Ch09 - Chapter 9 Sampling Sampling Distributions...

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Unformatted text preview: Chapter 9 Sampling Sampling Distributions Distributions 1 9.1 Introduction In real life calculating parameters of In populations is prohibitive because populations are very large. populations Rather than investigating the whole Rather population, we take a sample, calculate a statistic related to the parameter of statistic parameter interest, and make an inference. The sampling distribution of the statistic The sampling statistic is the tool that tells us how close is the statistic to the parameter. statistic 2 9.2 Sampling Distribution of 9.2 the Mean the An example A die is thrown infinitely many times. Let X die represent the number of spots showing on any throw. any The probability distribution of X is x p(x) 1 2 1/6 1/6 3 4 5 1/6 1/6 1/6 E(X) = 1(1/6) + 6 2(1/6) + 3(1/6)+ ………………….= 3.5 1/6 V(X) = (1-3.5)2(1/6) + (2-3.5)2(1/6) + 3 Throwing a die twice – sample mean Throwing Suppose we want to estimate µ Suppose from the mean x of a sample of size n = 2. size What is the distribution of x? 4 Throwing a die twice – sample mean Throwing Sample 1 2 3 4 5 6 7 8 9 10 11 12 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 Mean Sample Mean 1 13 3,1 2 1.5 14 3,2 2.5 2 15 3,3 3 2.5 16 3,4 3.5 3 17 3,5 4 3.5 18 3,6 4.5 1.5 19 4,1 2.5 2 20 4,2 3 2.5 21 4,3 3.5 3 22 4,4 4 3.5 23 4,5 4.5 4 24 4,6 5 Sample 25 26 27 28 29 30 31 32 33 34 35 36 Mean 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 3 3.5 4 4.5 5 5.5 3.5 4 4.5 5 5.5 6 5 Sample 1 2 3 4 5 6 7 8 9 10 11 12 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 Mean Sample Mean 1 13 3,1 2 1.5 14 3,2 2.5 2 15 3,3 3 2.5 16 3,4 3.5 3 17 3,5 4 3.5 18 3,6 4.5 1.5 19 4,1 2.5 2 20 4,2 3 2.5 21 4,3 3.5 3 22 4,4 4 3.5 23 4,5 4.5 4 24 4,6 5 Sample 25 26 27 28 29 30 31 32 33 34 35 36 Mean 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 3 3.5 4 4.5 5 5.5 3.5 4 4.5 5 5.5 6 The distribution of x when n = 2 The σ2 2 x Note µ x = µ x and σ x = : 2 E( x ) =1.0(1/36)+ 1.5(2/36)+….=3.5 6/36 5/36 V(X) = (1.03.5)2(1/36)+ (1.5-3.5)2(2/36)... = 1.46 4/36 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 x 6.0 6 Sampling Distribution of the Sampling Mean n= 5 µ x = 3.5 σ2 σ = .5833 = x ) ( 56 2 x n = 10 n = 25 µ x = 3.5 µ x = 3.5 σ2 2 σ x = .2917 = x ) ( 10 σ2 σ = .1167 = x ) ( 25 2 x 7 Sampling Distribution of the Sampling Mean n = 10 µ x = 3.5 σ2 2 σ x = .5833 = x ) ( 5 n = 25 µ x = 3.5 n= 5 µ x = 3.5 σ2 σ = .2917 = x ) ( 10 σ2 σ = .1167 = x ) ( 25 2 x 2 x 2 Notice that σ x is smaller than σ x. Notice that 2 is smaller than . The larger the sample size the The larger the sample size the 2 smaller Therefore, tends smaller σ x .. Therefore, x tends o fall closer to µ as the sample tto fall closer to µ ,, as the sample size increases. size increases. 8 Sampling Distribution of the Sampling Mean Demonstration: The variance of the sample mean is smaller than the variance of the population. Mean = 1.5 Mean = 2. Mean = 2.5 1.5 2.5 2 3 2 1.5 2.5 2 1.5 2 2.5 1.5 2 2.5 1.5 the variability of the population 2.5 2 Compare 1.5 2.5 1.5 2of the sample mean. 2.5 to the 1.5 variability 2 2.5 1.5 2.5 2 1.5 2.5 1.5 2 2.5 1.5 2 2.5 Let us take samples 1.5 2 2.5 Population 1 of two observations 9 Sampling Distribution of the Sampling 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 If 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 the sampling distribution of x will resemble a normal distribution. normal 11 Sampling Distribution of the Sample Sampling Mean Mean 1. µ x = µ x 2 σx 2 2. σ x = n 3. If x is normal, x