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Chapter 4 Section 4.5 Sampling Distribution Models of Sample Means and the Central Limit Theorem Another Population Parameter of Frequent Interest: the Population Mean µ To estimate the unknown value of µ, the sample mean x is often used. We need to examine the Sampling Distribution of the Sample Mean x (the probability distribution of all possible values of x based on a sample of size n). Example Professor Stickler has a large statistics class of over 300 students. He asked them the ages of their cars and obtained the following probability distribution: x 2 3 4 5 6 7 8 p(x) 1/14 1/14 2/14 2/14 2/14 3/14 3/14 SRS n=2 is to be drawn from pop. Find the sampling distribution of the sample mean x for samples of size n = 2. Solution 7 possible ages (ages 2 through 8) Total of 72=49 possible samples of size 2 All 49 possible samples with the corresponding sample means and probabilities are on the next slide All 49 possible samples of size n = 2 x p(x) 2 1/14 3 1/14 4 2/14 5 2/14 6 2/14 7 3/14 8 3/14 Sample 2,2 2,4 2,6 2,8 2,5 2,3 2,7 4,2 4,4 4,6 4,8 4,5 4,3 4,7 6,2 6,4 6,6 xbar 2 3 4 5 3.5 2.5 4.5 3 4 5 6 4.5 3.5 5.5 4 5 6 Prob 1 196 2 196 2 196 3 196 2 196 1 196 3 196 2 196 4 196 4 196 6 196 4 196 2 196 6 196 2 196 4 196 Sample 6,8 6,5 6,3 6,7 8,2 8,4 8,6 8,8 8,5 8,3 8,7 5,2 5,4 5,6 5,8 5,5 xbar 7 5.5 4.5 6.5 5 6 7 8 6.5 5.5 7.5 3.5 4.5 5.5 6.5 5 Prob 6 196 4 196 2 196 6 196 3 196 6 196 6 196 9 196 6 196 3 196 9 196 2 196 4 196 4 196 6 196 4 196 Sample 5,3 5,7 3,2 3,4 3,6 3,8 3,5 3,3 3,7 7,2 7,4 7,6 7,8 7,5 7,3 7,7 xbar 4 6 2.5 3.5 4.5 5.5 4 3 5 4.5 5.5 6.5 7.5 6 5 7 Prob 2 196 6 196 1 196 2 196 2 196 3 196 2 196 1 196 3 196 3 196 6 196 6 196 9 196 6 196 3 196 9 196 4 196 Population: ages of cars and their distribution Sample 2,2 2,4 2,6 2,8 2,5 2,3 2,7 4,2 4,4 4,6 4,8 4,5 4,3 4,7 6,2 6,4 6,6 xbar 2 3 4 5 3.5 2.5 4.5 3 4 5 6 4.5 3.5 5.5 4 5 6 Prob Probability Distribution of the Sample Mean Age of 2 Cars 1 2 2 3 2 1 3 2 4 4 6 4 2 6 2 4 4 196 196 196 196 196 196 196 196 196 196 196 196 196 196 196 196 196 Sample 6,8 6,5 6,3 6,7 8,2 8,4 8,6 8,8 8,5 8,3 8,7 5,2 5,4 5,6 5,8 5,5 xbar 7 5.5 4.5 6.5 5 6 7 8 6.5 5.5 7.5 3.5 4.5 5.5 6.5 5 Prob 6 4 2 6 3 6 6 9 6 3 9 2 4 4 6 4 196 196 196 196 196 196 196 196 196 196 196 196 196 196 196 196 Sample 5,3 5,7 3,2 3,4 3,6 3,8 3,5 3,3 3,7 7,2 7,4 7,6 7,8 7,5 7,3 7,7 xbar 4 6 2.5 3.5 4.5 5.5 4 3 5 4.5 5.5 6.5 7.5 6 5 7 Prob 2 6 1 2 2 3 2 1 3 3 6 6 9 6 3 9 196 196 196 196 196 196 196 196 196 196 196 196 196 196 196 196 Sample 2,2 2,4 2,6 2,8 2,5 2,3 2,7 4,2 4,4 4,6 4,8 4,5 4,3 4,7 6,2 6,4 6,6 xbar 2 3 4 5 3.5 2.5 4.5 3 4 5 6 4.5 3.5 5.5 4 5 6 1 196 Prob 2 196 2 196 3 196 2 196 1 196 3 196 2 196 4 196 4 196 6 196 4 196 2 196 6 196 2 196 4 196 4 196 Sample 6,8 6,5 6,3 6,7 8,2 8,4 8,6 8,8 8,5 8,3 8,7 5,2 5,4 5,6 5,8 5,5 xbar 7 5.5 4.5 6.5 5 6 7 8 6.5 5.5 7.5 3.5 4.5 5.5 6.5 5 6 196 Prob 4 196 2 196 6 196 3 196 6 196 6 196 9 196 6 196 3 196 9 196 2 196 4 196 4 196 6 196 4 196 Sample 5,3 5,7 3,2 3,4 3,6 3,8 3,5 3,3 3,7 7,2 7,4 7,6 7,8 7,5 7,3 7,7 xbar 4 6 2.5 3.5 4.5 5.5 4 3 5 4.5 5.5 6.5 7.5 6 5 7 2 196 Prob x 2 2.5 3 6 196 1 196 3.5 2 196 2 196 4 3 196 2 196 1 196 4.5 3 196 3 196 5 6 196 6 196 9 196 5.5 6 196 3 196 6 9 196 6.5 7 7.5 8 p(x) 1/196 2/196 5/196 8/196 12/196 18/196 24/196 26/196 28/196 24/196 21/196 18/196 9/196 Solution (cont.) Probability distribution of