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a) Find \( {\widehat{Y}}_{f} \) if \( {X}_{f} = {60} \) . | Solution: In this example, \( {\mathbf{x}}_{f} = {\left( 1,{X}_{f}\right) }^{T} \) since a constant is in the output above. Thus \( {\widehat{Y}}_{f} = {\widehat{\beta }}_{1} + {\widehat{\beta }}_{2}{X}_{f} = - {2.93739} + {1.12359}\left( {60}\right) = {64.478} \) . | Yes |
Proposition 2.6. Let \( {R}^{2} \) and \( {R}_{R}^{2} \) be the multiple coefficients of determination for the full and reduced models, respectively. Let \( \widehat{\mathbf{Y}} \) and \( {\widehat{\mathbf{Y}}}_{R} \) be the vectors of fitted values for the full and reduced models, respectively. Then the test statistic... | \[ {F}_{R} = \left\lbrack \frac{{SSE}\left( R\right) - {SSE}\left( F\right) }{d{f}_{R} - d{f}_{F}}\right\rbrack /{MSE}\left( F\right) = \] \[ \left\lbrack \frac{\parallel \widehat{\mathbf{Y}}{\parallel }^{2} - {\begin{Vmatrix}{\widehat{\mathbf{Y}}}_{R}\end{Vmatrix}}^{2}}{d{f}_{R} - d{f}_{F}}\right\rbrack /{MSE}\left( F... | Yes |
Example 2.10. For the Buxton (1920) data, \( n = {76} \) after 5 outliers and 6 cases with missing values are removed. Assume that the response variable \( Y \) is height, and the explanatory variables are \( {x}_{2} = \) bigonal breadth, \( {x}_{3} = \) cephalic index, \( {x}_{4} = \) finger to ground, \( {x}_{5} = \)... | Solution: The 4 step partial \( \mathrm{F} \) test is shown below. i) Ho: the reduced model is good Ha: use the full model ii) \n\n\[ \n{F}_{R} = \left\lbrack \frac{{SSE}\left( R\right) - {SSE}\left( F\right) }{d{f}_{R} - d{f}_{F}}\right\rbrack /{MSE}\left( F\right) = \left\lbrack \frac{{200623.0} - {34267.4}}{{73} - {... | Yes |
Now assume that the reduced model uses the constant, sternal height, finger to ground, and head length. Using the output below, test whether the reduced model is good. | The 4 step partial F test follows.\ni) Ho: the reduced model is good Ha: use the full model ii)\n\n\[ \n{F}_{R} = \left\lbrack \frac{{SSE}\left( R\right) - {SSE}\left( F\right) }{d{f}_{R} - d{f}_{F}}\right\rbrack /{MSE}\left( F\right) = \left\lbrack \frac{{35030.1.0} - {34267.4}}{{72} - {69}}\right\rbrack /{496.629} \]... | Yes |
a) Find a \( {95}\% \) confidence interval for \( {\beta }_{2} \) corresponding to population. | The CI is \( {\widehat{\beta }}_{k} \pm {t}_{n - p,1 - \delta /2}\operatorname{se}\left( {\widehat{\beta }}_{k}\right) \) . Since \( n = {26},{df} = n - p = {26} - 3 = {23} \) . From the \( t \) -table at the end of Chapter 14, intersect the \( \mathrm{{df}} = {23} \) row with the column that is labelled by \( {95}\% \... | Yes |
When a model depends on the predictors \( \mathbf{x} \) only through the linear combination \( {\mathbf{x}}^{T}\mathbf{\beta } \), then \( {\mathbf{x}}^{T}\mathbf{\beta } \) is called a sufficient predictor and \( {\mathbf{x}}^{T}\widehat{\mathbf{\beta }} \) is called an estimated sufficient predictor (ESP). For OLS th... | To illustrate the OLS criterion graphically, consider the Gladstone (1905) data where we used brain weight as the response. A constant, \( {x}_{2} = {age},{x}_{3} = {sex} \), and \( {x}_{4} = {\left( \operatorname{size}\right) }^{1/3} \) were used as predictors after deleting five \ | No |
Proposition 2.10. The OLS estimator \( \widehat{\mathbf{\beta }} \) is the unique minimizer of the OLS criterion if \( \mathbf{X} \) has full rank \( p \leq n \) . | Proof: Seber and Lee (2003, pp. 36-37). Recall that the hat matrix \( \mathbf{H} = \) \( \mathbf{X}{\left( {\mathbf{X}}^{T}\mathbf{X}\right) }^{-1}{\mathbf{X}}^{T} \) and notice that \( {\left( \mathbf{I} - \mathbf{H}\right) }^{T} = \mathbf{I} - \mathbf{H} \), that \( \left( {\mathbf{I} - \mathbf{H}}\right) \mathbf{H} ... | Yes |
Consider the Gladstone (1905) data described in Example 2.5. If the file of data sets lregdata is downloaded into \( R \), then the ANOVA \( F \) statistic for testing \( {\beta }_{2} = \cdots = {\beta }_{4} = 0 \) can be found with the following commands. The command \( {lsft} \) adds a column of ones to \( x \) which... | The ANOVA \( F \) test can also be found with the no intercept model by adding a column of ones to the \( R \) matrix \( x \) and then performing the partial \( F \) test with the full model and the reduced model that only uses the column of ones. Notice that the \ | No |
Example 3.6. Two varieties of cement that replace sand with coal waste products were compared to a standard cement mix. The response \( Y \) was the compressive strength of the cement measured after \( 7,{28},{60},{90} \), or 180 days of curing time \( = {x}_{2} \) . This cement was intended for sidewalks and barriers ... | Figure 3.6 shows the response plot and residual plots from this model. The standard cement mix uses the symbol + while the coal based mixes use an inverted triangle and square. OLS lines based on each mix are added as visual aids. The lines from the two coal based mixes do not intersect, suggesting that there may not b... | No |
Proposition 3.1. Suppose that a numerical variable selection method suggests several submodels with \( k \) predictors, including a constant, where \( 2 \leq k \leq p \) .\n\na) The model \( I \) that minimizes \( {C}_{p}\left( I\right) \) maximizes \( \operatorname{corr}\left( {r,{r}_{I}}\right) \) .\n\n\[ \text{b)}{C... | Proof. These results are a corollary of Proposition 3.2 below. | No |
Proposition 3.2. Suppose that every submodel contains a constant and that \( \mathbf{X} \) is a full rank matrix. | Proof: Recall that \( \mathbf{H} \) and \( {\mathbf{H}}_{I} \) are symmetric idempotent matrices and that \( \mathbf{H}{\mathbf{H}}_{I} = {\mathbf{H}}_{I} \) . The mean of OLS fitted values is equal to \( \bar{Y} \) and the mean of OLS residuals is equal to 0 . If the OLS line from regressing \( z \) on \( w \) is \( \... | Yes |
The pollution data of McDonald and Schwing (1973) can be obtained from STATLIB or the text’s website. The response \( Y = \) mort is the mortality rate, and most of the independent variables were related to pollution. A scatterplot matrix of the first 9 predictors and \( Y \) was made and then a scatterplot matrix of t... | <table><thead><tr><th>Response</th><th>\( = \) MORT</th><th></th><th></th><th></th></tr><tr><th>Label</th><th>Estimate</th><th>Std. Error</th><th>t-value</th><th>p-value</th></tr></thead><tr><td>Constant</td><td>1881.11</td><td>442.628</td><td>4.250</td><td>0.0001</td></tr><tr><td>DENS</td><td>0.00296</td><td>0.00397</... | Yes |
