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We wish to find a transformation of the form \( V = \left( {{Y}^{\lambda } - 1}\right) /\left( {\lambda {\dot{Y}}^{\lambda - 1}}\right) \) for \( \lambda \neq 0 \), or \( V = \dot{Y}\ln Y \) for \( \lambda = 0 \), that will allow a good first-order fit without need for second-order terms. | We first fit the model\n\n\[ V = {\beta }_{0} + {\beta }_{1}f + {\beta }_{2}p + {\beta }_{11}{f}^{2} + {\beta }_{22}{p}^{2} + {\beta }_{12}{pf} + \epsilon ,\]\n\n(13.3.2)\n\nwhere, as before, \( f \) is the filler level and \( p \) is the plasticizer level, for a series of chosen values of \( \lambda \) . [Fitting in c... | Yes |
The data in Table 14.1 show turkey weights \( \left( Y\right) \) in pounds, and ages \( \left( X\right) \) in weeks, of 13 Thanksgiving turkeys. Four of these turkeys were reared in Georgia \( \left( G\right) \), four in Virginia \( \left( V\right) \), and five in Wisconsin \( \left( W\right) \) . We would like to rela... | Suppose we first regress \( Y \) against \( X \) to give the fitted equation \( \widehat{Y} = {1.98} + {0.4167X} \) . The residuals from this fit are, in order, \( - {0.4}, - {1.4}, - {0.2}, - {0.8}, - {1.0}, - {0.8}, - {0.5} \) , \( - {1.0},{0.8},{1.0},{1.3},{1.5},{1.4} \) . When plotted in sets according to origin, t... | Yes |
Is the dummy variable scheme below a workable one for dealing with possible level differences among six groups? | The determinant value is -1 , so the system will work. Evaluating determinants is not difficult, and rules for quick reduction may be found in matrix algebra books. For our purposes, it is usually simplest to use the computer to find the determinant via the so-called eigenvalues, the values \( {\lambda }_{1},{\lambda }... | No |
Example 3. Another workable system in the same context as Example 2 would consist of columns \( {Z}_{0} = {X}_{0},{Z}_{i} = {X}_{0} + {X}_{i}, i = 1,2,\ldots ,5 \) . This would lead to the following scheme: | <table><thead><tr><th>Group</th><th>\( {Z}_{0} \)</th><th>\( {Z}_{1} \)</th><th>\( {Z}_{2} \)</th><th>\( {Z}_{3} \)</th><th>\( {Z}_{4} \)</th><th>\( {Z}_{5} \)</th></tr></thead><tr><td>1</td><td>1</td><td>2</td><td>1</td><td>1</td><td>1</td><td>1</td></tr><tr><td>2</td><td>1</td><td>1</td><td>2</td><td>1</td><td>1</td>... | Yes |
To the five data points \( \left( {X, Y}\right) = \left( {2,2}\right) ,\left( {3,3}\right) ,\left( {5,4}\right) ,\left( {9,6}\right) ,\left( {{12},7}\right) \), we fit a straight line. Minimizing (25.6.3), using the rreg option in MINITAB, for example, gives | \[ {\widehat{Y}}_{r} = {1.25} + {0.5X} \] The least squares fit (which uses weights of 1 for all the observations) is \[ \widehat{Y} = {1.3701} + {0.48870X} \] The best horizontal straight line fit is \( \widehat{Y} = \bar{Y} = {4.4} \) . (This is not a good fit but provides another fit to compare.) The weights \( \lef... | Yes |
Example 2. Steam data (Appendix 1A) with \( Y = {\beta }_{0} + {\beta }_{8}{X}_{8} + \epsilon \) . The two fits are as follows. | <table><thead><tr><th rowspan=\ | No |
Example 3. Full set of steam data (Appendix 1A) with \( Y = {\beta }_{0} + {\beta }_{2}{X}_{2} + \cdots + \) \( {\beta }_{10}{X}_{10} + \epsilon \) . The two fits are again quite similar. | <table><thead><tr><th rowspan=\ | No |
Example 4. Hald data (Appendix 15A), fitting \( Y = {\beta }_{0} + {\beta }_{1}{X}_{1} + {\beta }_{2}{X}_{2} + \epsilon \) . Again the two fits are quite similar. | <table><thead><tr><th rowspan=\ | No |
Definition 2.4. Suppose that the response variable \( Y \) and at least one predictor variable \( {x}_{i} \) are quantitative. Then the multiple linear regression (MLR) model is\n\n\[ \n{Y}_{i} = {x}_{i,1}{\beta }_{1} + {x}_{i,2}{\beta }_{2} + \cdots + {x}_{i, p}{\beta }_{p} + {e}_{i} = {\mathbf{x}}_{i}^{T}\mathbf{\bet... | In matrix notation, these \( n \) equations become\n\n\[ \n\mathbf{Y} = \mathbf{X}\mathbf{\beta } + \mathbf{e} \]\n\n(2.2)\n\nwhere \( \mathbf{Y} \) is an \( n \times 1 \) vector of dependent variables, \( \mathbf{X} \) is an \( n \times p \) matrix of predictors, \( \mathbf{\beta} \) is a \( p \times 1 \) vector of un... | Yes |
Proposition 2.1. Suppose that the regression estimator \( \mathbf{b} \) of \( \mathbf{\beta } \) is used to find the residuals \( {r}_{i} \equiv {r}_{i}\left( \mathbf{b}\right) \) and the fitted values \( {\widehat{Y}}_{i} \equiv {\widehat{Y}}_{i}\left( \mathbf{b}\right) = {\mathbf{x}}_{i}^{T}\mathbf{b} \) . Then in th... | Proof. The identity line in the response plot is \( Y = {\mathbf{x}}^{T}\mathbf{b} \) . Hence the vertical deviation is \( {Y}_{i} - {\mathbf{x}}_{i}^{T}\mathbf{b} = {r}_{i}\left( \mathbf{b}\right) \) . | Yes |