x: x 2 2.5 p(x) 1/196 2/196 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 5/196 8/196 12/196 18/196 24/196 26/196 28/196 24/196 21/196 18/196 1/196 This is the sampling distribution of x because it specifies the probability associated with each possible value of x From the sampling distribution above P(4 x 6) = p(4)+p(4.5)+p(5)+p(5.5)+p(6) = 12/196 + 18/196 + 24/196 + 26/196 + 28/196 = 108/196 Expected Value and Standard Deviation of the Sampling Distribution of x Example (cont.) Population probability dist. x 2 3 4 5 6 7 8 p(x) 1/14 1/14 2/14 2/14 2/14 3/14 3/14 Sampling dist. of x x p(x) 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 1/196 2/196 5/196 8/196 12/196 18/196 24/196 26/196 28/196 24/196 21/196 18/196 1/196 Population probability dist. x 2 3 4 5 6 7 8 p(x) 1/14 1/14 2/14 2/14 2/14 3/14 3/14 E(X)=2(1/14)+3(1/14)+4(2/14)+ … +8(3/14)=5.714 Sampling dist. of x Population mean E(X)= = 5.714 x 2 2.5 3 3.5 p(x) 1/196 2/196 5/196 8/196 4.5 4 5 5.5 6 6.5 7 7.5 8 12/196 18/196 24/196 26/196 28/196 24/196 21/196 18/196 1/196 E(X)=2(1/196)+2.5(2/196)+3(5/196)+3.5(8/196)+4(12/196)+4.5(18/196)+5(24/196) +5.5(26/196)+6(28/196)+6.5(24/196)+7(21/196)+7.5(18/196)+8(1/196) = 5.714 Mean of sampling distribution of x: E(X) = 5.714 Example (cont.) Population from which sample is selected: E ( X ) 2( 141 ) 3( 141 ) 4( 142 ) 8 143 5.714 2 Var ( X ) 3.4898 SD( X ) Var ( X ) 3.4898 1.8681 Sampling dist. of X : 1 2 E ( X ) 2( 196 ) 2.5( 196 ) 9 8( 196 ) 5.714 3.4898 Var ( X ) 2 2 Var ( X ) SD( X ) 1.8681 SD( X ) Var ( X ) 1.3209 2 2 2 Var ( X ) 1.7449 IMPORTANT Numerical Summaries of the Sampling Distribution of X are Related to the Numerical Summaries of the Population X from Which the Sample is Selected E ( X ) E ( X ) (the mean of the sampling distribution of X is always equal to the mean of the population from which the sample is selected) Var ( X ) Var ( X ) n Var ( X ) SD( X ) SD( X ) Var ( X ) n n the standard deviation of the sampling distribution of X is always equal to the standard deviation of the population from which the sample is selected, divided by the square root of the sample size n Sampling Distribution of the Sample Mean X: Example An example – A fair 6-sided die is thrown; let X represent the number of dots showing on the upper face. – The probability distribution Population mean : = E(X) = 1(1/6) +2(1/6) of X is x 1 2 3 4 5 6 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 + 3(1/6) +……… = 3.5. Population variance 2 2 =V(X) = (1-3.5)2(1/6)+ (2-3.5)2(1/6)+ ……… ………. = 2.92 Suppose we want to estimate from the mean x of a sample of size n = 2. What is the sampling distribution of x in this situation? Sample 1 2 3 4 5 6 7 8 9 10 11 12 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 Mean Sample Mean 1 13 3,1 2 1.5 14 3,2 2.5 2 15 3,3 3 2.5 16 3,4 3.5 3 17 3,5 4 3.5 18 3,6 4.5 1.5 19 4,1 2.5 2 20 4,2 3 2.5 21 4,3 3.5 3 22 4,4 4 3.5 23 4,5 4.5 4 24 4,6 5 Sample 25 26 27 28 29 30 31 32 33 34 35 36 Mean 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 3 3.5 4 4.5 5 5.5 3.5 4 4.5 5 5.5 6 Sample 1 2 3 4 5 6 7 8 9 10 11 12 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 Mean Sample Mean 1 13 3,1 2 1.5 14 3,2 2.5 2 15 3,3 3 2.5 16 3,4 3.5 3 17 3,5 4 3.5 18 3,6 4.5 1.5 19 4,1 2.5 2 20 4,2 3 2.5 21 4,3 3.5 3 22 4,4 4 3.5 23 4,5 4.5 4 24 4,6 5 Sample 25 26 27 28 29 30 31 32 33 34 35 36 Mean 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 3 3.5 4 4.5 5 5.5 3.5 4 4.5 5 5.5 6 Var ( X ) Note : E ( X ) E ( X ) and Var ( X ) 2 E( x) =1.0(1/36)+ 1.5(2/36)+….