is normal. If x is nonnormal x is approximately normally distributed for sufficiently large sample size. 12 Sampling Distribution of the Sample Mean Sample Example 9.1 The amount of soda pop in each bottle is normally The distributed with a mean of 32.2 ounces and a standard deviation of .3 ounces. standard Find the probability that a bottle bought by a Find customer will contain more than 32 ounces. customer Solution 0.7486 The random variable X is the The amount of soda in a bottle. amount P( x > 32) = P( x − µ 32 − 32.2 > ) σx .3 = P( z > −.67) = 0.7486 µ x = 32 = 32.2 13 Sampling Distribution of the Sample Mean Sample Find the probability that a carton of four bottles will Find have a mean of more than 32 ounces of soda per bottle. bottle. Solution Define the random variable as the mean amount of soda per Define bottle. bottle. P( x > 32) = P( x − µ 32 − 32.2 > ) σx .3 4 0.9082 = P( z > −1.33) = 0.9082 0.7486 x = 32 x = 32 µ = 32.2 µ x = 322 . 14 Sampling Distribution of the Sample Mean Sample Example 9.2 Dean’s claim: The average weekly income of Dean’s B.B.A graduates one year after graduation is $600. $600. Suppose the distribution of weekly income has a Suppose standard deviation of $100. What is the probability that 25 randomly selected graduates have an average weekly income of less than $550? average Solution x − µ 550− 600 P(x < 550 = P( ) < ) σx 100 25 = P(z < −2.5) = 0.0062 15 Sampling Distribution of the Sample Sampling Mean Mean Example 9.2– continued If a random sample of 25 graduates actually had If an average weekly income of $550, what would you conclude about the validity of the claim that the average weekly income is 600? the Solution With µ = 600 the probability of observing a sample mean With as low as 550 is very small (0.0062). The claim that the mean weekly income is $600 is probably unjustified. mean It will be more reasonable to assume that µ is smaller It than $600, because then a sample mean of $550 becomes more probable. becomes 16 Using Sampling Distributions for Using Inference Inference To make inference about population parameters we use To sampling distributions (as in Example 9.2). sampling The symmetry of the normal distribution along with the The sample distribution of the mean lead to: sample P( −1.96 ≤ z ≤ 1.96 ) = .95, or P( −1.96 ≤ x −µ ≤ 1.96 ) = .95 σn This can be written as σ σ P( −1.96 ≤ x − µ ≤ 1.96 ) = .95 n n which become P(µ −1.96 σ σ ≤ x ≤ µ +1.96 ) = .95 n n 17 Using Sampling Distributions for Using Inference Inference Standard normal distribution Z Normal distribution of x P(600 − 1.96 100 100 ≤ x ≤ 600 + 1.96 ) = .95 25 25 .95 .025 .025 .025 -1.96 0 Z -1.96 µ − 1..96 P(600 − 196 .95 .025 µ σ100 σ 100 µ= 6 0 0 Pµ600 +96 ( + 1. 1.96 25 n 25 n x 18 Using Sampling Distributions for Using Inference 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 There falls within the interval [560.8, 639.2] if the population mean is 600. population Since the sample mean was 550, the Since population mean is probably not 600. population 19 9.3 Sampling Distribution of 9.3 a Proportion Proportion The parameter of interest for nominal data The is the proportion of times a particular proportion outcome (success) occurs. outcome To estimate the population proportion ‘p’ To we use the sample proportion. we The number of successes ^ p= The estimate of p = The X n 20 9.3 Sampling Distribution of 9.3 a Proportion Proportion ^ Since X is binomial, probabilities about p Since can be calculated from the binomial distribution. distribution. ^ Yet, for inference about p we prefer to use Yet, normal approximation to the binomial. normal 21 Normal