xyFeb8Lec10 - Depic&ng Sampling Distribu&ons...

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Unformatted text preview: Depic&ng Sampling Distribu&ons Lecture 10: 2/8/12 Lecture 9 1 Diversifica&on A basic principle of investment is: " diversifica&on reduces risk". That is, buying several securi&es, (for example stocks), rather than just one, reduces the variability of the return on an investment. The following figures display two distribu&ons of returns in 1987. 2/8/12 Lecture 9 2 Distribu&on of returns for all 1815 stocks on the NYSE for the en&re year 1987. The mean return was 3.5% and the distribu&on shows a very wide spread. 2/8/12 Lecture 9 3 Distribu&on of returns for all possible porWolios that invested equal amounts in each of 5 stocks in 1987. The mean is s&ll 3.5%, but the variability is much less. 2/8/12 Lecture 9 4 Averages vs. Individuals The investment example shows that Averages are less variable than individual observa&ons; Averages are more normal than individual observa&ons. (Claimed by checking the Normal QQplot) Why? 2/8/12 Lecture 9 5 Sampling Distribu&ons Sta$s$cal Inferences: Answer ques&ons about the characteris&cs of popula'ons, like making statements about popula&on parameters (e.g. and ) Need to backup the statement with probability es&mates of the likelihood of being right (e.g. 95% or 99% likelihood) Are based on Random Samples Sampling Distribu$ons Probability 2/8/12 Lecture 9 6 Sampling Distribu&on The Sampling Distribu$on of a sta$s$c is a mass or density func&on that characterizes all the possible values that the sta&s&c can assume in repeated random samples from a popula&on or process Examples: Depends on the popula&on distribu&on, and the sample size. Sample mean X : Sampling distribu&on of X Sample standard devia&on s: Sampling distribu&on of s IQR: Sampling distribu&on of IQR Lecture 9 7 2/8/12 Approxima&ng the Sampling Distribu&on Repeatedly select a large number (say 1000) of random samples of size n from a given popula&on Calculate the value of the sta&s&c (e.g. sample mean ) for each sample and forming a histogram. The histogram can be used to describe the values of the sta&s&c that are likely to occur in any random sample Examples of sample sta&s&cs: Sample mean X, sample median , sample standard devia&on s, IQR X 2/8/12 Lecture 9 8 Example Suppose we draw 1000 random samples from the popula&on*, each sample is of size n=10 * The Popula&on: a normal r.v. with , If we calculate the mean of each sample, how many sample means do we get? Plot the histogram of the sample means 2/8/12 Lecture 9 9 The sampling distribu&on of X for samples of size 10 compared with the distribu&on of a single observa&on. 2/8/12 Lecture 9 10 2/8/12 Lecture 9 11 Sampling Distribu&on of X Suppose Xis the sample mean based on a sample of size n from a popula&on with mean and s.d. , then - the mean of the sampling distribu&on of X (denoted by X ) = , regardless of the sample size; - the spread of X 's sampling distribu&on (denoted by X ) = n Note: When a popula&on distribu&on is normal, the sampling distribu&on of is also normal, regardless of the sample X size!!! 2/8/12 Lecture 9 12 Sampling Distribu&on of X 2/8/12 Lecture 9 13 The distribu&ons of X for (a) 1 obs. (b) 2 obs. (c) 10 obs. (d) 25 obs. 2/8/12 Lecture 9 14 Soda Pop The amount of soda pop in each bojle is normally distributed with a mean of 32.2 ounces and a standard devia&on of .3 ounces. Find the probability that a bojle bought by a customer will contain more than 32 ounces. Answer: Let X denote the amount of soda in a bojle. X - 32.2 32 - 32.2 P( X > 32) = P( > ) 0.3 0.3 = P( z > -.67 ) = 0.7486 2/8/12 Lecture 10 x = 32 = 32.2 15 Soda Pop Find the probability that a box of four bojles will have a mean of more than 32 ounces of soda per bojle. Answer: The random variable here is the mean amount of soda per bojle, which is normally distributed with a mean of 32.2 and standard devia&on of 0.3/sqrt(4)=0.15. Then, X - 32.2 32 - 32.2 P ( X > 32) = P ( > ) 0.15 0.15 = P ( Z > -1.33) = 0.9082 2/8/12 Lecture 10 16 Weekly Income The average weekly income of graduates one year amer gradua&on is $600. Suppose the distribu&on of weekly income has a standard devia&on (SD) of $100. What is the probability that 25 randomly selected graduates have an average weekly income of less than or equal to $550? Answer: According to CLT, X approximately has a normal distribu&on with mean 600 and s.d. 100/sqrt(25)=20. X - 600 550 - 600 P ( X <= 550) = P ( <= ) 20 20 = P ( Z <= -2.5) = 0.0062. Based on the above result, we can claim: If the average is $600, then the probability to have a sample mean of $550 is very low (0.0062). The fact that there exists such a sample suggests that the average weekly income $600 is probably unjus&fied. 2/8/12 Lecture 10 17 Amer Class... Review Sec 5.4 Read Sec 5.5 and 5.6 Hw#4, due by 5pm next Monday. 2/8/12 Lecture 9 18 ...
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