hw6ans - Statistics 500 Fall 2009 Solutions to Homework 6...

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Statistics 500 – Fall 2009 Solutions to Homework 6 1 1. Economic growth and Democratic presidential vote (a) 1 4.149 ˆ 0.0108 385.256 β == , 0 8.288 32.18 ˆ 0.0108 0.4671 17 17 =− = (b) 0.0108 0 4.29 0.03653 15 385.26 t ( ) 17 2 4.29 0.001 p value P t −= >< There is very strong evidence that the linear slope relating growth rate to the Democratic vote is not 0. (c) () ˆ 0.4671 0.0108 5 0.521 Y =+ = ( ) 2 15,0.995 5 32.18 17 0.03653 1 ˆ 0.521 2.946 0.0143 0.479,0.563 15 17 385.26 Yt ⎛⎞ ±+ = ± = ⎜⎟ ⎝⎠ (d) ( ) 2 15,0.995 5 32.18 17 0.03653 1 ˆ 1 0.521 2.946 0.0514 0.37,0.67 15 17 385.26 + = ± = The prediction interval is much wider because it concerns a prediction for one year. (e) Source d.f. SS MS F Model 1 0.04468 0.04468 18.34 Error 15 0.03653 0.00244 Total 16 0.0812 Yes, the F statistic is, at least within round off error, 4.29 2 (f) R 2 = 0.04468 / 0.0812 = 0.55 or 55% Yes, this is a model with R 2 larger than 50%. (g) ( ) 15,0.975 0.03653 1 (0.4671 0.0108*1.9) 1 0.488 2.13 0.0508 0.38,0.60 15 17 t + = ± = I don’t think this is a very precise prediction. Comparing to my suggested criteria: The 95% prediction is almost as wide as the range of the data It is much wider than what is needed to predict close races. The prediction interval is over twice as wide as the desired 0.10 width.
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Statistics 500 – Fall 2009 Solutions to Homework 6 2 2. Snow gauge calibration a) X is density, Y is gain. The values of X are fixed by the experimenter’s choice of block densities. The values of Y are measured with error. The error variation is associated with gain, so gain is the Y variable. b) The proposed model was: ε β + + = density Gain 1 0 . My SAS code is: data snow; infile 'snow1.txt'; input density gain; /* I added a line with density = 0.2 and gain = . to the data file */ proc glm; model gain = density; estimate 'gain at 0.2' intercept 1 density 0.2; output out = resids r = resid p = yhat stdp = stderr stdi = stdobs lclm = lline uclm = uline lcl = lpred ucl = upred; proc print; where gain = .; run; I got: Standard Parameter Estimate Error t Value Pr > |t| Intercept 348.4059986 13.40911904 25.98 <.0001 density -579.9308681 33.49515224 -17.31 <.0001 So, the estimated intercept is 348 and the estimated slope is -579. Notice the wording of the question. A regression of gain on density is talking about a regression using Y=gain and X=density. Sometimes folks have misinterpreted the wording and reversed X and Y.
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hw6ans - Statistics 500 Fall 2009 Solutions to Homework 6...

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