6/1/11 Lecture 15 1 STOR 155 Introductory Statistics Lecture 15: Sampling Distributions for Sample Means The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL6/1/11 Lecture 15 2 Review • Sampling distribution for a sample count: – Binomial experiments – Binomial distribution – Normal approximation – Continuity correction • Sampling distribution for a sample proportion: – Exact calculation via binomial distribution – Normal approximation No continuity correction • How about a sample mean?6/1/11 Lecture 15 3 Diversification • A basic principle of investment is that diversification reduces risk. • That is, buying several securities, (for example stocks), rather than just one, reduces the variability of the return on an investment. • The following figures show two distributions of returns in 1987.6/1/11 Lecture 15 4 • Distribution of returns for all 1815 stocks on the NYSE for the entire year 1987. • The mean return was –3.5% and the distribution shows a very wide spread.6/1/11 Lecture 15 5 • Distribution of returns for all possible portfolios that invested equal amounts in each of 5 stocks in 1987. • The mean is still –3.5%, but the variability is much less.6/1/11 Lecture 15 6 Averages vs. Individuals The investment example shows that • Averages are less variable than individual observations; • Averages are closer to ``normal’’ than individual observations. Why?6/1/11 Lecture 15 7 • The sampling distribution of for samples of size 10 compared with the distribution of a single observation. X6/1/11 Lecture 15 86/1/11 Lecture 15 9 Sampling Distribution of X6/1/11 Lecture 15 10 The distributions of for (a) 1 obs. (b) 2 obs. (c) 10 obs. (d) 25 obs. --- correction for ``skewness’’ X6/1/11 Lecture 15 11 • The amount of soda pop in each bottle is normally distributed with a mean of 32.2 ounces and a standard deviation of 0.3 ounces. • Find the probability that a bottle bought by a customer will contain more than 32 ounces. • Answer: Let X denote the amount of soda in a bottle. 7486.0)67.()3.02.32323.02.32()32(zPXPXPm = 32.2 x = 32 Soda Pop6/1/11 Lecture 15 12 • Find the probability that a box of four bottles will have a mean of more than 32 ounces of soda per bottle. • Answer: The random variable here is the mean amount of soda per bottle, which is normally distributed with a mean of 32.2 and standard deviation of 0.3 / 2 = 0.15. Hence 9082.0)33.1()15.02.323215.02.32()32(ZPXPXPSoda Pop6/1/11 Lecture 15 13 • The average weekly income of graduates one year after graduation is $600. Suppose the distribution of weekly income has a standard deviation (SD) of $100. • What is the probability that 25 randomly selected graduates have an average weekly income of less than $550? • Answer: According to CLT, approximately has a normal distribution with mean 600 and SD 100 / 5 =20. .0062.0)5.2()2060055020600()550(ZPXPXPWeekly Income X6/1/11 Lecture 15 14 • Suppose we actually found a random sample of 25 graduates, with an average weekly income of $550. • What can you say about the validity of the claim that the average weekly income is $600? • Answer: – If the population mean is $600, then the probability to have observed a sample mean of $550 is very low (0.0062). The evidence provided by the sample suggests that the assumed average weekly income $600 is unjustified. – It’s more reasonable to believe that the population mean is actually smaller than $600. Then a sample mean of $550 becomes more probable. Weekly Income (continued)Sum of independent normal random variables • Fact: If X , Y are independent normal random variables and a, b are constants, then aX+bY also follows a normal distribution with mean E(aX+bY) = a E(X) + b E(Y) and variance Use this fact to solve Problem 5.60 (c) (d) on page 349. 6/1/11 Lecture 15 15 22222YXbYaXba6/1/11 Lecture 15 16 Take Home Message • Sampling distribution for a sample mean – Mean for a sample mean – Variance for a sample mean – Central Limit Theorem – Normal distribution
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