jeopardy_round2(1).pdf - Random Assessment Interpretation...

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Random Assessment Interpretation Misc. 100 100 100 100 200 200 200 200 300 300 300 300 400 400 400 400 500 500 500 500
Subject 1 100 TRUE/FALSE: The variance of a gamma random variable increases cubically with the mean? Question Answer Done! Home
Subject 1 100 FALSE - it increases quadratically. Question Answer Done! Home
Subject 1 200 I can be thought of as the sum of Bernoulli trials. That is, I’m a count out of a total. What distribution do I have? Question Answer Done! Home
Subject 1 200 Binomial Question Answer Done! Home
Subject 1 300 Which of the following situations motivate the use of a GLM with a Poisson random component? Check all that apply.
4. The response variable is a count of events within a set time span or spatial area. Question Answer Done! Home
Subject 1 300 1. The response tends to have a symmetric distribution around 0 2. The response variable is a count that can be thought of as a sum of m Bernoulli trials. 3. The response can be any non-negative integer. 4. The response variable is a count of events within a set time span or spatial area. Question Answer Done! Home
Subject 1 400 I’ve binned my data and calculated sample means and standard deviations. A plot of the log standard deviations by log means is shown below, along with the estimated slope of the line. Suggest an appropriate random component and back this up with facts. slope of line = 1.378 8.5 9.0 9.5 10.0 10.5 11.0 -1.0 -0.5 0.0 0.5 1.0 Log sample means Log sample standard deviations Question Answer Done! Home
Subject 1 400 For continuous random components, the mean-variance relationship can be written as: log ( Var ( Y i ) 1 / 2 ) = 1 2 log ( σ 2 ) + ρ 2 log ( E ( Y i )) . Our estimated slope is 1.378 Setting 1 . 378 = ρ 2 results in an optimal ρ of 2(1 . 378) = 2 . 756. This is much closer to 3 than 2 (or 0), suggesting an Inverse Gaussian random component might be a good candidate. (Since, for Inverse Gaussian, V ( Y ) = σ 2 E ( Y ) 3 , i.e., ρ = 3) Question Answer Done! Home
Subject 1 500 I’ve binned my data and calculated sample means and standard deviations. A plot of the log standard deviations by log means is shown below. Suggest an appropriate random component and back this up with facts. 0.9 1.0 1.1 1.2 1 2 3 Log sample means Log sample standard deviations Question Answer Done! Home
Subject 1 500 For continuous random components, the mean-variance relationship can be written as: log ( Var ( Y i ) 1 / 2 ) = 1 2 log ( σ 2 ) + ρ 2 log ( E ( Y i )) .

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