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deF : é/‘Fpss ' d’ﬁ’é 3 y’ 6 5 3‘. ? (Mr. 5, a) = 8.74]. Statistics 425 Final Exam December 10, 2007 Do all work in the exam solution booklet. Show all of your work unless told otherwise.
Closed book and notes. You are allowed 2 twosided 8.5inch x 11inch sheets of paper
with any information you like on them. Calculators are allowed. There is a ttable and an
F table on the last page of this exam. 1. A linear regression of y on x was fit to 11 observations with the following results:
Response: y Analysis of Variance Source DF Sum of Squares
Regression 1 8.9
Residual 9 13.8
Total 10
Coefficient Estimate Standard Error
(Intercept) 2.7 1.25 x 0.53 0.22 (a) Calculate R 2
(b) Calculate an unbiased estimate of the error variance. (0) What assumptions are needed for the t—test of the coefficient of x. Is the test
significant at level .05? ((1) Compute a 90% confidence interval for the slope of the regression line. (e) Estimate the expected response for x = 15. 2. Recall that the Akaike information criterion for a candidate model C has the
form AIC = n ln(RSSC /n) + 2 p‘C . (a) Using the results given in Problem 1, compute AIC for the simple linear regression of y on x
(with intercept). (b) Using the results given in Problem 1, compute AIC for the model that only includes the
intercept (so x is dropped from the model). (c) Based on the AIC criterion, which of the two models, (a) or (b), is preferred? Statistics 425 Final Exam December 10, 2007 3. Consider a simple linear regression model, yi = ,80 + ,lel. + ei, i = l,2,...,n, where the . . 2 .. _ l n
errors are uncorrelated w1th mean zero and variance 0' . Let x — 7 x. i=1 ' (a) Show that if It = 0 then the least squares estimates of intercept and slope are
uncorrelated, i.e., C0v(,80,,81) = 0. (b) Let ()cn+l , y"+1 ) be a new independent observation from the same model and let $2M] denote. the least squares prediction for y"+1 based on the original n observations. Show that if 76 = 0, then Var(yn+1 — 9”“) = 0'2 + Var(,30) + )chVar(ﬂA1 ). 4. In a study of endurance training methods, three different training plans A1, A2 and A3 were
compared. 15 female volunteers participated. The participants were divided into 5 matched
groups based on a preliminary 1 mile time trial. They were then randomized within groups to one
of the three training plans. After three months the participants ran another time trial. For each
participant the improvement in time (in seconds) was recorded. The results are given below. Training
A1 A2 A3 Group An analysis of variance table is given below. Based on the results, test whether there are significant differences between Training plans after adjustment for Groups. Is it significant at
level .05? What are the assumptions of the test? Response: Improvement in mile time Df Sum Sq
Group 4 1507.73
Training 2 19.73
Residual 8 92.27 Statistics 425 Final Exam December 10, 2007 5. The Fuel Consumption data provide information on fuel consumption and other population
level variables for the 50 states plus Washington DC (n = 51). The variable Fuel: fuel
consumption per 1000 population. Treating Fuel as the response, the full model considered is: E(Fuel I X) = [30 + ,BlTax + ﬂleic + ﬂ3Inc0me + ,84 log(miles) A nested sequence of models was fit to the data, giving the sequential ANOVA table below. Sequential analysis of variance Tax 1 19159 Income 1 61408 LogMiles 34573 Residual 193700 (a) Compute the residual sum of squares and residual degrees of freedom for the model that
includes only Dlic, Tax and Income, i.e. dropping LogMiles from the model. (b) Test the null hypothesis that both Income and LogMiles can be dropped from the model. Thus
we wish to test the model with [7’3 = ,84 = 0 against the full four variable model. Using the information given above, compute the appropriate F statistic for this hypothesis and give its
degrees of freedom. Does the test reject at significance level .05? 6. An experiment was done to study the effect of dissolved sulfur on the surface tension
of liquid copper. The predictor Sulfur is the weight proportion sulfur. The response
Tension is the decrease in surface tension in dynes per cm. Obs. Sulfur Tension
1 0.034 301
2 0.034 316
3 0.093 430
4 0.093 422
5 0.300 593
6 0.300 586
7 0.400 630
8 0.400 618
9 0.610 656
10 0.610 642
11 0.830 740
12 0.830 714 ykl A nonlinear model of the following form was fit to the data: yl. = ,30 + ,3, * + e, i = 1,2,. . . ,12 , where, as a working model, the errors are assumed to be independently normal Statistics 425 Final Exam December 10, 2007 with mean zero and unknown variance 0'2 . The results (using the R function nls) were as
follows: Formula: Tension ~ betaO + betal * (SulfurAlambda — l)/lambda Parameters: Estimate Std. Error
betaO 739.88406 12.35047
beta1 135.51828 19.50715
lambda 0.03442 0.07934 Residual standard error: 16.61 on 9 degrees of freedom Correlation of Parameter Estimates:
betaO betal betal 0.8664 lambda 0.7685 0.972 (a) Why is this model nonlinear? Explain brieﬂy.
