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Sampling Distributions Parameter & Statistic Parameter • Summary measure about population Sample Statistic • Summary measure about sample • P in Population & Parameter • S in Sample & Statistic Common Statistics & Parameters Sample Statistic Population Parameter Mean X Standard Deviation S Variance S2 2 Binomial Proportion ^ p p Sampling Distribution 1. Theoretical probability distribution 2. Random variable is sample statistic • Sample mean, sample proportion, etc. 3. Results from drawing all possible samples of a fixed size 4. List of all possible [x, p(x)] pairs • Sampling distribution of the sample mean Sampling from Normal Populations 定理1 if x ~ N ( , ) , 若 Y a bx , then Y ~ N ( a b , b ) 2 2 平均數 E (Y ) E ( a bx ) E ( a ) E ( bx ) a bE ( x ) a b 變異數 V (Y ) V ( a bx ) V ( a ) V ( bx ) 0 b V ( x ) b 2 2 2 2 定理2 if X ~ N ( x , x ) , Y ~ N ( Y , Y ) , X and Y are independent 2 2 if w aX bY , then w ~ N ( a X b Y , a X b Y ) 2 2 2 2 定理：Let Y1,Y2,…,Yn be a random sample of size n from a normal distribution with mean μand varianceσ2. Then 1 n Y Yi n i 1 is normally distribution with mean Y And variance Y 2 2 / n Proof: E (Yi ) and V (Yi ) Y 1 n Y n i 1 i 1 n (Y1 ) a1Y1 a 2 Y2 1 n 2 for i 1, 2 ,..., n (Y 2 ) 1 n (Y n ) a nYn w here a i 1 / n , i 1, 2, ..., n 1 1 E (Y ) E (Y1 ) (Y 2 ) n n 1 n ( ) 1 n ( ) (Y n ) n 1 1 n ( ) 1 1 V (Y ) V (Y1 ) (Y 2 ) n n 1 n 2 1 n 2 ( ) 2 1 n ( n ) 2 ( ) 2 2 (Y n ) n 1 1 n 2 n 2 ( ) 2 Properties of the Sampling Distribution of x Standard Error of the Mean 1. Standard deviation of all possible sample means, x ● Measures scatter in all sample means, x 2. Less than population standard deviation 3. Formula (sampling with replacement) x n Sampling from Normal Populations Central Tendency x Population Distribution = 10 Dispersion x n = 50 X Sampling Distribution Sampling with replacement n = 4 X = 5 n =16 X = 2.5 X- = 50 X Standardizing the Sampling Distribution of x Sampling Distribution X x X Z x n Standardized Normal Distribution X =1 X X =0 Z Thinking Challenge You’re an operations analyst for AT&T. Long-distance telephone calls are normally distribution with = 8 min. and = 2 min. If you select random samples of 25 calls, what percentage of the sample means would be between 7.8 & 8.2 minutes? © 1984-1994 T/Maker Co. Sampling Distribution Solution* X Sampling Distribution 7.8 8 Z .50 2 25 n X 8.2 8 Z .50 2 25 n X = .4 Standardized Normal Distribution =1 .3830 .1915 .1915 7.8 8 8.2 X –.50 0 .50 Z Sampling from Non-Normal Populations Developing Sampling Distributions Suppose There’s a Population ... Population Random Values size, N = 4 variable, x of x: 1, 2, 3, 4 Uniform distribution © 1984-1994 T/Maker Co. Population Characteristics Summary Measures Population Distribution N Xi i 1 .3 .2 .1 .0 2 .5 N N X i i 1 N P(x) x 1 2 1.12 2 3 4 All Possible Samples of Size n = 2 16 Samples 16 Sample Means 1st 2nd Observation Obs 1 2 3 4 1st 2nd Observation Obs 1 2 3 4 1 1,1 1,2 1,3 1,4 1 1.0 1.5 2.0 2.5 2 2,1 2,2 2,3 2,4 2 1.5 2.0 2.5 3.0 3 3,1 3,2 3,3 3,4 3 2.0 2.5 3.0 3.5 4 4,1 4,2 4,3 4,4 4 2.5 3.0 3.5 4.0 Sample with replacement Sampling Distribution of All Sample Means 16 Sample Means 1st 2nd Observation Obs 1 2 3 4 1 1.0 1.5 2.0 2.5 2 1.5 2.0 2.5 3.0 3 2.0 2.5 3.0 3.5 4 2.5 3.0 3.5 4.0 Sampling Distribution of the Sample Mean P(x) .3 .2 .1 .0 x 1.0 1.5 2.0 2.5 3.0 3.5 4.0 Summary Measures of All Sample Means N X Xi i 1 1 .0 1 .5 ... 4 .0 N 16 N X X i X 2 i 1 N (1.0 2.5) (1.5 2.5) ... (4.0 2.5) 2 2 .5 2 16 2 .79 Comparison Population .3 .2 .1 .0 Sampling Distribution P(x) x 1 2 2.5 1.12 3 4 P(x) .3 .2 .1 .0 x 1.0 1.5 2.0 2.5 3.0 3.5 4.0 x 2.5 x .79 Sampling Distribution of the Mean… A fair die is thrown infinitely many times, with the random variable X = # of spots on any throw. The probability distribution of X is: x 1 P(x) 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 …and the mean and variance are calculated as well: 9.25 Sampling Distribution of Two Dice A sampling distribution is created by looking at all samples of size n=2 (i.e. two dice) and their means… While there are 36 possible samples of size 2, there are only 11 values for , and some (e.g. =3.5) occur more frequently than others (e.g. =1). 