STAT 155 Introductory Statistics. Lecture 15 &16: Sampling Distributions for Counts and Proportions

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1 The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL STAT 155 Introductory Statistics Lecture 15 &16: Sampling Distributions for Counts and Proportions 10/26/06 Lecture 15&16 1

2 Statistic & Its Sampling Distribution A statistic is any numeric measure calculated from data. It is a random variable, its value varies from sample to sample. count/proportion: Ex: number/proportion of free throws made by a Tar Heel player who shoots 20 free throws in a practice sample mean Ex: average SAT score of a group of 10 students randomly selected from STAT 155 The probability distribution of a statistic is called its sampling distribution. It depends on the population distribution, and the sample size. 10/26/06 Lecture 15&16 2

3 Binomial Experiment n trials (with n fixed in advance). Each trial has two possible outcomes, success (S) and failure (F). The probability of success, p, remains the same from one trial to the next. The trials are independent, i.e. the outcome of each trial does not affect outcomes of other trials. 10/26/06 Lecture 15&16 3

4 Example The experiment: randomly draw n balls with replacement from an urn containing 10 red balls and 20 black balls. Let S represent {drawing a red ball} and F represent {drawing a black ball}. Then this is a binomial experiment with p =1/3. Q: Would it still be a binomial experiment if the balls were drawn without replacement? No! 10/26/06 Lecture 15&16 4

5 Binomial Distribution 10/26/06 Lecture 15&16 5

6 Do they follow binomial distributions (approximately)? X = number of stocks on the NY stock exchange whose prices increase today X = number of games the Tar Heel will win next season A couple decides to have children until they have a girl. X = number of boys the couple will have Answer: NO in all 3 cases. Why? 10/26/06 Lecture 15&16 6

7 Binomial Distribution If X ~ B(n, p), then µ X = np, σ 2 X = np(1-p). P(X=x) depends on n and p, which can be calculated using software or Table C (for some n and p), or a Binomial Formula (page 349) --- a simple argument given in class 10/26/06 Lecture 15&16 7

8 Binomial Table: for n 20, and certain values of p. Table C: Page T-6 10/26/06 Lecture 15&16 8

9 Credit Card Example Records show that 5% of the customers in a shoe store make their payments using a credit card. This morning 8 customers purchased shoes. Use the binomial table to answer the following questions. 1. Find the probability that exactly 6 customers did not use a credit card. X: number of customers who did not use a credit card. Then X ~ B(8, 0.95), which is not on the table. Y: number of customers who did use a credit card. Then Y ~ B(8, 0.05), which is on the table. P(X= 6) = P(Y = 2) = What is the probability that at least 3 customers used a credit card? (See the board ) 10/26/06 Lecture 15&16 9

10 Credit Card Example (continued) 3. What is the expected number of customers who used a credit card? µ Y = np = 8(0.05) = What is the standard deviation of the number of customers who used a credit card? σ 2 Y = np (1 p) = 8(0.05)(0.95) = The standard deviation is σy = 0.38 = /26/06 Lecture 15&16 10

11 Parking Example (bad impact?) Sarah drives to work everyday, but does not own a parking permit. She decides to take her chances and risk getting a parking ticket each day. Suppose A parking permit for a week (5 days) cost $ 30. A parking fine costs $ 50. The probability of getting a parking ticket each day is 0.1. Her chances of getting a ticket each day is independent of other days. She can get only 1 ticket per day. What is her probability of getting at least 1 parking ticket in one week (5 days)? What is the expected number of parking tickets that Sarah will get per week? Is she better off paying the parking permit in the long run? 10/26/06 Lecture 15&16 11

12 Sample Proportion If X ~ B(n, p), the sample proportion is defined as X pˆ = = n # of successes sample size mean & variance of a sample proportion:. µ pˆ = p σ, pˆ = p(1 p) / n. 10/26/06 Lecture 15&16 12

13 Example: Clinton's vote 43% of the population voted for Clinton in Suppose we survey a sample of size 2300 and see if they voted for Clinton or not in We are interested in the sampling distribution of the sample proportion pˆ, for samples of size What's the mean and variance of pˆ? 10/26/06 Lecture 15&16 13

14 Count & Proportion of Success A Tar Heel basketball player is a 95% free throw shooter. Suppose he will shoot 5 free throws during each practice. X: number of free throws he makes in a practice. pˆ : proportion of free throws made during practice. pˆ P(X=3) = P( =0.6). Why? 10/26/06 Lecture 15&16 14

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16 Normal Approximation for Counts and Proportions Let X ~ B(n, p) and If n is large, then p ˆ = X / n. X is approx. N ( np, np(1-p)) pˆ is approx. N ( p, p(1-p) / n ). Rule of Thumb: np 10, n(1 - p) /26/06 Lecture 15&16 16

17 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. Software tells us that the actual probability that no more than 9 of the switches in the sample fail inspection is P(X 9) = What will the normal approximation say? 10/26/06 Lecture 15&16 17

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19 Switches Inspection 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. µ X = np = ( 100)(.1) = 10, σ = np(1 p) = 100(.1)(.9) = 3. Using Table A, we have X P( X 9) = P( ) = P( Z 0.33) = 3 3 X The approximation.3707 to the binomial probability of is not very accurate. In this case np = /26/06 Lecture 15&16 19

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21 Continuity Correction The normal approximation is more accurate if we consider X = 9 to extend from 8.5 to 9.5, X = 10 to extend from 9.5 to 10.5, and so on. Example (Cont.): P( X 9) = P( X P( Z X ) = P( ) = ) 3 10/26/06 Lecture 15&16 21

22 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) replaced by P(X < 7.5) For large n the effects of the continuity correction factor is very small and will be omitted. 10/26/06 Lecture 15&16 22

23 Coin Tossing Example Toss a fair coin 200 times, what is the probability that the total number of heads is between 90 and 110? X = the total number of heads X ~ B(200, 0.5). Want: P(90 X 110). µ X = = 100, σ X = ( ) 1/2 = With continuity correction: P(90 X 110) = P(X 110) - P(X 89) P( Z ( )/7.07) - P (Z ( )/7.07) = P (Z 1.48) - P (Z -1.48)= /26/06 Lecture 15&16 23

24 Normal Approximation for Sample Proportions Let X ~ B(n, p) and If n is large, then p ˆ = X / n. pˆ is approx. N ( p, p(1-p) / n ). Rule of Thumb: np 10, n(1 - p) /26/06 Lecture 15&16 24

25 Example 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 =0.3. Also, np = 1000(0.3) > 10, n(1-p) = 1000(1-0.3) > 10. ˆ ( ˆ.32) p p P p > = P > = P( Z > 1.38) = (1 ) p p n /26/06 Lecture 15&16 25

26 Take Home Message Sampling distribution Binomial experiments Binomial distribution for count X: using Table C or the formula Sample proportion Normal approximation Continuity correction (for count X, not for proportion X/n) 10/26/06 Lecture 15&16 26