=3.5 6/36 5/36 V(X) = (1.0-3.5)2(1/36)+ (1.5-3.5)2(2/36)... = 1.46 4/36 3/36 2/36 1/36 1 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 x n5 E ( X ) 3.5 Var ( X ) .5833 ( Var ( X ) 5 n 10 ) E ( X ) 3.5 Var ( X ) .2917 ( Var ( X ) 10 n 25 ) E ( X ) 3.5 Var ( X ) .1167 ( 1 Var ( X ) 25 6 Notice that Var ( X ) is smaller 1 than Var(X). The larger the sample size the smaller is Var ( X ) . Therefore, x tends to fall closer to , as the sample size increases. 6 1 6 ) The variance of the sample mean is smaller than the variance of the population. Mean = 1.5 Mean = 2. Mean = 2.5 Population Let us take samples of two observations 1.5 2.5 22 3 1.5 2.5 22 1.5 2.5 1.5 2 2.5 1.5 2.5 Compare1.5 the variability population 2 of the 2.5 1.5 2.5 to the variability of 22the sample mean. 1.5 2.5 1.5 2.5 2 1.5 2.5 1.5 2 2.5 1.5 2 2.5 1.5 2 2.5 1 Also, Expected value of the population = (1 + 2 + 3)/3 = 2 Expected value of the sample mean = (1.5 + 2 + 2.5)/3 = 2 Properties of the Sampling Distribution of x 1. E ( x ) (the expected value of the sampling distribution of x = the expected value of the sampled population) SD( x) 2. SD( x ) n n where is the standard deviation of the population from which the sample is taken and n is the sample size. Unbiased l Confidence l Precision µ The central tendency is down the center BUS 350 - Topic 6.1 Handout 6.1, Page 1 6.1 - 14 Unbiased Biased µ Biased µ µ The central tendency is down the center BUS 350 - Topic 6.1 Handout 6.1, Page 2 6.1 - 15 Consequences 1. E ( x ) . This is why we use x to estimate an unknown population mean . The sampling dist. of x is "centered" at the parameter we are trying to estimate. 2. SD( x ) SD (nx ) ; the standard deviation of x is smaller than SD( x), the stand. dev. of the population from which the sample is taken. The values of x will cluster tightly around when n is large. A Billion Dollar Mistake “Conventional” wisdom: smaller schools better than larger schools Late 90’s, Gates Foundation, Annenberg Foundation, Carnegie Foundation Among the 50 top-scoring Pennsylvania elementary schools 6 (12%) were from the smallest 3% of the schools But …, they didn’t notice … Among the 50 lowest-scoring Pennsylvania elementary schools 9 (18%) were from the smallest 3% of the schools A Billion Dollar Mistake (cont.) Smaller schools have (by definition) smaller n’s. SD ( x ) When n is small, SD(x) = n is larger That is, the sampling distributions of small school mean scores have larger SD’s https://www.washingtonpost.com/news/ answer-sheet/wp/2014/06/09/howmuch-bill-gatess-disappointing-smallschools-effort-really-cost/ We Know More! We know 2 parameters of the sampling distribution of x : E(x) μ SD(x) SD(x) n The Central Limit Theorem tells us about the shape of the distribution of x when the sample size n is sufficiently large. THE CENTRAL LIMIT THEOREM The “World is Normal” Theorem But first,…Sampling Distribution of x- Normally Distributed Population Sampling distribution of x: N( , /10) n=10 /10 Population distribution: N( , ) Normal Populations Important Fact: If the population is normally distributed, then the sampling distribution of x is normally distributed for any sample size n. Previous slide Non-normal Populations What can we say about the shape