Lecture 10: Depicting Sampling Distributions of a Sample Proportion Chapter 5: Probability and Sampling Distributions 2/10/12 Lecture 10 1 Sample Proportion •  “1” is assigned to population members having a specified characteristic and “0” is assigned to those who don’t. The parameter of interest in this situation is p (or called π), the proportion of the population that has the characteristic of interest. •  Denote pˆ as the proportion of members having such characteristic (or say, successes), also called the sample proportion, in a random sample of size n. 2/10/12 Lecture 10 2 Sampling Distribution of Sample Proportion •  If X ~ B(n, p), the sample proportion is defined as X count of successes in sample pˆ = = . n size of sample •  Mean & variance of a sample proportion: µ pˆ = p, 2/10/12 σ pˆ = p(1 − p) / n . Lecture 10 3 2/10/12 Lecture 10 4 Example: Clinton's vote •  43% of the population voted for Clinton in 1992. •  Suppose we survey a sample of size 2300 and see if they voted for Clinton or not in 1992. •  We are interested in the sampling distribution of the sample proportion pˆ , for samples of size 2300. •  What's the mean and variance of pˆ ? 2/10/12 Lecture 10 5 Count & Proportion of “Success” •  A BoilerMaker basketball player is a 95% freethrow shooter. •  Suppose he will shoot 5 free-throws during each practice. •  X: number of free-throws he makes during practice. •  pˆ : proportion of made-free-throws during practice. •  P( pˆ =0.6) = ? •  P(X=3) = ? 2/10/12 Lecture 10 6 Normal Approximation for Counts and Proportions •  Let X ~ B(n, p) and pˆ = X / n is the sample proportion. •  If n is large*, then X is approx. N (np , np (1-p) ) pˆ is approx. N ( p, p (1-p) / n ). •  *Rule of Thumb: np ≥ 5, n(1 - p) ≥ 5. 2/10/12 Lecture 10 7 Switches Inspection •  A quality engineer selects an SRS of 100 switches from a large shipment for detailed inspection. •  Unknown to the engineer, 10% of the switches in the shipment fail to meet the specifications. •  What is the probability that at most 9 switches fail the standard test in the sample? •  Use Normal Approximation here. Try to use continuity correction as well. (Similar to ex. 43 in Hw4) 2/10/12 Lecture 10 8 Switches Inspection •  Express the probability in terms of X: P(X ≤ 9) •  The normal approximation to the probability of no more than 9 bad switches is the area to the left of X = 9 under the normal curve, which has µ X = np = (100)(.1) = 10, σ X = np(1 − p) = 100(.1)(.9) = 3. •  In this case np = 10, and n(1-p) = 90, both satisfying the condition for “rule of thumb”. •  Also, don’t forget the CONTINUITY CORRECTION when applying normal approximation to the Binomial distribution. 2/10/12 Lecture 10 9 Continuity Correction •  The normal approximation is more accurate if we consider X = 9 to extend from 8.5 to 9.5 •  Example (Cont.): X − 10 9.5 − 10 P ( X ≤ 9) = P ( X ≤ 9.5) = P ( ≤ ) 3 3 ≈ P( Z ≤ −.17) = .4325. 2/10/12 Lecture 10 10 Continuity Correction P(X ≤ 8) replaced by P(X < 8.5) P(X ≥ 14) replaced by P(X > 13.5) P(X < 8) = P(X<=7), then replaced by P(X < 7.5) For large n the effects of the continuity correction factor is very small and will be omitted. 2/10/12 Lecture 10 11 Coin Tossing Example •  Toss a fair coin 200 times, what is the probability that the total number of heads is between 90 and 110? Use the normal approximation and continuity correction for Binomial distribution to solve the problem. •  X= the total number of heads •  X ~ Bin(200, 0.5). Want: P(90 ≤ X ≤ 110). •  µX =200 × .5 = 100, σX = (200× .5× .5)1/2 = 7.07. •  With continuity correction: P(90 ≤ X ≤ 110) = P(89.5 ≤ X ≤ 110.5)   ≈ P( Z ≤ (110.5 - 100)/7.07) - P (Z ≤ (89 .5 - 100)/7.07)   = P (Z ≤ 1.48) - P (Z ≤ -1.48)= .8611. 2/10/12 Lecture 10 12 Example Note: If not required to use Binomial knowledge, we can use the sampling distribution for sample proportion as well. •  The Laurier company’s brand has a market share of 30%. In a survey 1000 consumers were asked which brand they prefer. What is the probability that more than 32% of the respondents say they prefer the Laurier brand? •  Solution: The number of respondents who prefer Laurier is binomial with n = 1000 and p = .30. Also, np = 1000(.3) >5, n(1-p) = 1000(1-.3) > 5. Therefore we can use normal approximation to sample proportion here: ⎛ ⎞ ˆ p − p . 32 − . 30 ⎟ = P( Z > 1.38) = .0838. P( pˆ > .32) = P⎜ > ⎜ p(1 − p) n .01449 ⎟ ⎝ ⎠ 2/10/12 Lecture 10 13 Summary •  Sampling Distribution •  Sampling distribution for a sample mean –  Mean and Variance/S.D. for a sample mean –  Central Limit Theorem –  Normal calculation •  Sampling distribution for a sample proportion –  Continuity correction for Binomial Distribution –  Rule of thumb, when n is large 2/10/12 Lecture 10 14 After Class… •  Review Sec 5.4 through 5.6 . •  Start Hw 4 – due by next Monday, Feb 14th •  Review all the notes up to today, go over the practice test. Come to Office Hours for Hw#1-#4 grading issues. •  About Exam 1. –  Start editing your own cheat-sheet –  Practice Test is posted. Try to do it in an hour 2/10/12 Lecture 10 15