185323959-Business-Stats-Ken-Black-Case-Answers.pdf

Colgate palmolive makes a total effort 1 two

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Colgate-Palmolive Makes a “Total” Effort 1. Two probabilities are given in this case. The first is the marginal probability that a U.S. household had purchased Total for the first time: P(T 1 ) = .21 The second is the conditional probability that a household purchased Total (for a second time) given that it had purchased it before: P(T 2 T 1 ) = .43 The percentage of U.S. household that purchased Total at least twice can be computed as: P(T 1 T 2 ) = P(T 1 ) · P(T 2 T 1 ) = (.21)(.43) = .0903 A little over nine percent (9.03%) of U.S. households purchased Total at least twice during this time period. 2. If age is independent of the willingness to try Total, then the marginal probability of an age category should equal the conditional probability that someone is in that age category given that they are willing to try Total. Here, for example, the probability that someone is in the 45-64 age category is: P(45-64) = .20 The conditional probability that someone is in the 45-64 age category given their willingness to try Total is: P(45-64 T) = .24 If age is independent of willingness to try Total, then the following must be true: P(45-64) = P(45-64 T) but .20 .24. Therefore, age is not independent of the willingness to try Total.
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Case Notes 10 3. The probability that a person is either in the 45-64 age category or purchased Total is a union probability and is computed as: P(45-64 T) = P(45-64) + P(T) - P(45-64 T) = .20 + .21 - .0903 = .3197 The probability that a person purchased Total given that they are in the 45-64 age category can be computed as: P(T 45-64) = 20 . 0903 . ) 64 45 ( ) 64 45 ( = - - P T P = .4515 4. Let S = saw the commercial, N = didn’t see the commercial, P i = prior probability P(S) = .32 P(N) = .68 P(T S) = .40 P(T NS) = .1206 Event Prior P(T P i ) P(T P i ) P(P i T) S .32 .40 (.32)(.40) = .128 .61 N .68 .1206 (.68)(.1206) = .082 .39 P(T) = .210
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Case Notes 11 Chapter 5 Fuji Film Introduces APS 1. n = 30, p = .40. The expected number is: μ = n · p = 30(.40) = 12 The probability that six or fewer purchase an APS camera is: Prob( x < 6 n = 30 and p = .40) = 30 0 0 30 25 5 5 30 24 6 6 30 ) 60 (. ) 40 (. ) 60 (. ) 40 (. ) 60 (. ) 40 (. C C C + + + = .0115 + .0041 + .0012 + .0003 + .0000 + .0000 + .0000 = .0171 If the market share actually is 40% ( p = .40), the expected number of purchasers from a sample of 30 is 12. Six or fewer are considerably less than the expected number (12). How often would one get six or fewer out of thirty when twelve is expect – less than 2% (.0171) of the time. Therefore, if six or fewer out of thirty actually purchase, it is likely that the market share is not 40%. 2. Let λ = 2.4 complaints/100,000 rolls. Shown below are some of the values for the Poisson distribution with λ = 2.4 from MINITAB: x Probability 0 .0907 1 .2177 2 .2613 3 .2090 4 .1254 5 .0602 6 .0241 7 .0083 8 .0025 9 .0007 10 .0002 11 .0000 12 .0000
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Case Notes 12 With a Poisson distributed average rate of 2.4 complaints/100,000 rolls, the probability of getting 7 complaints/100, 000 rolls by chance is .0083 or less than 1%. Often researchers like to use the cumulative probability of x > 7 on a problem like this. The probability of randomly getting 7 or more complaints/100,000 rolls when the average is only 2.4 is .
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