VIEWS: 0 PAGES: 16 POSTED ON: 3/1/2012 Public Domain
Chapter 7 Sampling Distribution I. Basic Definitions II. Sampling Distribution of Sample Mean III. Sampling Distribution of Sample Proportion IV. Odds and Ends I. Basic Definitions • Population and sample (N presidents and n presidents) • Parameter and statistic p.258&p.264 Based on all (N) possible values of X: , and p Based on n values of X: X , S and p • Statistic as a point estimator for parameter • Sample statistic becomes a new random variable • Sampling distribution: probability distribution of a statistic II. Sampling Distribution of Sample Mean X 1. Summary measures p.270 E( X ) x x x x - If N is infinite or N >> n p.271 n - n x If X is normally distributed, X is normally distributed. 2. Central Limit Theorem p.272 For any distribution of population, if sample size is large, the sampling distribution of sample mean is approximately a normal distribution. 3. Applications X P( X ) P( X ) X Example. P.278 #18 A population has a mean of 200 and a standard deviation of 50. A simple random sample of size 100 will be taken and the sample mean will be used to estimate the population mean. a. What is the expected value of X ? b. What is the standard deviation of X ? c. Show the sampling distribution of X ? d. What does the sampling distribution of X show? Answer. x 50 a. E ( X ) x = 200 b. x 5 n 100 c. X N(200, 5) d. Probability distribution of X 3. Applications (1) X P( X ) Procedure: X Z P(Z) P( X ) = P(Z) x x x z x x x x n Example. A population has a mean of 100 and a standard deviation of 16. What is the probability that a sample mean will be within 2 of the population mean for each of the following sample sizes? a. n = 50 b. n = 200 c. What is the advantage of larger sample size? Solution. Given x = 100, x = 16 a. n = 50, find P (98 X 102 ) x x 98 100 98 100 102 z1 .88 x 16 / 50 z1 0 z2 x x 102 100 z2 .88 x 16 / 50 From Z-Table: When z = -.88, the left tail is .1894. P(98 X 102 ) P(.88 Z .88 ) 1 (2)(. 1894 ) .6212 b. n = 200, find P (98 X 102 ) x x 98 100 z1 1.77 x 16 / 200 z2 = 1.77. Use Z-Table: P(98 X 102 ) 1 (2)(. 0384 ) .9232 c. Larger sample size results in a smaller standard error - a higher probability that the sample mean will be within 2 of x (2) P( X ) X Procedure: P ( X ) P(Z) Z X x Zx Example. The diameter of ping-pong balls is approximately normally distributed with a mean of 1.30 inches and a standard deviation of 0.04 inches. a. What is the probability that a randomly selected ping-pong ball will have a diameter between 1.28 and 1.30 inches? b. If a random sample of 16 balls are selected, 60% of the sample means would be between what two values that are symmetrically around population mean. Answer: Random variable X: diameter of a ping-pong ball. a. P(1.28 X 1.3) = .1915 z = (1.28-1.3)/.04 = .5 1.28 1.3 .5-.3085=.1915 z=-0.5 0 b. P( x1 X x2 ) .6 From Z table: When the left tail is .2005, x1 1.3 x2 z1 = -.84 z2 = .84 .2 .04 x1 1.3 (.84 )( ) 1.2916 z1 0 z2 16 .04 x2 1.3 (. 84 )( ) 1.3084 16 Homework: Sample Mean p.278 #18, #19 p.278 #25 III. Sampling Distribution of Sample Proportion p 1. Summary measures E ( p) p p.280 p (1 p ) p - If N is infinite or N >> n p.280 n - n p If n and p satisfy the rule of five, p is approximately normally distributed. P.282 2. Applications p P( p ) P( p ) p 2. Applications (1) p P( p ) Procedure: p Z P(Z) P( p ) = P(Z) p p p(1 p) z p p p p n Example. Assume that 15% of the items produced in an assembly line operation are defective, but that the firm’s production manager is not aware of this situation. Assume further that 50 parts are tested by the quality assurance department to determine the quality of the assembly operation. Let p be the sample proportion found defective by the quality assurance test. a. Show the sampling distribution for p. b. What is the probability that the sample proportion will be within .03 of the population proportion that is defective? c. If the test show p = .10 or more, the assembly line operation will be shut down to check for the cause of the defects. What is the probability that the sample of 50 parts will lead to the conclusion that the assembly line should be shut down? Answer. a. p ~ N ( p , p ) p p .15 p (1 p ) (. 15 )(1 .15 ) p .0505 n 50 b. P(.12 p .18) .12 .15 .12 .15 .18 z1 .59 .2776 .0505 .18 .15 z1 0 z2 z2 .59 .0505 P(.12 p .18) 1 (2)(.2776) .4448 c. P( p .10) .10 .15 z .99 .0505 P( p .10) 1 .1611 .8389 .10 .15 .1611 z 0 Homework: Sample Proportion p.283 #31 p.284 #35 p.285 #40 Reading: Compare wordings in Chapter 6 homework problems, homework problems for sample mean and homework problems for sample proportion: X? or X ? or p? IV. Odds and Ends 1. Three properties of a good estimator p.286&p.287 ˆ (1) Unbiasedness E ( ) Mean of a sample statistic is equal to the population parameter. ( E ( X ) , E ( p ) p ) (2) Efficiency The estimator with smaller standard deviation. ( X for , p for p ) (3) Consistency If n ˆ (The values of sample statistic tend to become closer to the population parameter. For example, x p (1 p ) x p n n ) 2. Sampling Methods (p.260) and fpc (finite population correction factor) p.271&p.280 (1) Infinite population, or finite population sampling with replacement, or n/N 0.05: x p (1 p ) x p n n (2) Finite population sampling without replacement and n/N > 0.05: N n fpc (fpc 1. If N >> n, fpc 1) N 1 x p (1 p ) x fpc p fpc n n