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ECO220Y Sampling Distributions: Central Limit Theorem Readings: Section 9.1 (pages 296-301, 305-307), 9.3, 9.4 Fall 2010 Lecture 18 (Fall 2010) Sampling Distributions Lecture 18 1 / 23 Sampling Distribution of Sample Mean The sample mean X̄ is a random variable that is determined by the particular items picked for the sample. If X̄ is a random variable, we can find it’s distribution and parameters of that distribution (we did it analytically in Lecture 17). We know that population has distribution with mean µX and variance σX2 . Sample is a sum of random variables, each of which is distributed with µX and σX2 . The sample mean, X̄ , is the sum of these random variables, divided n by the sample size n, or X̄ = X1 +X2 +...+X . n (Fall 2010) Sampling Distributions Lecture 18 2 / 23 Parameters of Sampling Distribution of Sample Mean Use Laws of Expectation and Variance to find parameters of sampling distribution of sample mean: h i n E [X̄ ] = E X1 +X2 +...+X = n1 E [X1 + X2 + . . . + Xn ] = n 1 n (E [X1 ] + E [X2 ] + . . . + E [Xn ]) = n1 (µx + µx + . . . + µx ) = n1 nµx = µx V [X̄ ] = V 1 n2 h X1 +X2 +...+Xn n i = (V [X1 ] + V [X2 ] + . . . + V [Xn ]) = 1 V [X1 n2 + X2 + . . . + Xn ] = 1 (σX2 +σX2 +. . .+σX2 ) n2 = 1 nσX2 n2 = σX2 n What must be true about the relationship between Xi0 s? (Fall 2010) Sampling Distributions Lecture 18 3 / 23 Parameters of Sampling Distribution of Sample Mean Sample mean X̄ is a random variable. Distributed with parameters: 1 Mean: µX̄ = µX 2 Variance: σX̄2 = σX2 n 3 St. dev.: σX̄ = σX √ n Standard deviation of the sampling distribution is called standard error. (Fall 2010) Sampling Distributions Lecture 18 4 / 23 Shape of Sampling Distribution of Sample Mean Use Central Limit Theorem to learn about the shape of sampling distribution of sample mean. Central Limit Theorem (CLT) implies that, no matter what the underlying distribution, the sample mean will tend to a normal distribution. CLT states that the sum of n independent, identically distributed random variables approaches a normal distribution as n increases. Recall that the sample mean is just a sum of n independent, identically distributed random variables divided by n. −→ Conditions σx for the CLT are satisfied −→ X̄ ∼ N(µx , √ ) n (Fall 2010) Sampling Distributions Lecture 18 5 / 23 Sufficiently Large Sample Size Rule of thumb: Sample size at least 30 (n ≥ 30) Normal approximation improves as n rises n < 30 sufficient for modest departures from normal If population is normal, then n ≥ 1 is sufficient (Fall 2010) Sampling Distributions Lecture 18 6 / 23 Which one is a sampling distribution? (Fall 2010) Sampling Distributions Lecture 18 7 / 23 What does sample size affect? (Fall 2010) Sampling Distributions Lecture 18 8 / 23 What happens to the standard error as sample size goes up? (Fall 2010) Sampling Distributions Lecture 18 9 / 23 (Fall 2010) Sampling Distributions Lecture 18 10 / 23 (Fall 2010) Sampling Distributions Lecture 18 11 / 23 Example North American adults watch TV on average 6 hours per day with a standard deviation of 1.5 hours. If we randomly sample 5 adults, what is the chance that on average they watch TV more than 7 hours a day? Notation: P(X̄ > 7kµx = 6, σx = 1.5, n = 5) =? Is it statistically