The FF and RR plots can be used as a diagnostic for whether a given numerical method is including too many variables. Gladstone (1905) attempts to estimate the weight of the human brain (measured in grams after the death of the subject) using simple linear regression with a variety of predictors including age in years,... | Figure 3.7 shows the response plots and residual plots for the full model and the final submodel that used a constant, \( {siz}{e}^{1/3} \), age, and sex. The five cases separated from the bulk of the data in each of the four plots correspond to five infants. These may be outliers, but the visual separation reflects th... | Yes |
Cook and Weisberg (1999a, pp. 261, 371) describe a data set where rats were injected with a dose of a drug approximately proportional to body weight. The data set is included as the file rat. lsp in the Arc software and can be obtained from the website (www.stat.umn.edu/arc/). The response \( Y \) is the fraction of th... | We deleted this case and again searched for submodels. The \( {C}_{p} \) statistic is less than one for all three simple linear regression models, and the RR and FF plots look the same for all submodels containing a constant. Figure 2.2 shows the RR plot where the residuals from the full model are plotted against \( Y ... | No |
Theorem 3.3. Let the \( {100}\left( {1 - \delta }\right) \) th percentile \( {D}_{1 - \delta }^{2} \) be a continuity point of the distribution of \( {D}^{2} \) . Assume that \( {D}_{\mathbf{\mu }}^{2}\left( {{T}_{n},{\mathbf{\sum }}_{T}}\right) \overset{D}{ \rightarrow }{D}^{2},{D}_{\mathbf{\mu }}^{2}\left( {{T}_{n},{... | Proof: i) Note that \( {D}_{\mathbf{\mu }}^{2}\left( {{T}_{n},{\widehat{\mathbf{\sum }}}_{T}}\right) = {D}_{{T}_{n}}^{2}\left( {\mathbf{\mu },{\widehat{\mathbf{\sum }}}_{T}}\right) \) . Thus the probability that \( {R}_{c} \) contains \( \mathbf{\mu } \) is \( P\left( {{D}_{\mathbf{\mu }}^{2}\left( {{T}_{n},{\widehat{\... | Yes |
It was expected that \( \log \left( S\right) \) may be the only predictor needed, along with a constant, since \( \log \left( S\right) \) and \( \log \left( M\right) \) are both \( \log \left( \text{mass}\right) \) measurements and likely highly correlated. Hence we want to test \( {H}_{0} : {\beta }_{2} = {\beta }_{3}... | Using the prediction region method with the full model gave an interval \( \left\lbrack {0,{2.930}}\right\rbrack \) with \( {D}_{0} = {1.641} \) . Note that \( \sqrt{{\chi }_{3,{0.95}}^{2}} = \) 2.795. So fail to reject \( {H}_{0} \) . Using the prediction region method with the \( {I}_{\min } \) variable selection mod... | Yes |
Consider the Gladstone (1905) data set where the variables are as in Problem 3.6. Output is shown below for the full model and the bootstrapped minimum \( {C}_{p} \) forward selection estimator. Note that the shorth intervals for length and sex are quite long. These variables are often in and often deleted from the boo... | large sample full model inference for Ex. 3.12\n\n<table><thead><tr><th colspan=\ | No |
Definition 3.13. The ith squared Mahalanobis distance is | \[ {D}_{i}^{2} = {D}_{i}^{2}\left( {T\left( \mathbf{W}\right) ,\mathbf{C}\left( \mathbf{W}\right) }\right) = {\left( {\mathbf{u}}_{i} - T\left( \mathbf{W}\right) \right) }^{T}{\mathbf{C}}^{-1}\left( \mathbf{W}\right) \left( {{\mathbf{u}}_{i} - T\left( \mathbf{W}\right) }\right) \] | Yes |
High leverage outliers are a particular challenge to conventional numerical MLR diagnostics such as Cook's distance, but can often be visualized using the response and residual plots. The following techniques are useful for detecting outliers when the multiple linear regression model is appropriate. | 1. Find the OLS residuals and fitted values and make a response plot and a residual plot. Look for clusters of points that are separated from the bulk of the data and look for residuals that have large absolute values. Beginners frequently label too many points as outliers. Try to estimate the standard deviation of the... | Yes |
Consider the OLS model \( \mathbf{Y} = \mathbf{X}\mathbf{\beta } + \mathbf{e} \) where the \( {e}_{i} \) are iid with mean 0 and variance \( {\sigma }^{2} \) . Then \( \mathbf{Y} \) and \( \mathbf{e} \) are random vectors while \( \mathbf{a} = \mathbf{X}\mathbf{\beta } \) is a constant vector. Notice that \( E\left( \m... | Thus\n\n\[ E\left( \mathbf{Y}\right) = \mathbf{X}\mathbf{\beta } + E\left( \mathbf{e}\right) = \mathbf{X}\mathbf{\beta }.\]\n\nSince the \( {e}_{i} \) are iid,\n\n\[ \operatorname{Cov}\left( \mathbf{Y}\right) = \operatorname{Cov}\left( \mathbf{e}\right) = {\sigma }^{2}{\mathbf{I}}_{n} \]\n\n(4.4)\n\nwhere \( {\mathbf{I... | Yes |
Proposition 4.1. a) \[ \mathbf{Z} = \mathbf{U}\mathbf{\beta } + \mathbf{\epsilon } \] (4.10) follows the OLS model since \( E\left( \mathbf{\epsilon }\right) = \mathbf{0} \) and \( \operatorname{Cov}\left( \mathbf{\epsilon }\right) = {\sigma }^{2}{\mathbf{I}}_{n} \) . | Proof. a) \( E\left( \mathbf{\epsilon }\right) = {\mathbf{K}}^{-1}E\left( \mathbf{e}\right) = \mathbf{0} \) and \[ \operatorname{Cov}\left( \mathbf{\epsilon }\right) = {\mathbf{K}}^{-1}\operatorname{Cov}\left( \mathbf{e}\right) {\left( {\mathbf{K}}^{-1}\right) }^{T} = {\sigma }^{2}{\mathbf{K}}^{-1}\mathbf{V}{\left( {\m... | Yes |
Proposition 4.2. a) \[ \mathbf{Z} = \mathbf{U}\mathbf{\beta } + \mathbf{\epsilon } \] follows the OLS model since \( E\left( \mathbf{\epsilon }\right) = \mathbf{0} \) and \( \operatorname{Cov}\left( \mathbf{\epsilon }\right) = {\sigma }^{2}{\mathbf{I}}_{n} \) . | Proof. a) \( E\left( \mathbf{\epsilon }\right) = {\mathbf{R}}^{-1}E\left( \mathbf{e}\right) = \mathbf{0} \) and \[ \operatorname{Cov}\left( \mathbf{\epsilon }\right) = {\mathbf{R}}^{-1}\operatorname{Cov}\left( \mathbf{e}\right) {\left( {\mathbf{R}}^{-1}\right) }^{T} = {\sigma }^{2}{\mathbf{R}}^{-1}\mathbf{V}{\left( {\m... | Yes |
Inference for the GLS model \( \mathbf{Y} = \mathbf{X}\mathbf{\beta } + \mathbf{e} \) can be performed by using the partial \( F \) test for the equivalent no intercept OLS model \( \mathbf{Z} = \mathbf{U}\mathbf{\beta } + \mathbf{\epsilon } \) . | Following Section 2.10, create \( \mathbf{Z} \) and \( \mathbf{U} \), fit the full and reduced model using the \ | No |
Suppose that the data from Example 4.2 has valid weights, so that WLS can be used instead of FWLS. The \( R \) commands below perform WLS. | > ls.print(lsfit(dsx, dsy, wt=dsw))\n\nResidual Standard Error=1.137\n\nR-Square=0.9209\n\nF-statistic \( \\left( {{df} = 1,{33}}\\right) = {384.4139}, p - \) value \( = 0 \)\n\nEstimate Std.Err t-value \( \\Pr \\left( { > \\left| t\\right| }\\right) \)\n\nIntercept -0.8891 0.3004 -2.9602 0.0057\n\nX 1.1648 0.0594 19.6... | Yes |
