 # 2007 會計資訊系統計學 ( 一 ) 上課投影片 9-1 Sampling Distributions Chapter 9.

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2007 會計資訊系統計學 ( 一 ) 上課投影片 9-1 Sampling Distributions Chapter 9

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-2 Introduction In real life calculating parameters of populations is prohibitive because populations are very large. Rather than investigating the whole population, we take a sample, calculate a statistic （統計量） related to the parameter （參數） of interest, and make an inference （推論）. The sampling distribution of the statistic （統計量 的抽樣分配） is the tool that tells us how close is the statistic to the parameter.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-3 Sampling Distributions （抽樣分配） 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.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-4 9.1 Sampling Distribution of the Mean An example –A fair die is thrown infinitely many times. –Let X represent the number of spots showing on any throw. –The probability distribution of X is x123456 P(x)1/6

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-5 Suppose we want to estimate  from the mean of a sample of size n = 2. What is the distribution of ? Throwing a die twice – sample mean

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-6 Throwing a die twice – sample mean 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).

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-7 Throwing a die twice – sample mean 1 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 6/36 5/36 4/36 3/36 2/36 1/36 6.0 2/365.5 3/365.0 4/364.5 5/364.0 6/363.5 5/363.0 4/362.5 3/362.0 2/361.5 1/361.0

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-8 Throwing a die twice – sample mean The sampling distribution of is shown below: 1/366.0 2/365.5 3/365.0 4/364.5 5/364.0 6/363.5 5/363.0 4/362.5 3/362.0 2/361.5 1/361.0 6/36 5/36 4/36 3/36 2/36 1/36

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-9 Compare Compare the distribution of X, with the sampling distribution of. As well, note that:

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-10 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 :

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-11 Sampling Distribution of the Mean

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-12 Sampling Distribution of the Mean Notice that is smaller than . The larger the sample size the smaller. Therefore, tends to fall closer to , as the sample size increases.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-13 Sampling Distribution of the Mean Demonstration: The variance of the sample mean is smaller than the variance of the population. 123 Mean = 1.5Mean = 2.5Mean = 2. Population 1.5 2.5 2 2 2 2 2 2 2 2 2 2 2 Compare the variability of the population to the variability of the sample mean. Let us take samples of two observations

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-14 Also, Expected value of the population = (1 + 2 + 3)/3 = 2 Expected value of the sample mean = (1.5 + 2 + 2.5)/3 = 2 Sampling Distribution of the Mean

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-15 The Central Limit Theorem If a random sample is drawn from any population, the sampling distribution of the sample mean is approximately normal （近似常態分配） for a sufficiently large sample size. The larger the sample size, the more closely the sampling distribution of will resemble a normal distribution.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-16 The Central Limit Theorem If the population is normal, then is normally distributed for all values of n. If the population is non-normal, then is approximately normal only for larger values of n. In many 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.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-17 Sampling Distribution of the Sample Mean 1. 2. 3. If X is normal, is normal. If X is nonnormal, 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)

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-18 Finite Populations Statisticians have shown that the mean of the sampling distribution is always equal to the mean of the population and that the standard error is equal to for infinitely large populations. However, if the population is finite the standard error is where N is the population size and is called the finite population correction factor. （有限母體校正因子） If the population size is large relative to the sample size the finite population correction factor is close to 1 and can be ignored.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-19 Finite Populations As a rule of thumb we will treat any population that is at least 20 times larger than the sample size as large. In practice most applications involve populations that qualify as large. As a consequence the finite population correction factor is usually omitted. There are several applications that deal with small populations. Section 12.5 introduces one of these applications.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-20 Example 9.1(a) The foreman of a bottling plant has observed that the amount of soda in each “ 32-ounce ” bottle is actually a normally distributed random variable, with a mean of 32.2 ounces, and a standard deviation of.3 ounce. If a customer buys one bottle, what is the probability that the bottle will contain more than 32 ounces?

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-21 The random variable X is the amount of soda in a bottle. where X is normally distributed and  =32.2 and  =.3  = 32.2 0.7486 x = 32 Example 9.1(a)-Solution “ There is about a 75% chance that a single bottle of soda contains more than 32oz. ”

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-22 Example 9.1(b) The foreman of a bottling plant has observed that the amount of soda in each “ 32-ounce ” bottle is actually a normally distributed random variable, with a mean of 32.2 ounces, and a standard deviation of.3 ounce. If a customer buys a carton of four bottles, what is the probability that the mean amount of the four bottles will be greater than 32 ounces?

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-23  = 32.2 0.7486 x = 32 Define the random variable as the mean amount of four bottles of soda. Then: X is normally distributed, therefore so will. 0.9082 Example 9.1(b)-Solution “ There is about a 91% chance the mean of the four bottles will exceed 32oz. ”

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-24 Graphically Speaking what is the probability that one bottle will contain more than 32 ounces? what is the probability that the mean of four bottles will exceed 32 oz? mean=32.2

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-25 Chapter-Opening Example The dean of the School of Business claims that the average salary of the school ’ s graduates one year after graduation is \$800 per week with a standard deviation of \$100. A second-year student would like to check whether the claim about the mean is correct. He does a survey of 25 people who graduated one year ago and determines their weekly salary. He discovers the sample mean to be \$750. To interpret his finding he needs to calculate the probability that a sample of 25 graduates would have a mean of \$750 or less when the population mean is \$800 and the standard deviation is \$100.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-26 Chapter-Opening Example-Solution Define the random variable as the mean weekly salary of 25 graduates. We want to compute Although X is likely skewed it is likely that is normally distributed. The mean of is The standard deviation is