(b) Estimate the expected Tension if the proportion weight of Sulfur = 0.8. (c) Test the null hypothesis that the expected Tension has no dependence on Sulfur. Is the test
significant at level .05? (It is ok to use an approximate test). ((1) Note that there are two observations for each value of Sulfur. How could we use this fact to
test for lack of fit? Explain what additional computations you would do and how you would
decide whether there was significant lack of fit at, say, the .05 level of significance. You don’t
have to do the additional calculations, just describe with formulas and give numerical values for
any degrees of freedom and cutoff values for test statistics. Statistics 425 Final Exam December 10, 2007 Upper critical values of Student's t distribution with l/degrees of freedom
Probability of exceeding the critical value
V 0.10 0.05 0.025 0.01 0.005 0.001 1. 3.078 6 314 12.706 31.821 63.657 318 313
2. 1.886 2 920 4.303 6.965 9.925 22.327
3. 1.638 2 353 3.182 4.541 5.841 10.215
4. 1.533 2 132 2.776 3.747 4.604 7.173
5. 1.476 2 015 2.571 3.365 4.032 5.893
6. 1.440 1 943 2.447 3.143 3.707 5.208
7. 1.415 1.895 2.365 2.998 3.499 4.782
8. 1.397 1.860 2.306 2.896 3.355 4.499
9. 1.383 1.833 2.262 2.821 3.250 4.296
10 1.372 1.812 2.228 2.764 3.169 4.143
11. 1.363 1 796 2.201 2.718 3.106 4.024
12. 1.356 1.782 2.179 2.681 3.055 3.929
13 1.350 1.771 2.160 2.650 3.012 3.852
14. 1.345 1 761 2.145 2.624 2.977 3.787
15. 1.341 1 753 2.131 2.602 2.947 3.733
16. 1.337 1 746 2.120 2.583 2.921 3.686
17. 1.333 1.740 2.110 2.567 2.898 3.646
18. 1.330 1 734 2.101 2.552 2.878 3.610
19. 1.328 1 729 2.093 2.539 2.861 3.579
20. 1.325 1.725 2.086 2.528 2.845 3.552 5% significance level (.05 probability of exceeding the critical value) lib(vhbb) \ VI 1 2 3 4 5 6 7 8 9 10
L3
1 161.448 199.500 215.707 224.583 230.162 233.986 236.768 238.882 240.543 241.882
2 18.513 19.000 19.164 19.247 19.296 19.330 19.353 19.371 19.385 19.396
3 10.128 9.552 9.277 9.117 9.013 8.941 8.887 8.845 8.812 8.786
4 7.709 6.944 6.591 6.388 6.256 6.163 6.094 6.041 5.999 5.964
5 6.608 5.786 5.409 5.192 5.050 4.950 4.876 4.818 4.772 4.735
6 5.987 5.143 4.757 4.534 4.387 4.284 4.207 4.147 4.099 4.060
7 5.591 4.737 4.347 4.120 3.972 3.866 3.787 3.726 3.677 3.637
8 5.318 4.459 4.066 3.838 3.687 3.581 3.500 3.438 3.388 3.347
9 5.117 4.256 3.863 3.633 3.482 3.374 3.293 3.230 3.179 3.137
10 4.965 4.103 3.708 3.478 3.326 3.217 3.135 3.072 3.020 2.978
11 4.844 3.982 3.587 3.357 3.204 3.095 3.012 2.948 2.896 2.854
12 4.747 3.885 3.490 3.259 3.106 2.996 2.913 2.849 2.796 2.753
13 4.667 3.806 3.411 3.179 3.025 2.915 2.832 2.767 2.714 2.671
14 4.600 3.739 3.344 3.112 2.958 2.848 2.764 2.699 2.646 2.602
15 4.543 3.682 3.287 3.056 2.901 2.790 2.707 2.641 2.588 2.544
16 4.494 3.634 3.239 3.007 2.852 2.741 2.657 2.591 2.538 2.494
17 4.451 3.592 3.197 2.965 2.810 2.699 2.614 2.548 2.494 2.450
18 4.414 3.555 3.160 2.928 2.773 2.661 2.577 2.510 2.456 2.412
19 4.381 3.522 3.127 2.895 2.740 2.628 2.544 2.477 2.423 2.378
20 4.351 3.493 3.098 2.866 2.711 2.599 2.514 2.447 2.393 2.348 ...
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 Regression Analysis

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