Consider using randomization to assign the following nine people (units) to three treatment groups.\n\nCarroll, Collin, Crawford, Halverson, Lawes,\n\nStach, Wayman, Wenslow, Xumong | Balanced designs have the group sizes the same: \( {n}_{i} \equiv m = n/p \) . Label the units alphabetically so Carroll gets \( 1,\ldots \), Xumong gets 9 . The \( R \) function sample can be used to draw a random permutation. Then the first 3 numbers in the permutation correspond to group 1 , the next 3 to group 2 , ... | Yes |
a) Find \( {\widehat{\mu }}_{1} \) . | a) \( {\widehat{\mathbf{\beta }}}_{1c} = {\widehat{\mu }}_{1} = {\bar{Y}}_{10} = {Y}_{10}/{n}_{1} = \mathop{\sum }\limits_{{j = 1}}^{{n}_{1}}{Y}_{1j}/{n}_{1} = \left( {{64} + {72} + {68} + {77} + {56} + {95}}\right) /6 = {432}/6 = {72}. \) | Yes |
The mean nitrogen content of clover depends on the strain of clover (3dok1, 3dok5, 3dok7, compos, 3dok4, 3dok13). Recall that means \( {\mu }_{1} \) and \( {\mu }_{2} \) are significantly different if you can conclude that \( {\mu }_{1} \neq {\mu }_{2} \) while \( {\mu }_{1} \) and \( {\mu }_{2} \) are not significantl... | Notice that the strain of clover 3dok1 appears to have the highest mean nitrogen content. There are 4 pairs of means that are not significantly different. The letter B suggests 3dok5 and 3dok7, the letter C suggests 3dok7 and compos, the letter D suggests compos and 3dok4, while the letter E suggests 3dok4 and 3dok13 a... | No |
For the random effects one way Anova, interest is in whether \( {\mu }_{i} \equiv \mu \) for every \( {\mu }_{i} \) in \( \Lambda \) where the population \( \Lambda \) is not necessarily finite. | Note that if \( {\sigma }_{\mu }^{2} = 0 \), then \( {\mu }_{i} \equiv \mu \) for all \( {\mu }_{i} \in \Lambda \) . | No |
a) State whether this is a random or fixed effects one way Anova. Explain briefly. | a) Random effects since the beer brands were selected at random from a population of brands. | Yes |
Cobb (1998, pp. 200-212) describes an experiment on weight gain for baby pigs. The response \( Y \) was the average daily weight gain in pounds for each piglet (over a period of time). Factor \( A \) consisted of \( 0\mathrm{{mg}} \) of an antibiotic or \( {40}\mathrm{{mg}} \) an antibiotic, while factor \( B \) consis... | The interaction plot shown in Figure 6.1 suggests that there is an interaction. If no vitamin B12 is given, then the pigs given the antibiotic have less mean weight gain than the pigs not given the antibiotic. For pigs given the diet with \( 5\mathrm{{mg}} \) of \( \mathrm{B}{12} \), the antibiotic was useful, with a m... | Yes |
Suppose there are \( b \) blocks and \( n = {kb} \) . The one way Anova design randomly assigns \( b \) of the units to each of the \( k \) treatments. Blocking places a constraint on the randomization, since within each block of units, exactly one unit is randomly assigned to each of the \( k \) treatments. | Hence a one way Anova design would use the \( R \) command sample \( \left( \mathrm{n}\right) \) and the first \( b \) units would be assigned to treatment 1, the second \( b \) units to treatment \( 2,\ldots \), and the last \( b \) units would be assigned to treatment \( k \) . For the completely randomized block des... | No |
a) From the graphical Anova in Figure 7.1, were the blocks useful? | Solution: a) In Figure 7.1, the top dot plot is for the scaled block deviations. The leftmost dot corresponds to blocks 4 and 1 , the middle dot to block 3 and the rightmost dot to block 1 (see output from the lregpack function ganova2). Yes, the blocks were useful since some (actually all) of the dots corresponding to... | Yes |
Which type of seed has the highest germination failure rate? | a) A since A is on the top for blocks 2-5 and second for block 1. | No |
a) There were 24 pigs in each block. How were they assigned to the \( {24} = \) \( 4 \times 3 \times 2 \) runs (a run is an L, M, P combination forming a pig diet)? | a) Randomly. | No |
c) Do an appropriate 4 step test. | c) i) \( {H}_{0} : {\mu }_{1} = {\mu }_{2} = {\mu }_{3} = {\mu }_{4}\;{H}_{A} \) : not \( {H}_{0} \)\nii) \( {F}_{0} = {4.87} \)\niii) pval \( = {0.0476} \)\niv) Reject \( {H}_{0} \) . The mean yield depends on the fertilizer level. | Yes |
Example 7.7. Following Cobb (1998, p. 255), suppose there is a rectangular plot divided into 5 rows and 5 columns to form 25 subplots. There are 5 treatments which are 5 varieties of a plant, labelled \( 1,2,3,4,5 \) ; and the response \( Y \) is yield. Adjacent subplots tend to give similar yields under identical trea... | Suppose \( 2,4,3,5,1 \) is the permutation drawn for rows. The middle Latin square with randomized rows has 1st row which is the 2nd row from the original unrandomized Latin square. The middle square has 2nd row that is the 4th row from the original, the 3rd row is the 3rd row from the original, the 4th row is the 5th ... | Yes |
a) Which two combinations of levels are the best? | a) The two lowest values of \( y \) are 0.8 and 1.2 which correspond to \( + + + \) and \( + + - \) . (Note that if the 1.2 was 4.2, then \( + + + \) corresponding to 0.8 would be a critical mix.) | No |
a) For fixed levels of \( R, T \), and \( C \), is the \( D + \) level or \( D - \) level of \( \mathrm{D} \) better (compare run 1 with run 9,2 with \( {10},\ldots ,8 with 16). | a) \( D + \) since for fixed levels of \( R, T \), and \( C \), the number of cracks is lower if \( D = + \) than if \( D = - \) . | Yes |
a) Find the \( A \) effect. | The \( A \) effect \( = \n\n\[ \n\frac{-{6.333} + {4.667} - 9 + {6.667} - {4.333} + {2.333} - {7.333} + {4.667}}{4} = - {8.665}/4 \n\] \n\n\( = - {2.16625} \) . Note that the appropriate + and - signs are obtained from the \( A \) column. | Yes |
a) What is the \( D \) effect? | 28.875, since the \( D \) effect is the \( {AB} \) effect. | No |
a) Estimate the \( B \) effect.\nb) Estimate the \( D \) effect. | Solution: a)\n\n\[ \frac{-{86.8} - {85.9} + {79.4} + {60} - {94.6} - {85.4} + {84.5} + {80.3}}{4} \]\n\n\( = - {48.5}/4 = - {12.125} \).\n\nb) Use Remark 8.3 to see that the \( D \) effect corresponds to the \( {AB} \) column. So the \( D \) effect \( = \)\n\n\[ \frac{{86.8} - {85.9} - {79.4} + {60} + {94.6} - {85.4} -... | Yes |