To use the response plot to visualize the conditional distribution of \( Y \mid {\mathbf{x}}^{T}\mathbf{\beta } \) , use the fact that the fitted values \( \widehat{Y} = {\mathbf{x}}^{T}\widehat{\mathbf{\beta }} \) . For example, suppose the height given fit \( = {1700} \) is of interest. Mentally examine the plot abou... | Cases 3, 44, and 63 are highlighted. The 3rd person was very tall while the 44th person was rather short. Beginners often label too many points as outliers. Mentally draw a box about the bulk of the data ignoring any outliers. Double the width of the box (about the identity line for the response plot and about the hori... | No |
Proposition 2.2. Suppose that \( \mathbf{X} \) is an \( n \times p \) matrix of full rank \( p \) . Then\n\na) \( \mathbf{H} \) is symmetric: \( \mathbf{H} = {\mathbf{H}}^{T} \) .\n\nb) \( \mathbf{H} \) is idempotent: \( \mathbf{H}\mathbf{H} = \mathbf{H} \) . | Proof. a) \( {\mathbf{X}}^{T}\mathbf{X} \) is symmetric since \( {\left( {\mathbf{X}}^{T}\mathbf{X}\right) }^{T} = {\mathbf{X}}^{T}{\left( {\mathbf{X}}^{T}\right) }^{T} = {\mathbf{X}}^{T}\mathbf{X} \) . Hence \( {\left( {\mathbf{X}}^{T}\mathbf{X}\right) }^{-1} \) is symmetric since the inverse of a symmetric matrix is ... | Yes |
Proposition 2.3. Assume that a constant is in the MLR model. Then \( {SSTO} = {SSE} + {SSR} \) . | Proof.\n\n\[ \n{SSTO} = \mathop{\sum }\limits_{{i = 1}}^{n}{\left( {Y}_{i} - {\widehat{Y}}_{i} + {\widehat{Y}}_{i} - \bar{Y}\right) }^{2} = {SSE} + {SSR} + 2\mathop{\sum }\limits_{{i = 1}}^{n}\left( {{Y}_{i} - {\widehat{Y}}_{i}}\right) \left( {{\widehat{Y}}_{i} - \bar{Y}}\right) . \n\] \n\nHence the result follows if \... | Yes |
For the Gladstone (1905) data, the response variable \( Y = \) brain weight, \( {x}_{1} \equiv 1,{x}_{2} = \) size of head, \( {x}_{3} = \operatorname{sex},{x}_{4} = \) breadth of head, \( {x}_{5} = \) circumference of head. Assume that the response and residual plots look good and test whether at least one of the nont... | Solution: i) Ho: \( {\beta }_{2} = \cdots = {\beta }_{5} = 0 \) Ha: not Ho\nii) \( {F}_{o} = {196.24} \) from output.\niii) The pval \( = {0.0} \) from output.\niv) The pval \( < \delta \) ( \( = {0.05} \) since \( \delta \) was not given). So reject Ho. Hence there is an MLR relationship between brain weight and the p... | Yes |
Theorem 2.7. Assume that the MLR model has a constant \( {\beta }_{1} \). | a)\n\n\[ \n{F}_{o} = \frac{MSR}{MSE} = \frac{{R}^{2}}{1 - {R}^{2}}\frac{n - p}{p - 1}.\n\]\n\nb) If the errors \( {e}_{i} \) are iid \( N\left( {0,{\sigma }^{2}}\right) \), and if Ho: \( {\beta }_{2} = \cdots = {\beta }_{p} = 0 \) is true, then \( {F}_{o} \) has an \( F \) distribution with \( p - 1 \) numerator and \(... | Yes |
For the Buxton (1920) data suppose that the response \( Y \) \( = \) height and the predictors were a constant, head length, nasal height, bigo-nal breadth, and cephalic index. Five outliers were deleted leaving 82 cases. Figure 2.3 shows a response plot of the fitted values versus the response \( Y \) with the identit... | The plot was made with the following \( R \) commands, using the lregpack function piplot.\n\n\[ \mathrm{x} < - \operatorname{buxx}\left\lbrack {-\mathrm{c}\left( {{61},{62},{63},{64},{65}}\right) ,}\right\rbrack \]\n\n\[ Y < - \operatorname{buxy}\left\lbrack {-c\left( {{61},{62},{63},{64},{65}}\right) }\right\rbrack \... | No |
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. | Solution: 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 /{4... | 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 |
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 |
Example 3.9. 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 f... | 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 \( {D}_{\mathbf{0}} = {1.641} \) . 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 sele... | 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. | 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... | No |
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)\n\n\[ \mathbf{Z} = \mathbf{U}\mathbf{\beta } + \mathbf{\epsilon } \]\n\n(4.10)\n\nfollows 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\n\n\[ \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( ... | Yes |
Proposition 4.2. a) \n\n\[ \mathbf{Z} = \mathbf{U}\mathbf{\beta } + \mathbf{\epsilon } \] \n\n(4.11) \n\nfollows 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 \n\n\[ \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(... | 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... | Yes |
