 = .9001 Standard devns from mean Normal (, ) Heights of students = 65 inches, = 2.7 inches Probability and area X ~ normal ( = 65 , = 2.7 ) P (X ≤ 67.7) = area Probability and area (cont.) Normal (, ) Exactly P(X within of ) = 0.683 P(X within 2 of ) = 0.954 P(X within 3 of ) = 0.997 70-95-100 rule approx 70% approx 95% approx 100% Finding approx probabilities Ht of college woman, X ~ normal ( = 65 , = 2.7 ) Prob (X ≤ 62 )? 1. Sketch normal density 2. Estimate area P (X ≤ 62) = area About 1/8 Translate question from X to Z X ~ Normal (, ) Find P(X ≤ x*) x* Translate to z-score: X Z Z ~ Normal ( = 0, = 1) -3 -2 z*-1 0 1 2 3 Finding probabilities Prob (height of randomly selected college woman ≤ 62 )? 62 65 P X 62 P Z 2.7 P Z 1.11 .1335 About 13%. Prob (X > value) Ht of college woman, X ~ normal ( = 65 , = 2.7 ) Prob (X > 68 inches)? 68 65 P X 68 P Z PZ 1.11 1 PZ 1.11 2.7 1 .8665 .1335 Finding upper quartile Blood Pressures are normal with mean 120 and standard deviation 10. What is the 75th percentile? Step 1: Solve for z-score Closest z* with area of 0.7500 (tables) z = 0.67 Step 2: Calculate x = z* + x = (0.67)(10) + 120 = 126.7 or about 127. Probabilities about means Blood pressure ~ normal ( = 120, = 10) 8 people given drug If drug does not affect blood pressure, Find P(average blood pressure > 130) P ( X > 130) ? X ~ normal ( = 120, = 10) n=8 10 X ~ normal X 120, X 3.54 8 130 120 z 2.83 3.54 prob = 0.0023 Very little chance! Distribution of sum X ~ distn with (, ) e.g. miles to kilometers aX ~ distn with (a, a) X ~ distn with , n X nX ~ distn with n, n Central Limit Theorem implies approx normal Probabilities about sum Profit in 1 day ~ normal (= $300, = $200) Prob(total profit in week < $1,000)? Total = X ~ normal 7 2,100, 1000 2100 z 2.08 529 Prob = 0.0188 7 529 Assumes independence Categorical data Most important parameter is = Prob (success) Corresponding summary statistic is p = Proportion (success) N.B. Textbook uses p and p^ Number of successes Easiest to deal with count of successes before proportion. If… 1. 2. 3. 4. n “trials” (fixed beforehand). Only “success” or “failure” possible for each trial. Outcomes are independent. Prob (success), remains same for all trials, . • Prob (failure) is 1 – . X = number of successes ~ binomial (n, ) Examples Binomial Probabilities nk n! k P X k 1 for k = 0, 1, 2, …, n k! n k ! You won’t need to use this!! Prob (win game) = 0.2 Plays of game are independent. What is Prob (wins 2 out of 3 games)? What is P(X = 2)? 32 3! 2 P X 2 .2 1 .2 2! 3 2! 3(.2)2 (.8)1 0.096 Mean & st devn of Binomial For a binomial (n, ) Mean n Standard deviation n 1 Extraterrestrial Life? 50% of large population would say “yes” if asked, “Do you believe there is extraterrestrial life?” Sample of n = 100 X = # “yes” ~ binomial (n = 100, = 0.5) Mean E X 100(.5) 50 Standard deviation 100(.5).5 5 Extraterrestrial Life? Sample of n = 100 X = # “yes” ~ binomial (n = 100, = 0.5) E X 100(.5) 50 100(.5).5 5 70-95-100 rule of thumb for # “yes” About 95% chance of between 40 & 60 Almost certainly between 35 & 65 Normal approx to binomial If X is binomial (n , ), and n is large, then X is also approximately normal, with Mean E X n Standard deviation n 1 Conditions: Both n and n(1 – ) are at least 10. (Justified by Central Limit Theorem) Number of H in 30 Flips X = # heads in n = 30 flips of fair coin X ~ binomial ( n = 30, = 0.5) Bell-shaped & approx normal. E X 30(.5) 15 30(.5).5 2.74 Opinion poll n = 500 adults; 240 agreed with statement If = 0.5 of all adults agree, what P(X ≤ 240) ? X is approx normal with E X 500(.5) 250 100(.5).5 11.2 240 250 P X 240 P Z P Z .89 .1867 11.2 Not unlikely to see 48% or less, even if 50% in population agree. Sample Proportion Suppose (unknown to us) 40% of a population carry the gene for a disease, ( = 0.40). Random sample of 25 people; X = # with gene. X ~ binomial (n = 25 , = 0.4) p = proportion with gene X p n Distn of sample proportion X ~ binomial (n , ) X n X n 1 X p n p p Large n: p is approx normal 1 n (n ≥ 10 & n (1 – ) ≥ 10) Examples Election Polls: to estimate proportion who favor a candidate; units = all voters. Television Ratings: to estimate proportion of households watching TV program; units = all households with TV. Consumer Preferences: to estimate proportion of consumers who prefer new recipe compared with old; units = all consumers. Testing ESP: to estimate probability a person can successfully guess which of 5 symbols on a hidden card; repeatable situation = a guess. Public opinion poll Suppose 40% of all voters favor Candidate A. Pollsters sample n = 2400 voters. Propn voting for A is approx normal p 0.4 p 1 n Simulation 400 times & theory. 0.4 0.6 0.01 2400 Probability from normal approx If 40% of voters favor Candidate A, and n = 2400 sampled p 0.4 p 0.01 Sample proportion, p, is almost certain to be between 0.37 and 0.43 Prob 0.95 of p being between 0.38 and 0.42