a) Which effects, if any, appear to be significant from the QQ plot? | a) The most significant effects are either in the top right or bottom left corner. Although the points do not all scatter closely about the line, the point in the bottom left is not significant. So none of the effects corresponding to the bottom left of the plot are significant. \( A \) is the significant effect with v... | Yes |
a) Perform the test corresponding to \( A \) . | Solution: a) Ho: \( {\mu }_{100} = \cdots = {\mu }_{800}\; \) Ha: not Ho \( {F}_{A} = {1.232} \)\npval \( = {0.3421} \)\nFail to reject Ho, the mean rate of germination does not depend on variety. (This test would make more sense if there was no variety * treatment interaction.) | No |
a) Perform the test corresponding to \( A \) . | a) Ho: \( {\mu }_{100} = \cdots = {\mu }_{300}\; \) Ha: not Ho\n\n\( {F}_{A} = {0.65} \)\n\npval \( = {0.5412} \)\n\nFail to reject Ho, the mean yield does not depend on variety. | Yes |
Example 9.3. Cobb (1998, pp. 200-212) describes an experiment on weight gain for baby pigs. The response \( Y \) was the average daily weight gain in pounds for each piglet (over a period of time). Factor \( A \) consisted of \( 0\mathrm{{mg}} \) of an antibiotic or \( {40}\mathrm{{mg}} \) of an antibiotic while factor... | Solution: \( A \) and \( B \) are crossed with each combination of \( \left( {A, B}\right) \) levels forming a diet. So the two way Anova (or \( {2}^{2} \) factorial) design was used. | Yes |
Briefly explain why this experiment did not give much information about how the software will behave on many images. | More images are needed, 1 image is not enough. | No |
Proposition 10.3. The conditional distribution of an MVN is MVN. If \( \mathbf{X} \sim {N}_{p}\left( {\mathbf{\mu },\mathbf{\sum }}\right) \), then the conditional distribution of \( {\mathbf{X}}_{1} \) given that \( {\mathbf{X}}_{2} = {\mathbf{x}}_{2} \) is multivariate normal with mean \( {\mathbf{\mu }}_{1} + {\math... | \[ {\mathbf{X}}_{1} \mid {\mathbf{X}}_{2} = {\mathbf{x}}_{2} \sim {N}_{q}\left( {{\mathbf{\mu }}_{1} + {\mathbf{\sum }}_{12}{\mathbf{\sum }}_{22}^{-1}\left( {{\mathbf{x}}_{2} - {\mathbf{\mu }}_{2}}\right) ,{\mathbf{\sum }}_{11} - {\mathbf{\sum }}_{12}{\mathbf{\sum }}_{22}^{-1}{\mathbf{\sum }}_{21}}\right) . \] | Yes |
Lemma 10.4. Let \( \mathbf{X} \) be a \( p \times 1 \) random vector with 1st moments; i.e. \( E\left( \mathbf{X}\right) \) exists. Let \( \mathbf{B} \) be any constant full rank \( p \times r \) matrix where \( 1 \leq r \leq p \) . Then \( \mathbf{X} \) is elliptically contoured iff for all such conforming matrices \(... | \[ E\left( {\mathbf{X} \mid {\mathbf{B}}^{T}\mathbf{X}}\right) = \mathbf{\mu } + {\mathbf{M}}_{B}{\mathbf{B}}^{T}\left( {\mathbf{X} - \mathbf{\mu }}\right) = {\mathbf{a}}_{B} + {\mathbf{M}}_{B}{\mathbf{B}}^{T}\mathbf{X} \] (10.11) where the \( p \times 1 \) constant vector \( {\mathbf{a}}_{B} \) and the \( p \times r \... | No |
Proposition 10.5. Let \( \mathbf{X} \sim E{C}_{p}\left( {\mathbf{\mu },\mathbf{\sum }, g}\right) \) and assume that \( E\left( \mathbf{X}\right) \) exists.\n\na) Any subset of \( \mathbf{X} \) is EC, in particular \( {\mathbf{X}}_{1} \) is EC. | Proof of a). Let \( \mathbf{A} \) be an arbitrary full rank \( q \times r \) matrix where \( 1 \leq r \leq q \ ).\n\nLet\n\[ \mathbf{B} = \left( \begin{array}{l} \mathbf{A} \\ \mathbf{0} \end{array}\right) \]\n\nThen \( {\mathbf{B}}^{T}\mathbf{X} = {\mathbf{A}}^{T}{\mathbf{X}}_{1} \), and\n\n\[ E\left\lbrack {\mathbf{X... | Yes |
Proposition 10.6. Let \( {\left( Y,{\mathbf{X}}^{T}\right) }^{T} \) be \( E{C}_{p + 1}\left( {\mathbf{\mu },\mathbf{\sum }, g}\right) \) where \( Y \) is a random variable.\n\na) Assume that \( E\left\lbrack {\left( Y,{\mathbf{X}}^{T}\right) }^{T}\right\rbrack \) exists. Then \( E\left( {Y \mid \mathbf{X}}\right) = \al... | Proof. a) The trick is to choose \( \mathbf{B} \) so that Lemma 10.4 applies. Let\n\n\[ \mathbf{B} = \left( \begin{matrix} {\mathbf{0}}^{T} \\ {\mathbf{I}}_{p} \end{matrix}\right) \]\n\nThen \( {\mathbf{B}}^{T}\mathbf{\sum }\mathbf{B} = {\mathbf{\sum }}_{XX} \) and\n\n\[ \mathbf{\sum }\mathbf{B} = \left( \begin{array}{... | Yes |
This example illustrates another application of Lemma 10.4. Suppose that \( \mathbf{X} \) comes from a mixture of two multivariate normals with the same mean and proportional covariance matrices. That is, let\n\n\[ \mathbf{X} \sim \left( {1 - \gamma }\right) {N}_{p}\left( {\mathbf{\mu },\mathbf{\sum }}\right) + \gamma ... | Since the multivariate normal distribution is elliptically contoured\n\n\[ E\left( {\mathbf{X} \mid {\mathbf{B}}^{T}\mathbf{X}}\right) = \left( {1 - \gamma }\right) \left\lbrack {\mathbf{\mu } + {\mathbf{M}}_{1}{\mathbf{B}}^{T}\left( {\mathbf{X} - \mathbf{\mu }}\right) }\right\rbrack + \gamma \left\lbrack {\mathbf{\mu ... | Yes |
The contours of constant density for the \( {N}_{p}\left( {\mathbf{\mu },\mathbf{\sum }}\right) \) distribution are hyperellipsoid boundaries of the form \( {\left( \mathbf{x} - \mathbf{\mu }\right) }^{T}{\mathbf{\sum }}^{-1}\left( {\mathbf{x} - \mathbf{\mu }}\right) = {a}^{2} \) . An \( \alpha \) -density region \( {R... | \[ \left\{ {\mathbf{x} : {\left( \mathbf{x} - \mathbf{\mu }\right) }^{T}{\mathbf{\sum }}^{-1}\left( {\mathbf{x} - \mathbf{\mu }}\right) \leq {\chi }_{p}^{2}\left( \alpha \right) }\right\} = \left\{ {\mathbf{x} : {D}_{\mathbf{x}}^{2}\left( {\mathbf{\mu },\mathbf{\sum }}\right) \leq {\chi }_{p}^{2}\left( \alpha \right) }... | Yes |
Theorem 11.6. If \( \mathbf{A} \) and \( \mathbf{B} \) are symmetric matrices and \( \mathbf{A}\mathbf{Y} ⫫ \mathbf{B}\mathbf{Y} \), then \( {\mathbf{Y}}^{T}\mathbf{A}\mathbf{Y} ⫫ {\mathbf{Y}}^{T}\mathbf{B}\mathbf{Y} \) . | Proof. Let \( g\left( {\mathbf{A}\mathbf{Y}}\right) = {\mathbf{Y}}^{T}{\mathbf{A}}^{T}{\mathbf{A}}^{ - }\mathbf{A}\mathbf{Y} = {\mathbf{Y}}^{T}\mathbf{A}{\mathbf{A}}^{ - }\mathbf{A}\mathbf{Y} = {\mathbf{Y}}^{T}\mathbf{A}\mathbf{Y} \), and let \( h\left( {\mathbf{B}\mathbf{Y}}\right) = {\mathbf{Y}}^{T}{\mathbf{B}}^{T}{\... | No |
Theorem 11.7. Let \( \mathbf{Y} \sim {N}_{n}\left( {\mathbf{\mu },\mathbf{\sum }}\right) \) . Let \( \mathbf{u} = \mathbf{A}\mathbf{Y} \) and \( \mathbf{w} = \mathbf{B}\mathbf{Y} \) . Then \( \mathbf{A}\mathbf{Y} ⫫ \mathbf{B}\mathbf{Y} \) iff \( \operatorname{Cov}\left( {\mathbf{u},\mathbf{w}}\right) = \mathbf{A}\mathb... | Proof. Note that\n\n\[ \left( \begin{array}{l} u \\ w \end{array}\right) = \left( \begin{array}{l} {AY} \\ {BY} \end{array}\right) = \left( \begin{array}{l} A \\ B \end{array}\right) Y \]\n\nhas a multivariate normal distribution. Hence \( \mathbf{A}\mathbf{Y} ⫫ \mathbf{B}\mathbf{Y} \) iff \( \operatorname{Cov}\left( {... | Yes |
