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Chapter 9 Sampling Distributions 1 9.1 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 related to the parameter of interest, and make an inference. The sampling distribution of the statistic is the tool that tells us how close is the statistic to the parameter. 2 9.2 Sampling Distribution of the Mean An example A die is thrown infinitely many times. Let X represent the number of spots showing on any throw. The probability distribution of X is E(X) = 1(1/6) + x 1 2 3 4 5 6 2(1/6) + 3(1/6)+ ………………….= 3.5 p(x) 1/6 1/6 1/6 1/6 1/6 1/6 V(X) = (1-3.5)2(1/6) + (2-3.5)2(1/6) + …………. …= 2.92 3 Throwing a die twice – sample mean Suppose we want to estimate m from the mean x of a sample of size n = 2. What is the distribution of x ? 4 Throwing a die twice – sample mean Sample Mean Sample Mean Sample Mean 1 1,1 1 13 3,1 2 25 5,1 3 2 1,2 1.5 14 3,2 2.5 26 5,2 3.5 3 1,3 2 15 3,3 3 27 5,3 4 4 1,4 2.5 16 3,4 3.5 28 5,4 4.5 5 1,5 3 17 3,5 4 29 5,5 5 6 1,6 3.5 18 3,6 4.5 30 5,6 5.5 7 2,1 1.5 19 4,1 2.5 31 6,1 3.5 8 2,2 2 20 4,2 3 32 6,2 4 9 2,3 2.5 21 4,3 3.5 33 6,3 4.5 10 2,4 3 22 4,4 4 34 6,4 5 11 2,5 3.5 23 4,5 4.5 35 6,5 5.5 12 2,6 4 24 4,6 5 36 6,6 6 5 Sample Mean Sample Mean Sample Mean 1 1,1 1 13 3,1 2 25 5,1 3 The distribution of x when n = 2 2 3 1,2 1,3 1.5 2 14 15 3,2 3,3 2.5 3 26 27 5,2 5,3 3.5 4 4 1,4 2.5 16 3,4 3.5 28 5,4 4.5 2 5 1,5 3 17 3,5 4 29 2 x 5,5 5 6 7 Note : m x m x and 1,6 2,1 3.5 1.5 18 19 3,6 4,1 4.5 2.5 x 30 31 5,6 6,1 5.5 3.5 8 9 2,2 2,3 2 2.5 20 21 4,2 4,3 3 3.5 32 33 2 6,2 6,3 4 4.5 10 2,4 3 22 4,4 4 34 6,4 5 11 2,5 3.5 23 4,5 4.5 35 6,5 5.5 12 2,6 4 24 4,6 5 36 6,6 6 E( x) =1.0(1/36)+ 6/36 1.5(2/36)+….=3.5 5/36 V(X) = (1.0-3.5)2(1/36)+ 4/36 (1.5-3.5)2(2/36)... = 1.46 3/36 2/36 1/36 1 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 x 6 Sampling Distribution of the Mean n5 n 10 n 25 m x 3.5 m x 3.5 m x 3.5 2 2 2 .5833 ( x ) 2 x x .2917 ( x ) 2 .1167 ( x ) 2 x 5 6 10 25 7 Sampling Distribution of the Mean n5 n 10 n 25 m x 3.5 m x 3.5 m x 3.5 2 2 2 2 .5833 ( x ) .2917 ( x ) 2 x .1167 ( x ) 2 x x 5 10 25 2 Notice that x is smaller than . x. 2 The larger the sample size the smaller x . Therefore, x tends 2 to fall closer to m, as the sample size increases. 8 Sampling Distribution of the Mean Demonstration: The variance of the sample mean is smaller than the variance of the population. Mean = 1.5 Mean = 2. Mean = 2.5 Population 1 1.5 2 2 2.5 3 1.5 1.5 2 2 2.5 2.5 1.5 2 2.5 Compare 2.5 1.5 the variability of the population 2 1.5 1.5 2 of the2.5 2.5 1.5 to the variability2 2.5 sample mean. 1.5 2 2.5 1.5 2.5 1.5 2 2.5 Let us take samples 1.5 1.5 2 2 2.5 2.5 of two observations 9 Sampling Distribution of the Mean Also, Expected value of the population = (1 + 2 + 3)/3 = 2 Expected value of the sample mean = (1.5 + 2 + 2.5)/3 = 2 10 The Central Limit Theorem If a random sample is drawn from any population, the sampling distribution of the sample mean 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. 11 Sampling Distribution of the Sample Mean 1. m x m x x 2 2. 2 x n 3. If x is normal, x is normal. If x is nonnormal x is approximately normally distribute d for sufficient ly large sample size. 12 Sampling Distribution of the Sample Mean Example 9.1 The amount of soda pop in each bottle is normally distributed with a mean of 32.2 ounces and a standard deviation of .3 ounces. Find the probability that a bottle bought by a customer will contain more than 32 ounces. Solution 0.7486 The random variable X is the amount of soda in a bottle. x m 32 32.2 P( x 32) P( ) x .3 x = 32 m = 32.2 P( z .67) 0.7486 13 Sampling Distribution of the Sample Mean Find the probability that a carton of four bottles will have a mean of more than 32 ounces of soda per bottle. Solution Define the random variable as the mean amount of soda per bottle. x m 32 32.2 P( x 32) P( ) 0.9082 x .3 4 P( z 1.33) 0.9082 0.7486 x = 32 x 32 m = 32.2 m x 32.2 14 Sampling Distribution of the Sample Mean Example 9.2 Dean’s claim: The average weekly income of B.B.A graduates one year after graduation is $600. Suppose the distribution of weekly income has a standard deviation of $100. What is the probability that 25 randomly selected