14 Sampling Distributions Part 3

# Takeaways from class 14 x 10 1019842 2 603 1001

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Takeaways from Class 14 x ̅ = 10 10+1.984*2 = 13.97 10-1.984*2 = 6.03 100(1- )% confidence interval for  % CI for : x̅ = 10; s = 20; n = 100;  = 0.05,  t(.05/2, df=99) = 1.984 Confidence interval,  unknown: 7 n s t x ) 2 / ( α + x n s t x ) 2 / ( α - 2 100 20 = = = n s s X

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a) Form a 95% confidence interval for μ d.f. = n 1 = , so Confidence interval: A manager at Harry-Winston, NYC wants to find the average purchase price of engagement rings for NYC’s rich and famous. A random sample of 55 resulted in a sample mean of \$50K and stdev of \$12.5K. b) If  = \$12.5K (instead of s), find a margin of error with 95% conf. 54 2.005 50 ± 2.005 * 12.5 / 7.42 = 50 ± 3.43 Example: Harry-Winston Jeweler 8 = /2 α t = ± n s /2 α t x x ± z α /2 σ n = 50 ± 1.96 × 12.5 55 = 50 ± 3.3
(1- α ) Confidence interval for proportion Population proportion is given by: Sample proportion is given by: Where x is the number of objects or individuals in the population with the attribute of interest Recall that, if samples are large enough, by Central Limit Theorem, sample proportions are normally distributed: Where x is the number of objects in the sample with the attribute of interest (the number of “yes”) Large enough ? : np > 5; n(1-p) > 5 9 N x p = n x p = ˆ ) ) 1 ( , ( ~ ˆ n p p p N p -

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Confidence interval for proportion Of course, don’t know p Estimate it with sample proportion, Use this to estimate standard error of sampling distribution Even though estimating standard error of sampling distribution, the distribution is still normal. So can use z-statistic 10 ( 29 ( 29 n p p z p ˆ 1 ˆ ˆ ) 2 / ( - ± α n p p s p ) ˆ 1 ( ˆ ˆ - = ˆ p
Confidence interval for proportion 0.33 .33+1.96*.047 = .422 .33-1.96*.047 = .238 100(1- )% confidence interval for p 95% CI for p: p̂ = .33; n=100;  = 0.05,  z(.05/2) = 1.96 11 n p p z p ) ˆ 1 ( ˆ ˆ ) 2 / ( - + α p ˆ n p p z p ) ˆ 1 ( ˆ ˆ ) 2 / ( - - α 047 . 0 100 ) 33 . 1 ( 33 . ) ˆ 1 ( ˆ ˆ = - = - = n p p s p

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Example: Weekly customer contacts An audit of recorded WeHelpU customer support phone conversations showed errors in the advice given to customers in 176 of 621 sampled conversations. 1. What is the point estimate of the error rate in advice to customers? 2. At a 95% confidence level, what is the margin of error? 3. Construct and interpret a 95% confidence interval for the population mean error rate in customer support.
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