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-27 Chapter-Opening Example-Solution The probability of observing a sample mean as low as \$750 when the population mean is \$800 is extremely small (0.0062). Because the event is quite unlikely, the claim that the mean weekly income is \$800 is probably unjustified. With  = 800 the probability of observing a sample mean as low as 750 is very small. It will be more reasonable to assume that  is smaller than \$800, because then a sample mean of \$750 becomes more probable.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-28 Using Sampling Distributions for Inference The sampling distribution can be used to make inferences about population parameters. In order to do so, the sample mean can be standardized to the standard normal distribution using the following formulation:

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-29 Another Way to State the Probability In Chapter 8 we saw that From the sampling distribution of the mean we have Substituting this definition of Z in the probability statement we produce

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-30 Another Way to State the Probability With a little algebra we rewrite the probability statement as

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-31 Another Way to State the Probability Similarly In general All are probability statements about, which we ’ ll use in statistical inference.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-32 Return to the Chapter-Opening Example Substituting  = 800,  = 100, n = 25, and  =.05, we get Conclusion –There is 95% chance that the sample mean falls within the interval [760.8, 839.2] if the population mean is 800. –It is unlikely that we would observe a sample mean as low as \$750 when the population mean is \$800. –Since the sample mean was 750, the population mean is probably not 800.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-33 Return to the Chapter-Opening Example -1.96 0.025 Standard normal distribution Z Normal distribution of.95 Z  

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-34 9.2 Sampling Distribution of a Proportion The parameter of interest for nominal data is the proportion of times a particular outcome (success) occurs. The estimator of a population proportion of successes “ p ” is the sample proportion “ ”. That is, we count the number of successes in a sample and compute: (read this as “ p-hat ” ). X is the number of successes, n is the sample size.  X is binomial random variable.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-35 9.2 Sampling Distribution of a Proportion Since X is binomial, probabilities about can be calculated from the binomial distribution. Yet, for inference about we prefer to use normal approximation to the binomial.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-36 Normal Approximation to Binomial Approximate the binomial probability P(X=10) when n = 20 and p =.5 with a normal approximation superimposed ( ).

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-37 Normal Approximation to Binomial Binomial distribution with n=20 and p=.5 with a normal approximation superimposed ( ) Where did these values come from?! From §7.6 we saw that:

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-38 Normal Approximation to Binomial To calculate P(X=10) using the normal distribution, we can find the area under the normal curve between 9.5 and 10.5. where Y is a normal random variable approximating the binomial random variable X continuity correction factor

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-39 10 9.510.5 P(X Binomial = 10) = ~ = P(9.5<Y<10.5)  = np = 20(.5) = 10;  2 = np(1 - p) = 20(.5)(1 -.5) = 5  = 5 1/2 = 2.24.176 Let us build a normal distribution to approximate the binomial P(X = 10). P(9.5<Y Normal <10.5) The approximation Normal approximation to the Binomial The approximation is quite good!

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-40  More examples of normal approximation to the binomial 4 4.5 14 13.5 P(X Binomial 14) P(X Binomial  14)  P(Y Normal < 4.5)  P(Y Normal > 13.5) Normal approximation to the Binomial P(X Binomial 4) P(X Binomial  4)

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-41 Normal Approximation to Binomial Normal approximation to the binomial works best when –the number of experiments, n, (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: (1) np ≥ 5 (2) n(1 － p) ≥ 5

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-42 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.)

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-43 Sampling Distribution of a Sample Proportion If both np > 5 and np(1-p) > 5, then sample proportions can be standardized to a standard normal distribution using this formulation: Z is approximately standard normally distributed.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-44 A state representative received 52% of the votes in the last election. One year later the representative wanted to study his popularity. If his popularity has not changed, what is the probability that more than half of a sample of 300 voters would vote for him? Example 9.2

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-45 The number of respondents who prefer the representative is binomial with n = 300 and p =.52. Thus, –np = 300(.52) = 156 and –n(1-p) = 300(1-.52) = 144 (both greater than 5) Example 9.2-Solution

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-46 9.3 Sampling Distribution of the Difference Between Two Means The final sampling distribution introduced is that of the difference between two sample means. Independent samples are drawn from each of two normal populations. We ’ re interested in the sampling distribution of the difference between the two sample means

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-47 9.3 Sampling Distribution of the Difference Between Two Means The distribution of is normal if –The two samples are independent, and –The parent populations are normally distributed. If the two populations are not both normally distributed, but the sample sizes are 30 or more, the distribution of is approximately normal.

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-48 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)

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-49 Sampling Distribution: Difference of two means 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:

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-50 Example 9.3 Starting salaries for MBA grads at two universities are normally distributed with the following means and standard deviations. Samples from each school are taken: University WLUUniversity UWO Mean  62,000 \$/yr60,000 \$/yr Std. Dev.  14,500 \$/yr18,300 \$/yr sample size n 5060 What is the probability that the sample mean starting salary of University WLU graduates will exceed that of the UWO grads?

2007 會計資訊系統計學 ( 一 ) 上課投影片 9-51 Example 9.3-Solution “ What is the probability that the sample mean starting salary of University WLU graduates will exceed that of the UWO grads? ” We are interested in determining. Converting this to a difference of means, what is: ? “There is about a 74% chance that the sample mean starting salary of U. WLU grads will exceed that of U. UWO”.

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