approximation to the Normal Binomial Binomial Normal approximation to Normal the binomial works best when when the number of the experiments (sample size) is large, and size) the probability of success, the p, is close to 0.5. p, For the approximation to For provide good results two conditions should be met: conditions np ≥ 5; n(1 - p) ≥ 5 22 Normal approximation to the Normal Binomial 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: µ = np; σ 2 = np(1 - p) 23 Normal approximation to the Normal Binomial Binomial Let us build a normal distribution to approximate the binomial P(X = 10). P(XBinomial = 10) .176 = µ = np = 20(.5) = 10; σ2 = np(1 - p) = 20(.5)(1 - .5) = 5 σ = 51/2 = 2.24 P(9.5<YNormal<10.5) The approximation =~ P(9.5<Y<10.5) 9.5 = P( 10 10.5 9.5− 10 105 − 10 . ≤ Z≤ ) = .1742 2.24 2.24 24 Normal approximation to the Normal Binomial Binomial More examples of normal More approximation to the binomial approximation P(X ≤ 4) ≅ P(Y< 4.5) P(X 4 P(X ≥ 14) ≅ P(Y > 13.5) P(X 14) 4.5 13.5 14 25 Approximate Sampling Distribution Approximate of a Sample Proportion of From the laws of expected value and variance, From it can be shown that E( p ) = p and V( p ) ˆ ˆ =p(1-p)/n =p(1-p)/n If both np > 5 and np(1-p) > 5, then ˆ p −p z= p(1−p) n Z iis approximately standard normally s distributed. distributed. 26 Example 9.3 A state representative received 52% of the state votes in the last election. votes One year later the representative wanted One to study his popularity. to If his popularity has not changed, what is If the probability that more than half of a sample of 300 voters would vote for him? 27 Example 9.3 Solution The The number of respondents who prefer the representative is binomial with n = 300 and p = .52. Thus, np = 300(.52) = 156 and .52. n(1-p) = 300(1-.52) = 144 (both greater than 5) ˆ P( p > .50) = P ˆ p− p p (1 − p ) n > = .7549 (.52)(1 − .52) 300 .50 − .52 28 9.4 Sampling Distribution of the 9.4 Difference Between Two Means Difference Independent samples are drawn from Independent each of two normal populations each We’re interested in the sampling We’re distribution of the difference between the two sample means x1 − x2 two 29 Sampling Distribution of the Sampling Difference Between Two Means Difference The distribution of x1 − x2 is normal if The two samples are independent, and The parent populations are normally The distributed. distributed. If the two populations are not both If not normally distributed, but the sample sizes are 30 or more, the distribution of x1 − x2 is approximately normal. 30 Sampling Distribution of the Sampling Difference Between Two Means Difference Applying the laws of expected value Applying and variance we have: and E(x1 − x2 ) = E(x1) − E(x2 ) = µ1 − µ 2 σ σ V(x1 − x2 ) = V(x1 ) + V(x2 ) = + n n 2 1 2 2 We can define: Z= ( x 1 − x 2 ) − ( µ1 − µ 2 ) 2 σ1 σ 2 +2 n1 n2 31 Sampling Distribution of the Difference Between Two Means Difference Example 9.4 The starting salaries of MBA students from The two universities (WLU and UWO) are $62,000 (stand.dev. = $14,500), and $60,000 (stand. dev. = $18,3000). dev. What is the probability that a sample mean of What WLU students will exceed the sample mean of UWO students? (nWLU = 50; nUWO = 60) UWO 32 Sampling Distribution of the Sampling Difference Between Two Means Difference Example 9.4 – Solution We need to determine P( x1 − x 2 > 0) µ 1 - µ 2 = 62,000 - 60,000 = $2,000 2 σ 12 σ 2 14,500 2 18,300 2 + = + = $3,128 n n 50 60 P(x1 − x2 > 0) = P( x1 − x2 − (µ1 - µ 2 ) 2 σ1 σ2 +2 n1 n2 = P(z > −.64 = .5+ .2389 .7389 ) = 0− 2000 > ) 3128 33 ...
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