graduates have an average weekly income of less than $550? Solution x m 550 600 P( x 550) P( ) x 100 25 P( z 2.5) 0.0062 15 Sampling Distribution of the Sample Mean Example 9.2– continued If a random sample of 25 graduates actually had an average weekly income of $550, what would you conclude about the validity of the claim that the average weekly income is 600? Solution With m = 600 the probability of observing a sample mean as low as 550 is very small (0.0062). The claim that the mean weekly income is $600 is probably unjustified. It will be more reasonable to assume that m is smaller than $600, because then a sample mean of $550 becomes more probable. 16 Using Sampling Distributions for Inference To make inference about population parameters we use sampling distributions (as in Example 9.2). The symmetry of the normal distribution along with the sample distribution of the mean lead to: xm P( 1.96 z 1.96 ) .95, or P( 1.96 1.96 ) .95 n - Z.025 Z.025 This can be written as P( 1.96 x m 1.96 ) .95 n n which become P(m 1.96 x m 1.96 ) .95 n n 17 Using Sampling Distributions for Inference Standard normal distribution Z Normal distribution of x 100 100 P(600 1.96 x 600 1.96 ) .95 25 25 .95 .95 .025 .025 .025 .025 Z x -1.96 0 -1.96 100 m m600 Pm600 9696 100 m 1..96 P(600 196 ( 1. 1. n 25 n 25 18 Using Sampling Distributions for Inference 100 100 P(600 1.96 x 600 1.96 ) .95 25 25 Which reduces to P(560.8 x 639.2) .95 Conclusion There is 95% chance that the sample mean falls within the interval [560.8, 639.2] if the population mean is 600. Since the sample mean was 550, the population mean is probably not 600. 19 9.3 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 = X n 20 9.3 Sampling Distribution of a Proportion Since X is binomial, probabilities about ^ p can be calculated from the binomial distribution. ^ Yet, for inference about p we prefer to use normal approximation to the binomial. 21 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) 5 22 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: m = np; 2 = np(1 - p) 23 Normal approximation to the Binomial Let us build a normal m = np = 20(.5) = 10; distribution to approximate the 2 = np(1 - p) = 20(.5)(1 - .5) = 5 = 51/2 = 2.24 binomial P(X = 10). P(XBinomial = 10) =.176 P(9.5<YNormal<10.5) The approximation ~ P(9.5<Y<10.5) = 9.5 10 10.5 9.5 10 10.5 10 P( Z ) .1742 2.24 2.24 24 Normal approximation to the Binomial More examples of normal approximation to the binomial P(X 4) @ P(Y< 4.5) 4 4.5 P(X 14) @ P(Y > 13.5) 13.5 14 25 Approximate Sampling Distribution of a Sample Proportion From the laws of expected value and variance, ˆ ˆ it can be shown that E( p ) = p and V( p ) =p(1-p)/n If both np > 5 and np(1-p) > 5, then ˆ pp z p(1 p) n Z is approximately standard normally distributed. 26 Example 9.3 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? 27 Example 9.3 Solution 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) p p ˆ .50 .52 P( p .50) P ˆ .7549 p(1 p) n (.52)(1 .52) 300 28 9.4 Sampling Distribution of the Difference Between Two 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 x 1 x 2 29 Sampling Distribution of the Difference Between Two Means The distribution of x 1 x 2 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 x 1 x 2 is approximately normal. 30 Sampling Distribution of the Difference Between Two Means Applying the laws of expected value and variance we have: E( x1 x 2 ) E( x1 ) E( x 2 ) m1 m 2 1 2 2 V( x 1 x 2 ) V( x 1 ) V( x 2 ) 2 n n We can define: ( x 1 x 2 ) ( m1 m 2 ) Z 1 2 2 2 n1 n2 31 Sampling Distribution of the Difference Between Two Means Example 9.4 The starting salaries of MBA students from two universities (WLU and UWO) are $62,000 (stand.dev. = $14,500), and $60,000 (stand. dev. = $18,3000). What is the probability that a sample mean of WLU students will exceed the sample mean of UWO students? (nWLU = 50; nUWO = 60) 32 Sampling Distribution of the Difference Between Two Means Example 9.4 – Solution We need to determine P( x1 x 2 0) m1 - m2 = 62,000 - 60,000 = $2,000 12 2 2 14,5002 18,3002 $3,128 n n 50 60 x1 x 2 (m1 - m 2 ) 0 2000 P( x1 x 2 0) P( ) 1 2 2 2 3128 n1 n2 P( z .64) .5 .2389 .7389 33

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