Week_8_Practice_Problems_Solution

Week_8_Practice_Problems_Solution - Week 8 Practice...

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Week 8 Practice Problems Here we have a bunch of problems all involving confidence intervals with a prediction interval tossed in for fun. 1. Flaws occur in glass sheet manufactured in high volume for automobile windshields. In an experiment to try to reduce the number of flaws, the manufacturer varies the temperature of the molten glass before it is poured into a mould. Thirty windshields are produced at each of two temperatures and the number of flaws is counted The data are: temperature 1 temperature 2 1 1 2 0 0 0 0 2 1 1 0 0 0 0 1 1 0 0 0 2 0 0 1 0 0 1 2 4 4 0 2 3 1 0 1 1 1 1 1 1 0 3 0 2 1 0 2 4 2 1 3 1 1 0 1 2 1 0 0 1 Suppose that we can model the number of flaws in each windshield by a Poisson random variable with mean 1 λ for temperature 1 and 2 for temperature 2. a) Which technique for producing a confidence interval will work here? Since ˆ 1 = 1 n y i i = 1 n 1 , ˆ 2 = 1 n y i i = 1 n 2 , then we can use a technique based on central limited theorem. b) Find a 95% confidence interval for 12 θ λλ = . Be sure to include all of the steps since this is a practice exercise. The confidence interval for = is ˆˆ ()( cSE ) −± where comes from the G(0,1) table. 1.96 c = For a Poisson model, the maximum-likelihood estimate of is 1 1 ˆ n i i y n = = 13 5 1 (35) , (26) 30 30 30 30 35 26 ˆˆˆ 0.3 30 θλλ == =−= = 2 6 2 To find the standard error, 1 = ±±± and
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~( , / 3 0 ) ,1 , approx ii i Gi λλ λ = ± 2 so 12 () /3 0 0 SD θλ =+ ±± ± and ˆˆ ˆ ( ) / 30 / 30 0.26 SE = The confidence interval is 0.3 1.96 0.26 ± × , that is (-0.21,0.81). c) Based on the confidence interval, does temperature effect the number of flaws? Based on the confidence interval, we know temperature do not affect the number of flaws because 0 is included in the confidence interval. The two means may be the same. 2. We have seen that a likelihood interval with ( ) 0.15 R θ corresponds (approximately) to a 95% confidence interval. What levels correspond to a 90% and a 99% interval? R ( μ ) = L ( ) L ( ˆ ) and 2ln( L ( ) L ( ˆ ) )~ χ 1 2 For 90% confidence interval, we need to find all such that P ( 1 x ) = 0.9 ,look up chi squared table, L ( ) L ( ˆ ) ) 2.706, so R ( ) = L ( ) L ( ˆ ) e 2.706 2 = 0.258 For 99% confidence interval, we need to find all such that P ( 1 x ) = 0.99 ,look up chi squared table, L ( ) L ( ˆ ) ) 6.635, so R ( ) = L ( ) L ( ˆ ) e 6.635 2 = 0.036 3. A marketing firm wants to compare four different programs (here called A,B,C,D) for promoting a soft drink. The plan is to pick one of the programs and use it across all convenience stores in Canada. In a preliminary trial, 40 convenience stores are selected in Toronto and a store is randomly assigned one of the four promotions, 10 stores to each. Then the total sales (in $) is measured over a one
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This note was uploaded on 03/20/2011 for the course STATISTICS 231 taught by Professor Mckay during the Spring '10 term at Waterloo.

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Week_8_Practice_Problems_Solution - Week 8 Practice...

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