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12 Sampling Distributions Part 1 (1)

You have a sample of customers who purchased family

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market. You have a sample of customers who purchased family sedans in 2003. Assume that purchase price has a normal distribution with a mean of $25,000 and a standard deviation of $5,000. You draw a random sample of n=25 customers. 34
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1. What is the probability that the sample mean is between $24K and $26K? 3. What is the probability that one customer’s purchase is between $24K and$26K? x ~ N(25,000, 5,000/√25) = N(25,000,1,000) ̅ P(2 4 ,0 0 0 < x < 26,000) = 1 ̅ 2*P(x < ̅ 24,000) = 1 – 2 *NORM.DIS T(2 4 0 0 0 , 2 5 0 0 0 , 1 0 0 0 , TRUE) = 0 .6 8 2 7 x ~ N(25000, 5000) P(2 4 ,0 0 0 < x < 26,000) = 1 2*P(x < 24,000) = 1 – 2 *NORM.DIS T(2 4 0 0 0 , 2 5 0 0 0 , 5 0 0 0 , TRUE) = 0 .1 5 8 5 = 1 – 2 *NORM.S .DIS T(-1 ,TRUE) = .6 8 2 7 Example: Family Sedan Purchase Price 35
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Example: Homeowner’s insurance Liberty Mutual decided to use data to calculate the parameters of interest for insurance cost and sampling distribution of the insurance cost for accounting purposes. In addition to this, they are interested in the proportion of insurance costs that are greater than a certain number for risk estimation. Because of the time and budget limitations they can not analyze the whole data. Therefore, they will use a small sample from the population. Assume that the mean cost of homeowner’s insurance is $754 and the standard deviation is σ = $227. Suppose you draw a simple random sample of n=100 insurance policies. 36
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1. What is the sampling distribution of the mean cost of homeowner’s insurance? 2. What is the probability that the random sample of HO insurance policies will have a sample mean of more than $20 greater than the population mean? 3. If we increase n to 400, what is the sampling distribution of the mean cost of homeowner’s policies? 4. What is the probability that a random sample of n=400 will have a sample mean more than $20 greater than the population mean? 5. What is the advantage of the larger sample size? By CLT: x ~ N(754, 227/√100) = N(754, ̅ 22.7) P (x > 774) = 1 ̅ NORM.DIST(774, 754, 22.7, TRUE) = . 1891 By CLT: x ~ N(754, 227/√400) = N(754, ̅ 11.35) P(x > 774) = 1 ̅ NORM.DIST(774, 754, 11.35, TRUE) = . 0390 Example: Homeowner’s insurance 37
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Sampling distribution of sample proportion Facts: If sample large enough, Central Limit Theorem applies here, too. What is large enough? np > 5 and n(1-p) > 5 If so, 38 p ˆ ) ) 1 ( , ( ~ ˆ n p p p N p - n p p p p p ) 1 ( ˆ ˆ - = = σ μ
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Example: Sampling distribution of proportions The Vice President of sales for Chemicals, Inc. believes that 30% of orders come from new customers. A random sample, n=100, orders is drawn from past orders. 1. Assume the VP is correct. What is the sampling distribution of p̂? 2. What is the probability that the sample proportion, p , ̂ will be less than 0.275? 3. What is the probability that the sample proportion, p̂, will be greater than 0.24? By CLT: p ~ N(0.30, √(.3*.7/100)) = N(0.30, ̂ 0.0458) P(p > .24) = 1- NORMDIST(.24, .30, .0458, TRUE) = ̂ 0.905 P(p < .275) = NORM.DIST(.275, .30, .0458, TRUE) = ̂ 0.293 39
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Key learning points: Sampling distributions Means Sampling distribution of sample means has: When population is normally distributed, distribution of sample means is normal regardless of n Central Limit Theorem: When population not normal, if sample size n
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