Theorem 11.9. Let \( \mathbf{A} = {\mathbf{A}}^{T} \) be symmetric.\na) Let \( \mathbf{Y} \sim {N}_{n}\left( {\mathbf{0},\mathbf{I}}\right) \) . Then \( {\mathbf{Y}}^{T}\mathbf{A}\mathbf{Y} \sim {\chi }_{r}^{2} \) iff \( \mathbf{A} \) is idempotent of rank \( r = \operatorname{tr}\left( \mathbf{A}\right) \) since then ... | Note that a) is a special case of c). To see that b) holds, note \( \mathbf{Z} = \mathbf{Y}/\sigma \sim \) \( {N}_{n}\left( {\mathbf{0},\mathbf{I}}\right) \) . Hence by a)\n\n\[ \frac{{\mathbf{Y}}^{T}\mathbf{A}\mathbf{Y}}{{\sigma}^{2}} = {\mathbf{Z}}^{T}\mathbf{A}\mathbf{Z} \sim {\chi }_{r}^{2} \]\n\niff \( \mathbf{A} ... | No |
Theorem 11.11. If \( \mathbf{Y} \sim {N}_{n}\left( {\mathbf{\mu },\mathbf{\sum }}\right) \) with \( \mathbf{\sum } > 0 \), then the population squared Mahalanobis distance \( {\left( \mathbf{Y} - \mathbf{\mu }\right) }^{T}{\mathbf{\sum }}^{-1}\left( {\mathbf{Y} - \mathbf{\mu }}\right) \sim {\chi }_{n}^{2} \) . | Proof. Let \( \mathbf{Z} = {\mathbf{\sum }}^{1/2}\left( {\mathbf{Y} - \mathbf{\mu }}\right) \sim {N}_{n}\left( {\mathbf{0},\mathbf{I}}\right) \) . Then \( \mathbf{Z} = {\left( {Z}_{1},\ldots ,{Z}_{n}\right) }^{T} \) where the \( {Z}_{i} \) are iid \( N\left( {0,1}\right) \) . Hence \( {\left( \mathbf{Y} - \mathbf{\mu }... | Yes |
Theorem 11.16. Let \( \mathbf{\theta } = \mathbf{X}\mathbf{\eta } \in C\left( \mathbf{X}\right) \) where \( {Y}_{i} = {\mathbf{x}}_{i}^{T}\mathbf{\eta } + {r}_{i}\left( \mathbf{\eta }\right) \) and the residual \( {r}_{i}\left( \mathbf{\eta }\right) \) depends on \( \mathbf{\eta } \) . The least squares estimator \( \w... | Proof. Following Seber and Lee (2003, pp. 36-36), let \( \widehat{\mathbf{Y}} = \widehat{\mathbf{\theta }} = \mathbf{P}\mathbf{X}\mathbf{Y} \in \) \( C\left( \mathbf{X}\right) ,\mathbf{r} = \left( {\mathbf{I} - \mathbf{P}\mathbf{X}}\right) \mathbf{Y} \in {\left\lbrack C\left( \mathbf{X}\right) \right\rbrack }^{ \bot } ... | Yes |
Theorem 11.17. Let \( \mathbf{Y} = \mathbf{X}\mathbf{\beta } + \mathbf{e} = \widehat{\mathbf{Y}} + \mathbf{r} \) where \( \mathbf{X} \) is full rank, and \( \mathbf{Y} \sim {N}_{n}\left( {\mathbf{X}\mathbf{\beta },{\sigma }^{2}\mathbf{I}}\right) \) . Then the MLE of \( \mathbf{\beta } \) is the least squares estimator ... | Proof. The likelihood function is\n\n\[ L\left( {\mathbf{\beta },{\sigma }^{2}}\right) = {\left( 2\pi {\sigma }^{2}\right) }^{-n/2}\exp \left( {\frac{-1}{2{\sigma }^{2}}\parallel \mathbf{y} - \mathbf{X}\mathbf{\beta }{\parallel }^{2}}\right) .\n\]\n\nFor fixed \( {\sigma }^{2} \), maximizing the likelihood is equivalen... | Yes |
Theorem 11.19. Second way to compute \( \widehat{\beta } \) : Suppose that \( {\mathbf{w}}_{i} = \) \( {\left( {Y}_{i},{\mathbf{u}}_{i}^{T}\right) }^{T} \) are iid random vectors such that \( {\sigma }_{Y}^{2},{\mathbf{\sum }}_{\mathbf{u}}^{-1} \), and \( {\mathbf{\sum }}_{\mathbf{u}Y} \) exist. Then \( {\widehat{\beta... | \[ {\widehat{\mathbf{\beta }}}_{S} = \frac{n}{n - 1}{\widehat{\mathbf{\sum }}}_{\mathbf{u}}^{-1}\widetilde{\mathbf{\sum }}{\mathbf{u}}_{Y} = {\widetilde{\mathbf{\sum }}}_{\mathbf{u}}^{-1}\widetilde{\mathbf{\sum }}{\mathbf{u}}_{Y} = {\widehat{\mathbf{\sum }}}_{\mathbf{u}}^{-1}\widehat{\mathbf{\sum }}{\mathbf{u}}_{Y}\ove... | Yes |
Theorem 11.20. Let \( \mathbf{Y} = \mathbf{X}\mathbf{\beta } + \mathbf{e} = \widehat{\mathbf{Y}} + \mathbf{r} \) where \( \mathbf{X} \) is full rank, \( E\left( \mathbf{e}\right) = \) \( \mathbf{0} \), and \( \operatorname{Cov}\left( \mathbf{e}\right) = {\sigma }^{2}\mathbf{I} \) . Let \( \mathbf{P} = \mathbf{P}\mathbf... | Proof. i) \( {\mathbf{X}}^{T}\mathbf{r} = {\mathbf{X}}^{T}\left( {\mathbf{I} - \mathbf{P}}\right) Y = \mathbf{0}\mathbf{Y} = \mathbf{0} \), while ii) and iii) are immediate. \[ \text{iv)}\operatorname{Cov}\left( {\mathbf{r},\widehat{\mathbf{Y}}}\right) = E\left( {\left\lbrack {\mathbf{r} - E\left( \mathbf{r}\right) }\r... | Yes |
Theorem 11.21. Gauss Markov Theorem. Let \( \mathbf{Y} = \mathbf{X}\mathbf{\beta } + \mathbf{e} \) where \( \mathbf{X} \) is full rank, \( E\left( \mathbf{e}\right) = \mathbf{0} \), and \( \operatorname{Cov}\left( \mathbf{e}\right) = {\sigma }^{2}\mathbf{I} \) . Then \( {\mathbf{a}}^{T}\widehat{\mathbf{\beta }} \) is t... | Proof. (Guttman (1982, pp. 141-142):) Note that \( {\mathbf{a}}^{T}{\left( {\mathbf{X}}^{T}\mathbf{X}\right) }^{-1}{\mathbf{X}}^{T}\mathbf{Y} = \) \( {\mathbf{a}}^{T}\widehat{\mathbf{\beta }} = {\mathbf{d}}^{T}\mathbf{Y} \) is linear, and by Theorem 11.20 vi), \( V\left( {{\mathbf{a}}^{T}\widehat{\mathbf{\beta }}}\righ... | Yes |
Theorem 11.22. Suppose \( \mathbf{Y} = \mathbf{X}\mathbf{\beta } + \mathbf{e} \) where \( \mathbf{X} \) is full rank, \( \mathbf{e} \sim \) \( {N}_{n}\left( {\mathbf{0},{\sigma }^{2}\mathbf{I}}\right) \), and \( \mathbf{Y} \sim {N}_{n}\left( {\mathbf{X}\mathbf{\beta },{\sigma }^{2}\mathbf{I}}\right) \). a) \( \widehat{... | Proof. Let \( \mathbf{P} = \mathbf{P}\mathbf{X} \). a) Since \( \mathbf{A} = {\left( {\mathbf{X}}^{T}\mathbf{X}\right) }^{-1}{\mathbf{X}}^{T} \) is a constant matrix,\n\n\[ \widehat{\mathbf{\beta }} = \mathbf{A}\mathbf{Y} \sim {N}_{p}\left( {\mathbf{A}E\left( \mathbf{Y}\right) ,\mathbf{A}\operatorname{Cov}\left( \mathb... | Yes |
Theorem 11.24. If \( {W}_{n} \sim {F}_{r,{d}_{n}} \) where the positive integer \( {d}_{n} \rightarrow \infty \) as \( n \rightarrow \infty \), then \( r{W}_{n}\overset{D}{ \rightarrow }{\chi }_{r}^{2} \) . | Proof. If \( {X}_{1} \sim {\chi }_{{d}_{1}}^{2} ⫫ {X}_{2} \sim {\chi }_{{d}_{2}}^{2} \), then\n\n\[ \frac{{X}_{1}/{d}_{1}}{{X}_{2}/{d}_{2}} \sim {F}_{{d}_{1},{d}_{2}} \]\n\nIf \( {U}_{i} \sim {\chi }_{1}^{2} \) are iid, then \( \mathop{\sum }\limits_{{i = 1}}^{k}{U}_{i} \sim {\chi }_{k}^{2} \) . Let \( {d}_{1} = r \) a... | Yes |
Theorem 12.2. \( {\mathbf{S}}_{r} = {\mathbf{\sum }}_{\mathbf{\epsilon }} + {O}_{P}\left( {n}^{-1/2}\right) \) and \( \frac{1}{n}\mathop{\sum }\limits_{{i = 1}}^{n}{\mathbf{\epsilon }}_{i}{\mathbf{\epsilon }}_{i}^{T} = {\mathbf{\sum }}_{\mathbf{\epsilon }} + \) \( {O}_{P}\left( {n}^{-1/2}\right) \) if the following thr... | Proof. Note that \( {\mathbf{y}}_{i} = {\mathbf{B}}^{T}{\mathbf{x}}_{i} + {\mathbf{\epsilon }}_{i} = {\widehat{\mathbf{B}}}^{T}{\mathbf{x}}_{i} + {\widehat{\mathbf{\epsilon }}}_{i} \) . Hence \( {\widehat{\mathbf{\epsilon }}}_{i} = {\left( \mathbf{B} - \widehat{\mathbf{B}}\right) }^{T}{\mathbf{x}}_{i} + \) \( {\epsilon... | Yes |