of the sampling distribution of x when the population from which the sample is selected is not normal? Baseball Salaries 600 490 500 Frequency 400 300 200 100 53 102 72 35 21 26 17 8 10 0 Salary ($1,000's) 2 3 1 0 0 1 The Central Limit Theorem (for the sample mean x) If a random sample of n observations is selected from a population (any population), then when n is sufficiently large, the sampling distribution of x will be approximately normal. (The larger the sample size, the better will be the normal approximation to the sampling distribution of x.) The Importance of the Central Limit Theorem When we select simple random samples of size n, the sample means x will vary from sample to sample. We can model the distribution of these sample means with a probability model that is … N , n How Large Should n Be? For the purpose of applying the Central Limit Theorem, we will consider a sample size to be large when n > 30. Baseball Salaries 600 Frequency ← Even if the population from ← which the sample is ← selected looks like this … 490 500 400 300 200 100 53 102 72 35 21 26 17 8 10 2 3 1 0 0 1 0 Salary ($1,000's) … the Central Limit → Theorem tells us that a → good model for the sampling → distribution of the sample mean x is … Summary Population: mean ; stand dev. ; shape of population dist. is unknown; value of is unknown; select random sample of size n; Sampling distribution of x: mean ; stand. dev. /n; always true! By the Central Limit Theorem: the shape of the sampling distribution is approx normal, that is x ~ N(, /n) The Central Limit Theorem (for the sample proportion p ) If x “successes” occur in a random sample of n observations selected from a population (any population), then when n is sufficiently large, the sampling distribution of p =x/n will be approximately normal. (The larger the sample size, the better will be the normal approximation to the sampling distribution of p.) The Importance of the Central Limit Theorem When we select simple random samples of size n from a population with “success” probability p and observe x “successes”, the sample proportions p =x/n will vary from sample to sample. We can model the distribution of these sample proportions with a probability model that is… p(1 p) N p, n How Large Should n Be? For the purpose of applying the central limit theorem, we will consider a sample size n to be large when np ≥ 10 and n(1-p) ≥ 10 Population, "success" proportion = p 0.7 p __ 0.6 p 0.5 0.4 0.3 1-p 0.2 0.1 0 0 1 … the Central Limit → Theorem tells us that a → good model for the sampling → distribution of the sample x proportion pˆ n is … ← If the population from ← which the sample is ← selected looks like this … Population Parameters and Sample Statistics Population parameter Value Sample statistic used to estimate p proportion of population with a certain characteristic Unknown p̂ µ mean value of a population variable Unknown x The value of a population parameter is a fixed number, it is NOT random; its value is not known. The value of a sample statistic is calculated from sample data The value of a sample statistic will vary from sample to sample (sampling distributions) Example A random sample of n =64 observations is drawn from a population with mean =15 and standard deviation =4. SD( X ) 4 a. E ( X ) 15; SD( X ) 0.5 8 n b. The shape of the sampling distribution model for x is approx. normal (by the CLT) with