possible that the average time watching television by a random sample of 16 adults is less than 4.5 hours? Notation: P(X̄ < 5kµx = 6, σx = 1.5, n = 16) =? (Fall 2010) Sampling Distributions Lecture 18 12 / 23 TV Example µx̄ = µx = 6 (a) σx̄ = σx √ n = 1.5 √ 5 (b) σx̄ = σx √ n = 1.5 √ 16 (Fall 2010) = 0.67 = 0.375 Sampling Distributions Lecture 18 13 / 23 TV Example P(X̄ > 7) = P =P Z > 7−6 0.67 X̄ −µx̄ σx̄ > 7−µx̄ σx̄ = P(Z > 1.49) = 1 − P(Z < 1.49) = 1 − 0.9319 = 0.0681 (Fall 2010) Sampling Distributions Lecture 18 14 / 23 TV Example P(X̄ < 4.5) = P =P Z < 4.5−6 0.375 X̄ −µx̄ σx̄ < 4.5−µx̄ σx̄ = P(Z < −4) Do we need table to find P(Z < −4)? (Fall 2010) Sampling Distributions Lecture 18 15 / 23 Comparing Means of Two Populations Suppose we want to compare the average salary of male and female employees in the same job in the same industry. Population one is the women, and population two is the men. Population one has mean salary µ1 and standard deviation σ1 . Population two has mean salary µ2 and standard deviation σ2 . A random sample of size n1 from population one and n2 from population two gives sample means of X¯1 and X¯2 and standard errors σX¯1 and σX¯2 We are interested in µ2 − µ1 = 0? (Fall 2010) Sampling Distributions Lecture 18 16 / 23 Comparing Means of Two Populations We need to know the sampling distribution of difference between two sample means. Sample means are random variables, difference between two sample means is a linear combination of two random variables. A linear combination of independent normal random variables yields a normal variable. We can standardize the difference! (Fall 2010) Sampling Distributions Lecture 18 17 / 23 Mean and Variance of Sampling Distribution of Difference σ22 σ12 X̄2 ∼ N µ2 , n2 X̄1 ∼ N µ1 , n1 E [X̄1 − X̄2 ] = E [X̄1 ] − E [X̄2 ] = µ1 − µ2 V [X̄1 − X̄2 ] = 12 V [X̄1 ] + (−1)2 V [X̄2 ] = q 2 σ X̄1 − X̄2 ∼ N µ1 − µ2 , n11 + σ22 n2 σ12 n1 + σ22 n2 What is assumed about the relationship between two sample means? (Fall 2010) Sampling Distributions Lecture 18 18 / 23 (Fall 2010) Sampling Distributions Lecture 18 19 / 23 Example The manager of a restaurant believes that waitresses who introduce themselves by telling customers their names will get larger tips than those who don’t. Specifically, she claims that the average tip for the former is 18% whereas that for the later is only 15% with a standard deviation 3%. What is the chance that the average tip in a random sample of 10 waitresses who introduce themselves will exceed that of the random sample of 10 waitresses who don’t? Notation: P X¯1 − X¯2 > 0 =? (Fall 2010) Sampling Distributions Lecture 18 20 / 23 Example Mean = E [X̄1 − X̄2 ] = 18 − 15 = 3 Standard Error = q σ12 n1 + σ22 n2 = q 9 10 9 10 + = 1.34 X̄1 − X̄2 ∼ N(3, 1.34) P(X̄1 − X̄2 > 0) = P Z > 0−3 1.34 = P(Z > −2.24) = 1 − P(Z < −2.24) = 1 − 0.0125 = 0.9875 (Fall 2010) Sampling Distributions Lecture 18 21 / 23 Example (Fall 2010) Sampling Distributions Lecture 18 22 / 23 Example What is the chance that in a random sample of those who do not introduce themselves, the average tip will exceed that of those who tell the customers their names? X̄1 − X̄2 ∼ N(3, 1.34) P(X̄1 − X̄2 < 0) =? P(X̄1 − X̄2 < 0) = P Z < (Fall 2010) 0−3 1.34 = P(Z < −2.24) = 0.0125 Sampling Distributions Lecture 18 23 / 23