Theorem 12.3. Let \( a > 0 \) and assume that \( \left( {{\widehat{\mathbf{\mu }}}_{n},{\widehat{\mathbf{\sum }}}_{n}}\right) \) is a consistent estimator of \( \left( {\mathbf{\mu }, a\mathbf{\sum }}\right) \). a) \( {D}_{\mathbf{x}}^{2}\left( {{\widehat{\mathbf{\mu }}}_{n},{\widehat{\mathbf{\sum }}}_{n}}\right) - \fr... | Proof. Let \( {B}_{n} \) denote the subset of the sample space on which \( {\widehat{\mathbf{\sum }}}_{n} \) has an inverse. Then \( P\left( {B}_{n}\right) \rightarrow 1 \) as \( n \rightarrow \infty \). Now \[ {D}_{\mathbf{x}}^{2}\left( {{\widehat{\mathbf{\mu }}}_{n},{\widehat{\mathbf{\sum }}}_{n}}\right) = {\left( \m... | Yes |
Theorem 12.4. Suppose \( {\mathbf{y}}_{i} = E\left( {{\mathbf{y}}_{i} \mid {\mathbf{x}}_{i}}\right) + {\mathbf{\epsilon }}_{i} = {\widehat{\mathbf{y}}}_{i} + {\widehat{\mathbf{\epsilon }}}_{i} \) where \( \operatorname{Cov}\left( {\mathbf{\epsilon }}_{i}\right) = \) \( {\mathbf{\sum }}_{\mathbf{\epsilon }} > 0 \), and ... | \[ {D}_{i}^{2} \equiv {D}_{i}^{2}\left( {{\widehat{\mathbf{y}}}_{f},{\widehat{\mathbf{\sum }}}_{\mathbf{\epsilon }}}\right) = {\left( {\widehat{\mathbf{z}}}_{i} - {\widehat{\mathbf{y}}}_{f}\right) }^{T}{\widehat{\mathbf{\sum }}}_{\mathbf{\epsilon }}^{-1}\left( {{\widehat{\mathbf{z}}}_{i} - {\widehat{\mathbf{y}}}_{f}}\r... | Yes |
Theorem 12.5. For multivariate linear regression, when least squares is used to compute \( {\widehat{\mathbf{y}}}_{f},{\mathbf{S}}_{r} \), and the pseudodata \( {\widehat{\mathbf{z}}}_{i} \), prediction region (12.6) is the Olive (2013a) nonparametric prediction region applied to the \( {\widehat{\mathbf{z}}}_{i} \) . | Proof. Multivariate linear regression with least squares satisfies Theorem 12.4 by Su and Cook (2012). (See Theorem 12.7.) Let \( \left( {T,\mathbf{C}}\right) \) be the sample mean and sample covariance matrix (defined above Definition 10.7) applied to the \( {\widehat{\mathbf{z}}}_{i} \) . The sample mean and sample c... | Yes |
Theorem 12.6. The Hotelling-Lawley trace statistic\n\n\[ U\left( \mathbf{L}\right) = \frac{1}{n - p}{\left\lbrack \operatorname{vec}\left( \mathbf{L}\widehat{\mathbf{B}}\right) \right\rbrack }^{T}\left\lbrack {{\widehat{\mathbf{\sum }}}_{\mathbf{\epsilon }}^{-1} \otimes {\left( \mathbf{L}{\left( {\mathbf{X}}^{T}\mathbf... | Proof. Using the Searle (1982, p. 333) identity\n\n\( \operatorname{tr}\left( {\mathbf{A}{\mathbf{G}}^{T}\mathbf{D}\mathbf{G}\mathbf{C}}\right) = {\left\lbrack \operatorname{vec}\left( \mathbf{G}\right) \right\rbrack }^{T}\left\lbrack {\mathbf{C}\mathbf{A} \otimes {\mathbf{D}}^{T}}\right\rbrack \left\lbrack {\operatorn... | Yes |
Theorem 12.8. If assumption D1 holds and if \( {H}_{0} \) is true, then \( \left( {n - p}\right) U\left( \mathbf{L}\right) \overset{D}{ \rightarrow }{\chi }_{rm}^{2} \) | Proof. By Theorem 12.7, \( \sqrt{n}\operatorname{vec}\left( {\widehat{\mathbf{B}} - \mathbf{B}}\right) \overset{D}{ \rightarrow }{N}_{pm}\left( {\mathbf{0},{\mathbf{\sum }}_{\mathbf{\epsilon }} \otimes \mathbf{W}}\right) \). Then under \( {H}_{0},\sqrt{n}\operatorname{vec}\left( {\mathbf{L}\widehat{\mathbf{B}}}\right) ... | Yes |
a) Do the MANOVA \( F \) test. | a) i) \( {H}_{0} \) : the nontrivial predictors are not needed in the mreg model \( {H}_{1} \) : at least one of the nontrivial predictors is needed\nii) \( {F}_{0} = {295.071} \niii) pval \( = 0 \niv) Reject \( {H}_{0} \), the nontrivial predictors are needed in the mreg model. | Yes |
a) First use the multivariate location and dispersion model for this data. | Figure 12.1 shows a scatterplot matrix of the data and Figure 12.2 shows a DD Plot of the Mussels Data, MLD Model. These plots suggest that the data may come from an elliptically contoured distribution that is not multivariate normal. The semiparametric and nonparametric 90% prediction regions consist of the cases belo... | No |
Theorem 13.1. Assume that \( Y \) is binary and that \( \operatorname{Cov}\left( \mathbf{x}\right) = \mathbf{\sum }\mathbf{x} \) is nonsingular. Let \( \left( {{\widehat{\alpha }}_{OLS},{\widehat{\mathbf{\beta }}}_{OLS}}\right) \) be the OLS estimator found from regressing \( Y \) on a constant and \( \mathbf{x} \) (us... | Proof. From Theorem 11.19, \[ {\widehat{\mathbf{\beta }}}_{OLS} = {\widetilde{\mathbf{\sum }}}_{\mathbf{x}}^{-1}{\widetilde{\mathbf{\sum }}}_{\mathbf{x}Y}\overset{P}{ \rightarrow }{\mathbf{\beta }}_{OLS}\text{ as }\mathrm{n} \rightarrow \infty \] \[ \text{and}\widetilde{\mathbf{\sum }}{\mathbf{x}}_{\mathbf{Y}Y} = \frac... | Yes |
Schaaffhausen (1878) gives data on skulls at a museum. The 1st 47 skulls are humans while the remaining 13 are apes. The response variable ape is 1 for an ape skull. The response plot in Figure 13.1 a) uses the predictor face length. The model fits very poorly since the probability of a 1 decreases then increases. | The response plot in Figure 13.1b) uses the predictor head height and perfectly classifies the data since the ape skulls can be separated from the human skulls with a vertical line at ESP \( = 0 \) . | Yes |
Definition 13.11. The Poisson regression (PR) model states that \( {Y}_{1},\ldots ,{Y}_{n} \) are independent random variables with \( {Y}_{i} \sim \operatorname{Poisson}\left( {\mu \left( {\mathbf{x}}_{\mathrm{i}}\right) }\right) \) where \( \mu \left( {\mathbf{x}}_{i}\right) = \exp \left( {\alpha + {\mathbf{\beta }}^... | To see that the PR model is a GLM, assume that \( Y \) is a Poisson \( \left( \mu \right) \) random variable. For a one parameter family, take \( a\left( \phi \right) \equiv 1 \) . Then the pmf of \( Y \) is\n\n\[ f\left( y\right) = P\left( {Y = y}\right) = \frac{{e}^{-\mu }{\mu }^{y}}{y!} = \underset{k\left( \mu \righ... | Yes |
For the crab data, the response \( Y \) is the number of satellites (male crabs) near a female crab. The sample size \( n = {173} \) and the predictor variables were the color, spine condition, caparice width, and weight of the female crab. Agresti (2002, pp. 126-131) first uses Poisson regression, and then uses the NB... | Figure 13.5a suggests that there is one case with an unusually large value of the ESP. The lowess curve does not track the exponential curve all that well. Figure 13.5b suggests that overdispersion is present since the vertical scale is about 10 times that of the horizontal scale and too many of the plotted points are ... | No |