mean E(X) 15 and SD( X ) 0.5 (The answer depends on the sample size n since SD( X ) SD ( X ) n 4 64 84 0.5) Example (cont.) c. x 15.5; z x SD ( X ) 15.5.515 .5.5 1 This means that x =15.5 is one standard deviation above the mean E ( X ) 15 Example 2 The probability distribution of 6-month incomes of account executives has mean $20,000 and standard deviation $5,000. a) A single executive’s income is $20,000. Can it be said that this executive’s income exceeds 50% of all account executive incomes? ANSWER No. P(X<$20,000)=? No information given about shape of distribution of X; we do not know the median of 6-month incomes. Example 2(cont.) b) n=64 account executives are randomly selected. What is the probability that the sample mean exceeds $20,500? answer E(X) = $20, 000 SD(X) = $5, 000 E ( X ) $20, 000 SD ( X ) SD ( x ) n 5,000 64 625 By CLT, X ~ N (20, 000, 625) P ( X 20, 500) P X 20,000 625 20,500 20,000 625 P ( z .8) 1 .7881 .2119 Example 3 A sample of size n=16 is drawn from a normally distributed population with E(X)=20 and SD(X)=8. X ~ N (20, 8); X ~ N (20, 816 ) a ) P ( X 24) P ( X 220 24 2 20 ) P ( z 2) 1 .9772 .0228 b) P (16 X 24) P 16 220 z 24 2 20 P ( 2 z 2) .9772 .0228 .9544 Example 3 (cont.) c. Do we need the Central Limit Theorem to solve part a or part b? NO. We are given that the population is normal, so the sampling distribution of the mean will also be normal for any sample size n. The CLT is not needed. Example 4 Battery life X~N(20, 10). Guarantee: avg. battery life in a case of 24 exceeds 16 hrs. Find the probability that a randomly selected case meets the guarantee. E ( x ) 20; SD( x ) 10 P ( X 16) P( 2.04 X 20 .1 .0250 .9750 24 2.04. X ~ N (20, 2.04) 16 20 2.04 ) P( z 1.96) Example 5 Cans of salmon are supposed to have a net weight of 6 oz. The canner says that the net weight is a random variable with mean =6.05 oz. and stand. dev. =.18 oz. Suppose you take a random sample of 36 cans and calculate the sample mean weight to be 5.97 oz. Find the probability that the mean weight of the sample is less than or equal to 5.97 oz. Population X: amount of salmon in a can E(x)=6.05 oz, SD(x) = .18 oz X sampling dist: E(x)=6.05 SD(x)=.18/6=.03 By the CLT, X sampling dist is approx. normal P(X 5.97) = P(z [5.97-6.05]/.03) =P(z -.08/.03)=P(z -2.67)= .0038 How could you use this answer? Suppose you work for a “consumer watchdog” group If you sampled the weights of 36 cans and obtained a sample mean x 5.97 oz., what would you think? Since P( x 5.97) = .0038, either – you observed a “rare” event (recall: 5.97 oz is 2.67 stand. dev. below the mean) and the mean fill E(x) is in fact 6.05 oz. (the value claimed by the canner) – the true mean fill is less than 6.05 oz., (the canner is lying ). Example 6 X: weekly income. E(X)=1050, SD(X) = 100 n=64; X sampling dist: E(X)=1050 SD(X)=100/8 =12.5 P(X 1022)=P(z [1022-1050]/12.5) =P(z -28/12.5)=P(z -2.24) = .0125 Suspicious of claim that average is $1050; evidence is that average income is less. Example 7 12% of students at NCSU are left-handed. What is the probability that in a sample of 100 students, the sample proportion that are lefthanded is less than 11%? .12*.88 ˆ ˆ E ( p) p .12; SD( p) .032 100 np 100 .12 12 10; n(1 p) 100 .88 88 10; So By the CLT, pˆ ~ N (.12,.032) Example 7 (cont.) pˆ .12 .11 .12 ˆ P( p .11) P .032 .032 P( z .31) .3783 P ( pˆ .11) .3783 p̂ pˆ .11 P( z .31) .3783 z .31