9.26 Sampling Distribution of Two Dice… 6/36 P( ) 1.5 2/36 2.0 3/36 2.5 4/36 3.0 5/36 3.5 6/36 4.0 5/36 4.5 4/36 5.0 3/36 5.5 2/36 6.0 1/36 5/36 ) 1/36 4/36 P( 1.0 3/36 2/36 1.0 1/36 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 9.27 Compare… Compare the distribution of X… 1 2 3 4 5 6 1.0 1.5 …with the sampling distribution of 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 . As well, note that: 9.28 Sampling from Non-Normal Populations Law of Large Numbers The law of large numbers states that, under general conditions, will be near with very high probability when n is large. The conditions for the law of large numbers are Yi , i=1, …, n, are i.i.d. The variance of Yi , , is finite. Central Limit Theorem… The sampling distribution of the mean of a random sample drawn from any population is approximately normal for a sufficiently large sample size. The larger the sample size, the more closely the sampling distribution of X will resemble a normal distribution. 9.32 Central Limit Theorem… If the population is normal, then X is normally distributed for all values of n. If the population is non-normal, then X is approximately normal only for larger values of n. In most practical situations, a sample size of 30 may be sufficiently large to allow us to use the normal distribution as an approximation for the sampling distribution of X. 9.33 Sampling from Non-Normal Populations Central Tendency x Population Distribution s = 10 Dispersion x n m = 50 X Sampling Distribution Sampling with replacement n = 4 X=5 n =30 X = 1.8 mX- = 50 X Central Limit Theorem As sample size gets large enough (n 30) ... x n x sampling distribution becomes almost normal. X Central Limit Theorem Example The amount of soda in cans of a particular brand has a mean of 12 oz and a standard deviation of .2 oz. If you select random samples of 50 cans, what percentage of the sample means would be less than 11.95 oz? SODA Central Limit Theorem Solution* X 11.95 12 Z 1.77 .2 n 50 Sampling Distribution Standardized Normal Distribution X = .03 .0384 =1 .4616 11.95 12 X –1.77 0 Shaded area exaggerated Z Example One survey interviewed 25 people who graduated one year ago and determines their weekly salary. The sample mean to be $750. To interpret the finding one needs to calculate the probability that a sample of 25 graduates would have a mean of $750 or less when the population mean is $800 and the standard deviation is $100. After calculating the probability, he needs to draw some conclusions. 9.40 Example We want to find the probability that the sample mean is less than $750. Thus, we seek X P ( X 750 ) The distribution of X, the weekly income, is likely to be positively skewed, but not sufficiently so to make the distribution of nonnormal. As a result, we may assume that X is normal with mean x 800 and standard deviation x / n 100 / 25 20 9.41 Example Thus, P ( X 750 ) X x 750 800 P 20 x P ( Z 2 .5 ) . 5 . 4938 . 0062 The probability of observing a sample mean as low as $750 when the population mean is $800 is extremely small. Because this event is quite unlikely, we would have to conclude that the dean's claim is not justified. 9.42 Using the Sampling Distribution for Inference Here’s another way of expressing the probability calculated from a sampling distribution. P(-1.96 < Z < 1.96) = .95 Substituting the formula for the sampling distribution P ( 1 . 96 X / 1 . 96 ) . 95 n With a little algebra P ( 1 . 96 n X 1 . 96 ) . 95 n 9.43 Using the Sampling Distribution for Inference Returning to the chapter-opening example where µ = 800, σ = 100, and n = 25, we compute P ( 800 1 . 96 100 X 800 1 . 96 25 100 ) . 95 25 or P ( 760 . 8 X 839 . 2 ) . 95 This tells us that there is a 95% probability that a sample mean will fall between 760.8 and 839.2. Because the sample mean was computed to be $750, we would have to conclude that the dean's claim is not supported by the statistic. 9.44 Using the Sampling Distribution for Inference For example, with µ = 800, σ = 100, n = 25 and α= .01, we produce P ( z .005 P (800 2 . 575 n X z .005 100 ) 1 . 01 n X 800 2 . 575 25 100 ) . 99 25 P ( 748 . 5 X 851 . 5 ) . 99 9.45 Sampling Distributions The Proportion The proportion of the population having some characteristic is denoted π. p X n number of items in the sample having sample size the characteri stic of interest Sampling Distributions The Proportion Standard error for the proportion: (1 ) σp n Z value for the proportion: Z p σp p (1 ) n Sampling Distributions The Proportion: Example If the true proportion of voters who support Proposition A is π = .4, what is the probability that a sample of size 200 yields a sample proportion between .40 and .45? In other words, if π = .4 and n = 200, what is P(.40 ≤ p ≤ .45) ? Sampling Distributions The Proportion: Example Find σ p : Convert to standardize d normal: σp (1 ) n .4(1 .4) .03464 200 .45 .40 .40 .40 P(.40 p .45) P Z .03464 .03464 P(0 Z 1.44) Sampling Distributions The Proportion: Example Use cumulative normal table: P(0 ≤ Z ≤ 1.44) = P(Z ≤ 1.44) – 0.5 = .4251 Standardized Normal Distribution Sampling Distribution .4251 Standardize .40 .45 p 0 1.44 Z