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solutions2 - Stat 665(Spring 2011 Kaizar Solutions 2...

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Stat 665 (Spring 2011) Kaizar Solutions 2 Exercises 40 points total 1. (8 points) Agresti 2.16 (a) Response: lung cancer Explanatory: have smoked (or not) (b) case-control study (c) No. This would be looking at the probability of cancer conditional on the smoking status , whereas we fixed the number of cancer/no cancer participants in our study. Therefore, we can estimate the probability of smoking conditional on cancer, but not the other way around. If we were to attempt to use Bayes rule to turn the conditioning around: P ( C | S ) = P ( S | C ) P ( C ) P ( S ) we would need to know the probability of having cancer – a value that we can not estimate with this design. (d) The odds ratio is: ˆ θ = n 11 n 22 n 12 n 21 = 688 * 59 21 * 650 = 2 . 97 Since the odds ratio is interpretable in either the row conditional on column, or column conditional on row direction, I interpret the odds ratio using the explanatory/response variables from part (a). The estimated odds of cancer conditional on smoking are 2.97 times larger than the odds of cancer conditional on not smoking. Thus, smoking is quite dangerous in terms of increasing the risk of cancer incidence. 2. (8 points) Effect Size Measures and Tests of Independence This question relates to data reported in: Nasar J, Hecht P and Wener R. ‘Call if You Have Trouble’: Mobile Phones and Safety among College Students. International Journal of Urban and Regional Research 2007; 31(4): 863-73. For this research, a survey center at The Ohio State University randomly sampled 264 students who carry cell phones to answer a survey (consider this number to be fixed). Of these, 45.9% happened to be female. The results section reported: A higher percentage of females (42.1%) than males (27.9%) reported walking [with their cell phone] somewhere after dark they would not normally go [if they did not have their cell phone with them]. 1
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In this question, I would like you to analyze the data to determine if there is a relationship between gender and whether or not the students would walk somewhere they would not normally go. (a) What kind of sampling distribution does this study’s data approximatly follow? This study is approximately a multinomial study, since only the total sample size is fixed. (b) From the reported conditional and marginal sample proportions, reconstruct the approxi- mate 2x2 table of counts for gender and walking. (Note that due to rounding error, this will not be exact, but your table should only contain whole numbers!) I can estimate the following proportions using the given data: π f + = 0 . 459 a marginal probability for gender π f = 0 . 421 the conditional probability of walking given female π m = 0 . 279 the conditional probability of walking given male Using these, I can also derive the probabilities: π m + = 1 - π f + = 0 . 541 a marginal probability for gender π fw = P ( W | F ) P ( F ) = π f π f + = 0 . 193 the joint probability of walking and female π fn = P ( F ) - P ( W, F ) = π f + - π fw = 0 . 266 the joint probability of not walking and female π mw = P ( W | M ) P ( M ) = π m π m + = 0 . 151 the joint probability of walking and male π mn = P ( M ) - P ( W, M ) = π m + - π mw = 0 . 390 the joint probability of not walking and male
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