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Ch 9 實習 Copyright © 2009 Cengage Learning Sampling Distributions… A sampling distribution is created by, as the name suggests, sampling. The method we will employ on the rules of probability and the laws of expected value and variance to derive the sampling distribution. For example, consider the roll of one and two dice… Copyright © 2009 Cengage Learning 9.2 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 P(x) 1 2 3 4 5 6 1/6 1/6 1/6 1/6 1/6 1/6 …and the mean and variance are calculated as well: Copyright © 2009 Cengage Learning 9.3 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). Copyright © 2009 Cengage Learning 9.4 Sampling Distribution of Two Dice… The sampling distribution of 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 5/36 ) 6/36 4/36 P( 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 P( ) 3/36 2/36 1/36 1.0 Copyright © 2009 Cengage Learning is shown below: 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 9.5 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: Copyright © 2009 Cengage Learning 9.6 Generalize… We can generalize the mean and variance of the sampling of two dice: …to n-dice: The standard deviation of the sampling distribution is called the standard error: Copyright © 2009 Cengage Learning 9.7 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. Copyright © 2009 Cengage Learning 9.8 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. Copyright © 2009 Cengage Learning 9.9 Sampling Distribution of the Sample Mean 1. 2. 3. If X is normal, X is normal. If X is nonnormal, X is approximately normal for sufficiently large sample sizes. Note: the definition of “sufficiently large” depends on the extent of nonnormality of x (e.g. heavily skewed; multimodal) Copyright © 2009 Cengage Learning 9.10 Sampling Distribution of sample mean Standard Deviation of sample mean Finite Population N n x ( ) n N 1 Infinite Population x n • A finite population is treated as being infinite if n/N < .05. • ( N n) / ( N 1) is the finite correction factor. • x is referred to as the standard error of the mean. Copyright © 2009 Cengage Learning Example 1 • The amount of time the university professors devote to their jobs per week is normally distributed with a mean of 52 hours and a standard deviation of 6 hours • a. What is the probability that a professor works for more than 60 hours per week? • b. Find the probability that the mean amount of work per week for three randomly selected professors is more than 60 hours. • c. Find the probability that if three professors are randomly selected, all three work for more than 60 hours per week. Copyright © 2009 Cengage Learning Solution • a. • b. • c. Copyright © 2009 Cengage Learning 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 / n 1.96 ) .95 With a little algebra P( 1.96 n Copyright © 2009 Cengage Learning X 1.96 ) .95 n 9.14 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. Copyright © 2009 Cengage Learning 9.15 Sampling Distribution of a Proportion • The parameter of interest for nominal data is the proportion of times a particular outcome (success) occurs. • To estimate the population proportion ‘p’ we use the sample proportion. The number of successes ^ = The estimate of p = p Copyright © 2009 Cengage Learning X n Sampling Distribution of a Proportion ^ can • Since X is binomial, probabilities about p be calculated from the binomial distribution. ^ we prefer to use • Yet, for inference about p normal approximation to the binomial. Copyright © 2009 Cengage Learning Normal approximation to the Binomial –Normal approximation to the binomial works best when •the number of experiments (sample size) is large, and •the probability of success, p, is close to 0.5. –For the approximation to provide good results two conditions should be met: np 5; n(1 - p) Copyright © 2009 Cengage Learning 5 Normal approximation to the Binomial Example Approximate the binomial probability P(x=10) when n = 20 and p = .5 The parameters of the normal distribution used to approximate the binomial are: = np; 2 = np(1 - p) Copyright © 2009 Cengage Learning Normal approximation to the Binomial Let us build a normal distribution to approximate the binomial P(X = 10). = np = 20(.5) = 10; 2 = np(1 - p) = 20(.5)(1 - .5) = 5 = 51/2 = 2.24 P(9.5<YNormal<10.5) The approximation P(XBinomial = 10) =.176 ~= P(9.5<Y<10.5) 9.5 10 10.5 9.5 10 10.5 10 P( Z ) .1742 2.24 2.24 Copyright © 2009 Cengage Learning Normal approximation to the Binomial More examples of normal approximation to the binomial P(X 4) @ P(Y< 4.5) 4 P(X 14) @ P(Y > 13.5) Copyright © 2009 Cengage Learning 4.5 13.5 14 Sampling Distribution of a Sample Proportion… Using the laws of expected value and variance, we can determine the mean, variance, and standard deviation of . (The standard deviation of is called the standard error of the proportion.) Sample proportions can be standardized to a standard normal distribution using this formulation: Copyright © 2009 Cengage Learning 9.22 Sampling Distribution of • Standard Deviation of p Finite Population p p p(1 p) N n n N 1 Infinite Population p p(1 p) n – p is referred to as the standard error of the proportion. Copyright © 2009 Cengage Learning Example 2 • A psychologist believes that 80% of male drivers when lost continue to drive, hoping to find the location they seek rather than ask directions. To examine this belief, he took a random sample of 350 male drivers and asked each what they did when lost. If the belief is true, determine the probability that less than 75% said they continue driving. Copyright © 2009 Cengage Learning Solution Copyright © 2009 Cengage Learning Sampling Distribution: Difference of two means The final sampling distribution introduced is that of the difference between two sample means. This requires: independent random samples be drawn from each of two normal populations If this condition is met, then the sampling distribution of the difference between the two sample means, i.e. will be normally distributed. (note: if the two populations are not both normally distributed, but the sample sizes are “large” (>30), the distribution of is approximately normal) Copyright © 2009 Cengage Learning 9.26 Sampling Distribution: Difference of two means The expected value and variance of the sampling distribution of are given by: mean: standard deviation: (also called the standard error if the difference between two means) Copyright © 2009 Cengage Learning 9.27 Example 9.3… Since the distribution of is normal and has a mean of and a standard deviation of We can compute Z (standard normal random variable) in this way: Copyright © 2009 Cengage Learning 9.28 Example 3 • The manager of a restaurant believes that waiters and waitresses who introduce themselves by telling customers their names will get larger tips than those who don’t. In fact, she claims that the average tip for the former group is 18% whereas that of the latter is only 15%. If tips are normally distributed with a standard deviation of 3 %, what is the probability that in a random sample of 10 tips recorded from waiters and waitresses who introduce themselves and 10 tips from waiters and waitresses who don’t, the mean of the former will exceed that of the latter? Copyright © 2009 Cengage Learning Solution Copyright © 2009 Cengage Learning