a) Predict \( \widehat{\rho }\left( \mathbf{x}\right) \) if height \( = {x}_{1} = {65} \) and size \( = {x}_{2} = {3500} \). | Solution: a) \( {ESP} = \widehat{\alpha } + {\widehat{\beta }}_{1}{x}_{1} + {\widehat{\beta }}_{2}{x}_{2} = - {6.26111} - {0.0536078}\left( {65}\right) + \) \( {0.0028215}\left( {3500}\right) = {0.1296} \) . So\n\n\[ \widehat{\rho }\left( \mathbf{x}\right) = \frac{{e}^{ESP}}{1 + {e}^{ESP}} = \frac{1.1384}{1 + {1.1384}}... | Yes |
Suppose that \( \mathrm{Y} \) is a 1 or 0 depending on whether the person is or is not credit worthy. Let \( {x1} \) through \( {x6} \) be the predictors and use the following output to perform a 4 step deviance test. | Solution: i) Ho: \( {\beta }_{1} = \cdots = {\beta }_{6}\; \) Ha: not Ho\n\nii) \( {G}^{2}\left( {0 \mid F}\right) = {1221.73} - {1158.22} = {63.51} \)\n\niii) Now \( {df} = {999} - {993} = 6 \), and comparing 63.51 with \( {\chi }_{6,{0.999}}^{2} = {22.46} \) shows that the pval \( = 0 < 1 - {0.999} = {0.001} \).\n\ni... | Yes |
a) Predict \( \widehat{\rho }\left( x\right) \) if \( x = {40.0} \) . | a) \( \exp \left\lbrack {ESP}\right\rbrack = \exp \left\lbrack {\widehat{\alpha } + \widehat{\beta }\left( {40}\right) }\right\rbrack = \exp \left\lbrack {-{5.84211} + {0.103606}\left( {40}\right) }\right\rbrack = \) \( \exp \left\lbrack {-{1.69787}}\right\rbrack = {0.1830731} \) . So\n\n\[ \widehat{\rho }\left( \mathb... | Yes |
a) Predict \( \widehat{\mu }\left( \mathbf{x}\right) \) if bombload \( = {x}_{1} = {7.0} \), exper \( = {x}_{2} = {80.2} \), and type \( = {x}_{3} = {1.0} \) . | Solution: a) \( {ESP} = \widehat{\alpha } + {\widehat{\beta }}_{1}{x}_{1} + {\widehat{\beta }}_{2}{x}_{2} + {\widehat{\beta }}_{3}{x}_{3} = - {0.406023} + {0.165426}\left( 7\right) - \) \( \begin{matrix} {0.0135223}\left( {80.2}\right) \; + \;{0.568773}\left( 1\right) \; = \;{0.2362}.\;\text{So}\;\widehat{\mu }\left( x... | Yes |
The following output is for forward selection and backward elimination. All models use a constant. For forward selection, the min AIC model uses \( \{ \mathrm{F}\} \mathrm{{LOC}} \), TYP, AGE, CAN, SYS, PCO, and PH. Model \( {I}_{I} \) uses \{F\}LOC, TYP, AGE, CAN, and SYS. Let model \( I \) use \{F\}LOC, TYP, AGE, and... | Forward Selection comment\n\nBase terms: (\{F\}LOC TYP)\n\nDeviance Pearson X2 | k AIC > min AIC + 7\n\nAdd:AGE 141.873 187.84 | 5 151.873\n\nBase terms: (\{F\}LOC TYP AGE)\n\nDeviance Pearson X2| k AIC < min AIC + 7\n\nAdd:CAN 134.595 170.367 | 6 146.595\n\n(\{F\}LOC TYP AGE CAN) could be a good model\n\nBase terms: (... | No |
Which two models are the best candidates for the final submodel? Explain briefly why each of the other 2 submodels should not be used. | Solution: B2 and B3 are best. B1 has too many predictors with rather large p-values. For B4, the AIC is too high and the corr and p-value are too low. | No |
Which model is the best candidate for the final submodel? Explain briefly why each of the other 3 submodels should not be used. | Solution: P2 is best. P1 has too many predictors with large pvalues and more predictors than the minimum AIC model. P3 and P4 have corr and pvalue too low and AIC too high. | No |
For the ICU data of Example 13.13, a binary generalized additive model was fit with unspecified functions for AGE, SYS, and HRA, and linear functions for the remaining 16 variables. Output suggested that functions for SYS and HRA are linear but the function for AGE may be slightly curved. Several cases had \( \widehat{... | A binary logistic regression was also fit, and Figure 13.11 shows the plot of EAP versus ESP. The plot shows that the near zero and near one probabilities are handled differently by the GAM and GLM, but the estimated success probabilities for the two models are similar: \( \widehat{\rho }\left( {ESP}\right) \approx \wi... | Yes |
For binary data, Kay and Little (1987) suggest examining the two distributions \( x\left| {Y = 0\text{and}x}\right| Y = 1 \) . Use predictor \( x \) if the two distributions are roughly symmetric with similar spread. Use \( x \) and \( {x}^{2} \) if the distributions are roughly symmetric with different spread. Use \( ... | The Gladstone (1905) data is useful for illustrating these suggestions. The response was gender with \( Y = 1 \) for male and \( Y = 0 \) for female. The predictors were age, height, and the head measurements circumference, length, and size. When the GAM was fit without \( \log \left( \text{age}\right) \) or \( \log \l... | Yes |
Wood (2006, pp. 82-86) describes heart attack data where the response \( Y \) is the number of heart attacks for \( {m}_{i} \) patients suspected of suffering a heart attack. The enzyme \( {ck} \) (creatine kinase) was measured for the patients and it was determined whether the patient had a heart attack or not. A bino... | Hence a GLM with the single predictor \( \log \left( {ck}\right) \) was fit. Figure 13.13 shows the EE plot, and Figure 13.14 shows the response plot where the \( {Z}_{i} = {Y}_{i}/{m}_{i} \) track the logistic curve closely. There was no evidence of overdispersion and the model had AIC \( = {33.45} \) . The GAM using ... | No |
The species data is from Cook and Weisberg (1999a, pp. 285-286) and Johnson and Raven (1973). The response variable is the total number of species recorded on each of 29 islands in the Galápagos Archipelago. Predictors include area of island, areanear \( = \) the area of the closest island, the distance to the closest ... | A negative binomial regression suggested that only \( \log \left( \text{endem}\right) \) was needed in the model, and had a deviance of 26.12 on 27 degrees of freedom. The residual plot for this model was roughly ellipsoidal. The negative binomial GAM with \( \log \left( \text{endem}\right) \) had an \( \widehat{S} \) ... | Yes |
Consider the \( m = {25} \) residuals given in Table 1.2. We first arrange these in ascending order, giving due regard to sign: | To obtain a full normal plot of these we set \( m = {25} \) in Eq. (2A.2) and successively set \( i = 1,2,\ldots, m \) to give the ordinate values:\n\n\[ 2,6,{10},{14},{18},{22},{26},{30},{34},{38},{42},{46},{50},{54},{58},{62},{66},{70},{74},{78},{82},{86},{90},{94},{98}\text{.} \]\n\nThese ordinate values are associa... | No |
