1Stat 511 Exam 1 February 23, 2004 Prof. Vardeman 1. Consider an instance of the linear model for 5n= observations, 112123234345510 011111 111111 1yyyyyεβεβεβεε  −−  =+− −    a) This is a full rank model. One way to easily see this is to argue that rows 1, 2, and 3 of the model matrix are linearly independent (and so the rank is at least 3). Put these three rows into a 33× matrix M and show this matrix is non-singular by arguing that =Mc 0 implies that =c0. Notice that the columns of X are perpendicular, so that ′XX is diagonal. b) In a Gauss-Markov version of this model, which of the parameters 12 3, , or βββ can be estimated with the greatest precision? Explain carefully. c) Compute a matrix XP that projects any element of 5ℜ onto ()C X (in a perpendicular fashion).2d) In a Gauss-Markov version of this model, which row of the X matrix represents a set of conditions under which Ey can be estimated with the best precision? Explain carefully. For the next two parts of this question (parts e) and f)), suppose that Y is such that 3SSE = and ()OLS5, 6, 2′=b . Consider an analysis under the normal version of the Gauss Markov model. e) In the future, two new observations, new1 new 2 and yy are going to be observed under the conditions described respectively by the 1st and 2nd rows of the X matrix. Give 95% two-sided prediction limits for new1 new 2yy− . (Plug correct numbers into correct formulas, but do not take time to do arithmetic.) f) Write the hypothesis 01 2 1 3H:EE and EEyy yy== in testable form 0H: =Cβ 0 for an appropriate matrix C (write out such a matrix) and compute an F statistic for testing this (you need not do the arithmetic, but plug correct numbers into a correct formula).3g) Consider an Aitken version of the model on page 1, where ()diag 1, ,1,1,δδ=V for δ small. Generalized least squares estimation of β under these circumstances will essentially force 21232ˆyyβββ=−−≈ and 51235ˆyyβββ=++≈. This is ()125/2yyβ≈+ and ()23 52/2yyββ+≈ − . Take these approximations as given and find estimates of 23and ββ if ()15, 4, 6,8,10′=Y . 2. Attached to this exam is a printout of an R session for a time series analysis (via an ordinary linear model) of 6 years worth of quarterly retail sales data (for the JC Penney Company). For consecutive 3-month periods that we will simply label as 1, 2, , 24t=… we’ll model sales in period tyt= as roughly linearly increasing in t , but with different “effects” for the 4 quarters of the year. That is, with ()1 if period is from the th quarter of the year0 otherwise itiqt= for 1, 2,3, 4i = we consider a model ()()()()0 1 11 22 33 44t ty t qt qt qt qtββγγγγε=++ + + + + for 1, 2, , 24t = … the values 0 11234,,,,,ββγγ γ γ unknown constants and the 12 24,,,εεε… iid normal ()20,σ. (Period 1t = is a first quarter period.) You may use the printout to answer the following questions. Refer very carefully to where you find anything you take from the printout (give page and location on the page). a) Is the parametric function 12γγ− estimable in this model? Argue this very carefully. (Write the X matrix for the first 5 periods below and use it in your argument.)4Since the model as originally posed is not full rank, a call to R’s lm() function introduces a restriction in order to produce a full rank version. The restriction used by R in this case is to set to 0 the coefficient for the last column of the model matrix entered in the function call. That is, R fits the model ()()()** * * *0 1 11 22 33ttytqtqtqtββγ γ γ ε=+ + + + + b) In this model, find 90% two-sided confidence limits for σ. (No need to simplify after plugging in.) c) Give 95% two-sided confidence limits for **12γγ−. (No need to simplify after plugging in.) d) Give 95% prediction limits for 28y (the retail sales in the 4th quarter of the year after the end of the data in hand) based on this model. (Plug correct numbers into a correct formula, but you need not do arithmetic.)5Stat 511 Exam I Spring 2004 Printout > data sales t q1 q2 q3 q4 [1,] 4452 1 1 0 0 0 [2,] 4507 2 0 1 0 0 [3,] 5537 3 0 0 1 0 [4,] 8157 4 0 0 0 1 [5,] 6481 5 1 0 0 0 [6,] 6420 6 0 1 0 0 [7,] 7208 7 0 0 1 0 [8,] 9509 8 0 0 0 1 [9,] 6755 9 1 0 0 0 [10,] 6483 10 0 1 0 0 [11,] 7129 11 0 0 1 0 [12,] 9072 12 0 0 0 1 [13,] 7339 13 1 0 0 0 [14,] 7104 14 0 1 0 0 [15,] 7639 15 0 0 1 0 [16,] 9661 16 0 0 0 1 [17,] 7528 17 1 0 0 0 [18,] 7207 18 0 1 0 0 [19,] 7538 19 0 0 1 0 [20,] 9573 20 0 0 0 1 [21,] 7522 21 1 0 0 0 [22,] 7211 22 0 1 0 0 [23,] 7729 23 0 0 1 0 [24,] 9542 24 0 0 0 1 > JCPenney<-lm(sales~t+q1+q2+q3+q4) > summary(JCPenney) Call: lm(formula = sales ~ t + q1 + q2 + q3 + q4) Residuals: Min 1Q Median 3Q Max -1232.1 -274.0 157.3 332.1 853.9 Coefficients: (1 not defined because of singularities) Estimate Std. Error t value Pr(>|t|) (Intercept) 7858.76 331.26 23.724 1.40e-15 *** t 99.54 16.93 5.878 1.16e-05 *** q1 -2274.21 331.12 -6.868 1.49e-06 *** q2 -2564.58 328.94 -7.796 2.45e-07 *** q3 -2022.79 327.63 -6.174 6.22e-06 *** q4 NA NA NA NA --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 Residual standard error: 566.7 on 19 degrees of freedom Multiple R-Squared: 0.8683, Adjusted R-squared: 0.8405 F-statistic: 31.3 on 4 and 19 DF, p-value: 4.014e-086> vcov(JCPenney) (Intercept) t q1 q2 q3 (Intercept) 109733.448 -4014.6383 -65572.4262 -61557.7879 -57543.1495 t -4014.638 286.7599 860.2796 573.5198 286.7599 q1 -65572.426 860.2796 109637.8613 55249.0705 54388.7908 q2 -61557.788 573.5198 55249.0705 108204.0619 54102.0310 q3 -57543.150 286.7599 54388.7908 54102.0310 107343.7823 > predict(JCPenney) 1 2 3