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 B12, the antibiotic was useful, with a mean gain near 1.6... | 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 designs, ... | 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)? | Solution: 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? | a) 28.875, since the \( D \) effect is the \( {AB} \) effect. | Yes |
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? | 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 valu... | 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.) | Yes |
a) Perform the test corresponding to \( A \) . | Solution: 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 |
Let \( p = 2 \) and let \( {\left( Y, X\right) }^{T} \) have a bivariate normal distribution. That is,\n\n\[ \left( \begin{array}{l} Y \\ X \end{array}\right) \sim {N}_{2}\left( {\left( \begin{array}{l} {\mu }_{Y} \\ {\mu }_{X} \end{array}\right) ,\left( \begin{matrix} {\sigma }_{Y}^{2} & \operatorname{Cov}\left( {Y, X... | Also \( {aX} + {bY} \) is univariate normal with mean \( a{\mu }_{X} + b{\mu }_{Y} \) and variance\n\n\[ {a}^{2}{\sigma }_{X}^{2} + {b}^{2}{\sigma }_{Y}^{2} + {2ab}\operatorname{Cov}\left( {X, Y}\right) . \] | 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 \( ... | A useful fact is that \( {\mathbf{a}}_{B} \) and \( {\mathbf{M}}_{B} \) do not depend on \( g \): \[ {\mathbf{a}}_{B} = \mathbf{\mu } - {\mathbf{M}}_{B}{\mathbf{B}}^{T}\mathbf{\mu } = \left( {{\mathbf{I}}_{p} - {\mathbf{M}}_{B}{\mathbf{B}}^{T}}\right) \mathbf{\mu }, \] and \[ {\mathbf{M}}_{B} = \mathbf{\sum }\mathbf{B}... | 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.\na) Assume that \( E\left\lbrack {\left( Y,{\mathbf{X}}^{T}\right) }^{T}\right\rbrack \) exists. Then \( E\left( {Y \mid \mathbf{X}}\right) = \alph... | 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 |
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\[ \mathbf{X} \sim \left( {1 - \gamma }\right) {N}_{p}\left( {\mathbf{\mu },\mathbf{\sum }}\right) + \gamma {N}_{p}\left( {\mathbf{\mu }, c\mathbf{\sum }}\right) \]\n\n... | 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... | The above theorem can be proved using Lemma 11.18 and using the fact that the sample covariance matrices are consistent estimators of the population covariance matrices. | No |
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, \[ \widehat{\mathbf{\beta }} = \mathbf{A}\mathbf{Y} \sim {N}_{p}\left( {\mathbf{A}E\left( \mathbf{Y}\right) ,\mathbf{A}\operatorname{Cov}\left( \mathbf{Y... | 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... | 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 ... | 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 |
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 |
For the popcorn data of Myers et al. (2002, p. 154), the response variable \( Y \) is the number of inedible popcorn kernels. The sample size was \( n = {15} \) and the predictor variables were temperature (coded as 5, 6, or 7), amount of oil (coded as 2,3, or 4), and popping time \( ({75},{90} \), or 105). One batch o... | then the Poisson regression model would suggest that temperature and time are important predictors, and overdispersion diagnostics such as the deviance would be greatly inflated. However, we probably need to delete the high temperature, low oil, and long popping time combination, to conclude that the response is indepe... | Yes |
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 |
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 |
The survival of 200 patients following admission to an intensive care unit was studied with logistic regression. The response variable was STA \( \left( {0 = \text{Lived,}1 = \text{Died}}\right) \). Predictors were AGE, SEX \( \left( {0 = \text{Male,}1 = \text{Female}}\right) \), RACE \( (1 = \) White, \( 2 = \) Black,... | Factors LOC and RACE had two indicator variables to model the three levels. The response plot in Figure 13.7 shows that the logistic regression model using the 19 predictors is useful for predicting survival, although the output has \( \widehat{\rho }\left( \mathbf{x}\right) = 1 \) or \( \widehat{\rho }\left( \mathbf{x... | 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 \( \widehat{\rho }\left( {AP}\right) \) equal to zero or one, but the response plot in Figure 13.10 suggests that the full model is useful for predicting survival. Note that the ten slice step function closely tracks the logistic curve. To visualize the model with the response plot, use \( Y \mid \mat... | 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 ... | Yes |
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