The two fits to all the data \( \left( {X = \text{heelstick,}Y = \text{catheter}}\right) \) are \( \widehat{Y} = {2.786} + {0.8805X} \) , and \( \widehat{X} = {4.210} + {0.7870Y} \), which we can invert to the form \( \widehat{Y} = - {5.349} + {1.2706}\widehat{X} \) . The geometric mean functional relationship is thus ... | Both individual regressions indicate that the second observation is influential, however, and a plot of the data indicates we might consider dropping it. The two fits to the remaining 19 observations give\n\n\[ \widehat{Y} = - {1.628} + {1.1147X} \]\n\nand\n\n\[ \widehat{X} = {5.482} + {0.70462Y} \]\n\nwhich we can inv... | Yes |
Find the sum of squares due to the least squares estimates \( \mathbf{b} = {\left( {\mathbf{X}}^{\prime }\mathbf{X}\right) }^{-1}{\mathbf{X}}^{\prime }\mathbf{Y} \) in the nonsingular case. | Here \( {\mathbf{C}}^{\prime } = {\left( {\mathbf{X}}^{\prime }\mathbf{X}\right) }^{-1}{\mathbf{X}}^{\prime } \), so that \( \mathbf{z} \) is any solution to\n\n\[ \n{\left( {\mathbf{X}}^{\prime }\mathbf{X}\right) }^{-1}{\mathbf{X}}^{\prime }\mathbf{X}{\left( {\mathbf{X}}^{\prime }\mathbf{X}\right) }^{-1}\mathbf{z} = {... | No |
Find \( \operatorname{SS}\left( \bar{Y}\right) \) | We see that \( \bar{Y} = \sum {Y}_{i}/n = {\mathbf{c}}^{\prime }\mathbf{Y} \), where\n\n\[{\mathbf{c}}^{\prime } = \left( {\frac{1}{n},\frac{1}{n},\ldots ,\frac{1}{n}}\right)\]\n\nThus \( {\mathbf{c}}^{\prime }\mathbf{c} = 1/n \) and \( \operatorname{SS}\left( \bar{Y}\right) = {\bar{Y}}^{2}/\left( {1/n}\right) = n{\bar... | Yes |
For a straight line fit, find \( \operatorname{SS}\left( {b}_{1}\right) \) . | We write \( {b}_{1} = {\mathbf{c}}^{\prime }\mathbf{Y} \), where the \( i \) th element of \( {\mathbf{c}}^{\prime } \) is \( \left( {{X}_{i} - \bar{X}}\right) /{S}_{XX} \) . Thus \( {\mathbf{c}}^{\prime }\mathbf{c} = 1/{S}_{XX} \) and \( \operatorname{SS}\left( {b}_{1}\right) = {S}_{XX}{b}_{1}^{2} = {S}_{XY}^{2}/{S}_{... | Yes |
For a straight line fit, find \( \operatorname{SS}\left( {b}_{0}\right) \) . | Now \( {b}_{0} = \bar{Y} - {b}_{1}\bar{X} \) so that the \( i \) th element of \( {\mathbf{c}}^{\prime } \) is\n\n\[ \n{c}_{i} = \frac{1}{n} - \frac{\bar{X}\left( {{X}_{i} - \bar{X}}\right) }{{S}_{XX}} \]\n\nand\n\n\[ \n{\mathbf{c}}^{\prime }\mathbf{c} = \mathop{\sum }\limits_{{i = 1}}^{n}\left\{ {\frac{1}{{n}^{2}} - \... | Yes |
The residuals from a straight line fit to \( n = {50} \) pairs of values of \( \left( {X, Y}\right) \) gave rise to a \( d \) -statistic of value \( d = {0.625} \) . For a two-sided test of \( {H}_{1} : \rho = 0 \) against the two-sided alternative \( \rho \neq 0 \) we first compare \( d \) and \( 4 - d = {3.375} \) ag... | From the \( \alpha = {0.01} \) table, with \( k = 1 \) and \( n = \) 50 , we find\n\n\[ d = {0.625} < {d}_{L} = {1.32}\text{.} \]\n\nIt follows, from applying procedure (3), that we reject \( {H}_{0} \) at the \( {2\alpha } = {0.02} \) level and conclude that there does appear to be serial correlation, of the type test... | Yes |
Example 2. Model: \( E\left( Y\right) = {\beta }_{0} + {\beta }_{1}{X}_{1} + {\beta }_{2}{X}_{2} + \cdots + {\beta }_{k}{X}_{k} \). | \[ {H}_{0} : {\beta }_{1} = 0 \] \[ {\beta }_{2} = 0 \] \[ {\beta }_{k} = 0\;\left( {k\text{ linear functions, all independent }}\right) . \] | No |
Example 4 (General Case). Model: \( E\left( Y\right) = {\beta }_{0} + {\beta }_{1}{X}_{1} + {\beta }_{2}{X}_{2} + \cdots + {\beta }_{k}{X}_{k} \). | \[ {H}_{0} : {c}_{10}{\beta }_{0} + {c}_{11}{\beta }_{1} + {c}_{12}{\beta }_{2} + \cdots + {c}_{1k}{\beta }_{k} = 0, \] \[ {c}_{20}{\beta }_{0} + {c}_{21}{\beta }_{1} + {c}_{22}{\beta }_{2} + \cdots + {c}_{2k}{\beta }_{k} = 0, \] \[ {c}_{m0}{\beta }_{0} + {c}_{m1}{\beta }_{1} + {c}_{m2}{\beta }_{2} + \cdots + {c}_{mk}{... | Yes |
Suppose we postulate the model\n\n\[ E\left( Y\right) = {\beta }_{0} + {\beta }_{1}X \]\n\nbut the model\n\n\[ E\left( Y\right) = {\beta }_{0} + {\beta }_{1}X + {\beta }_{11}{X}^{2} \]\n\nis actually the true response function, unknown to us. If we use observations of \( Y \) at \( X = - 1,0 \), and 1 to estimate \( {\... | The true model, in terms of the observations, is\n\n\[ E\left( \mathbf{Y}\right) = E\left\lbrack \begin{array}{l} {Y}_{1} \\ {Y}_{2} \\ {Y}_{3} \end{array}\right\rbrack = \left\lbrack \begin{array}{rrr} 1 & \mathbf{X} & {\mathbf{X}}^{2} \\ 1 & - 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \end{array}\right\rbrack \left\lbrack \beg... | Yes |
Suppose the postulated model is\n\n\[ E\left( Y\right) = {\beta }_{0} + {\beta }_{1}X \]\n\nbut the true model is actually\n\n\[ E\left( Y\right) = {\beta }_{0} + {\beta }_{1}X + {\beta }_{11}{X}^{2} + {\beta }_{111}{X}^{3}. \]\n\nWhat biases are induced by taking observations at\n\n\[ X = - 3, - 2, - 1,0,1,2,3\text{?}... | We find\n\n\[ {\mathbf{X}}_{1} = \left\lbrack \begin{array}{rr} 1 & - 3 \\ 1 & - 2 \\ 1 & - 1 \\ 1 & 0 \\ 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{array}\right\rbrack ,\;{\mathbf{X}}_{2} = \left\lbrack \begin{array}{rr} 9 & - {27} \\ 4 & - 8 \\ 1 & - 1 \\ 0 & 0 \\ 1 & 1 \\ 4 & 8 \\ 9 & {27} \end{array}\right\rbrack \]\n\n\[ {\left... | Yes |
For the steam data fit of \( Y \) on \( {Y}_{8} \) in Chapter 1: Max \( {\widehat{Y}}_{i} = {11.38} \) , Min \( {\widehat{Y}}_{i} = {7.50};n = {25}, p = 2,{s}^{2} = {0.7923} \) . So the ratio is | \[ \left( {{11.38} - {7.50}}\right) /\{ 2\left( {0.7923}\right) /{25}{\} }^{1/2} = {3.88}/{0.252} = {15.4}\text{.} \] Thus the range of predicted values is over 15 times the average standard error of prediction, clearly satisfactory. This is obviously consistent with the fact that the \( F \) -ratio of 57.54 is 13.4 ti... | Yes |
For the steam data fit of \( Y \) on \( {X}_{8} \) and \( {X}_{6} \) in Section 6.2, the ratio (11.1.2) is\n\n\[ \left( {{11.56} - {6.29}}\right) /\{ 3\left( {0.4377}\right) /{25}{\} }^{1/2} = {5.27}/{0.2292} = {23.0}\text{.} \] | Alternatively, the \( F \) ratio of 61.90 is 17.99 times as great as the \( {95}\% \) point \( F\left( {2,{22},{0.95}}\right) = {3.44} \) . Both numbers indicate a useful predictive ability for this fit. | Yes |
We wish to find a transformation of the form \( \left( {{Y}^{\lambda } - 1}\right) /\left( {\lambda {\dot{Y}}^{\lambda - 1}}\right) \) for \( \lambda \neq 0 \), or \( \dot{Y}\ln Y \) for \( \lambda = 0 \), which will provide a good first-order fit to the data and leave satisfactory residuals. | Table 13.3 shows selected values of \( S\left( {\lambda ,\mathbf{V}}\right) \) for various \( \lambda \) . [An initial set of calculations for \( \lambda = 2\left( {0.1}\right) 2 \) was followed by a finer division \( \lambda = - {0.2}\left( {0.01}\right) {0.1} \) near the bottom of the curve.] A smooth curve through t... | Yes |
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