exam1at430sol

STAT 425 Exam 1 @ 4:30 pm October 4, 2023, 4:30pm Name: SOLUTIONS Netid: _________________________ This is an 80 minute handwritten exam. There are 5 problems, each worth 10 points. Do not start working until your proctor tells you to start. Your head must be visible to your proctor on Zoom along with your screen and the work in front of you. You may use a calculator but not the computer or R or the internet for your work. To work on the exam you can either: 1. Print out the exam and do your handwritten work on the exam itself; or 2. View the exam on your screen and do work on separate sheets of paper. Clearly label your work as to which problem number (1,2,3..) and part (a,b,c..) you are solving; or 3. Do work on a blank file or pdf of the exam on a tablet, and upload your work file from the tablet. In this case the proctor must be able to see your tablet. Scanning and uploading your exam: After you finish, scan/photograph each page and upload into Moodle in the same way you upload assignment files. You are allowed two one-sided 8.5 by 11 inch sheets of notes for yourself. Scan and upload after the exam. 1

Problem 1. (3 parts) Data will be collected in the form ( x 1 , y 1 ) , ( x 2 , y 2 ) , . . . , ( x n , y n ) , where x i is the i th value of a fixed, nonrandom explanatory variable, and y i is the corresponding random response. Consider the model y i = β 0 + β 1 x i + e i , i = 1 , . . . , n for unknown parameters β 0 and β 1 and random errors e 1 , . . . , e n that are independent with mean zero and variances equal to an unknown constant σ 2 . The ordinary least squares estimators for β 0 and β 1 are given by ˆ β 0 = ¯ y − ¯ x ˆ β 1 and ˆ β 1 = QQQQQQQ n i =1 ( x i − ¯ x )( y i − ¯ y ) QQQQQQQ n i =1 ( x i − ¯ x ) 2 , where ¯ x = 1 n n YYYYYYY i =1 x i and ¯ y = 1 n n YYYYYYY i =1 y i . (a) (3 pts) If x 1 = 3 . 5 and x 2 = 5 . 0 , find E ( y 2 ) − E ( y 1 ) in terms of the model parameters. E ( y 2 ) − E ( y 1 ) = ( β 0 + 5 β 1 ) − ( β 0 + 3 . 5 β 1 ) = 1 . 5 β 1 (b) (3 pts) Find an explicit expression for E ( ¯ y ) in terms of the model parameters and predictor variables. E (¯ y ) = E 1 n n YYYYYYY i =1 y i = 1 n n YYYYYYY i =1 E ( y i ) = 1 n n YYYYYYY i =1 ( β 0 + β 1 x i ) = β 0 + β 1 1 n n YYYYYYY i =1 x i = β 0 + β 1 ¯ x (c) (4 pts) After the data are collected we find n = 20 , QQQQQQQ 20 i =1 ( x i − ¯ x ) 2 = 50 , QQQQQQQ 20 i =1 ( y i − ¯ y ) 2 = 33 , and QQQQQQQ 20 i =1 ( y i − ˆ y i ) 2 = 9 . 6 , where ˆ y 1 , . . . , ˆ y 20 are the fitted values for LS regression of y on x . Based on these results, show how to calculate the standard error for ˆ β 1 , plugging in all the relevant numbers. You do not have to complete the calculation. se ( ˆ β 1 ) = ˆ σ rrrrrrr QQQQQQQ 20 i =1 ( x i − ¯ x ) 2 = wwwwwww vvvvvvv vvvvvvv uuuuuuu QQQQQQQ 20 i =1 ( y i − ˆ y i ) 2 / (20 − 2) QQQQQQQ 20 i =1 ( x i − ¯ x ) 2 = ttttttt 9 . 6 / 18 50 2

(c) (3 pts) Consider the F-Statistic test results given on the last line of the summary. State the null hypothesis H 0 and alternative hypothesis H A for this test. Express the hypotheses in terms of the unknown parameters β 0 , β 1 , β 2 , β 3 , σ 2 . H 0 : β 1 = β 2 = β 3 = 0 H A : at least one of β 1 , β 2 , β 3 ̸ = 0 Equivalently, H A : β 1 ̸ = 0 or β 2 ̸ = 0 or β 3 ̸ = 0 . (d) (3 pts) Based on the model summary and mathematical notation above, provide the t value and p-value for testing the null hypothesis H 0 : β 1 = 0 against the alternative H a : β 1 ̸ = 0 . Also give the degrees of freedom for this test. This is the coefficient t test for Tax . From the model summary we have t value = − 1 . 84 , p-value = 0 . 07175 The degrees of freedom = 47 = degrees of freedom for residual standard error. 4

Problem 3. (3 parts) Consider a model of the form y = X β + e , where X is an n × p full rank matrix (its columns are linearly independent), y and e are n × 1 , and β is p × 1 . Assume X is a fixed (non-random) matrix, E ( e ) = 0 , and cov ( e ) = σ 2 I . The least square estimator of β is ˆ β = ( X T X ) − 1 X T y . The projection matrix or “hat” matrix is H = X ( X T X ) − 1 X T . (a) (3 pts) Show that X ˆ β = Hy . X ˆ β = X ( X T X ) − 1 X T y = Hy (b) (4 pts) If ˆ y is the vector of least square fitted values, show that Cov (ˆ y ) = σ 2 H . Method 1: Cov (ˆ y ) = Cov ( X ˆ β ) = X Cov ( ˆ β ) X T = X ( σ 2 ( X T X − 1 )) X T = σ 2 X ( X T X ) − 1 X T = σ 2 H Method 2: Cov (ˆ y ) = Cov ( Hy ) = H Cov ( y ) H T = H ( σ 2 I ) H ( H is symmetric ) = σ 2 HH = σ 2 H (c) (3 pts) Explain why var ( ˆ y i ) = σ 2 h i for i = 1 , 2 , . . . , n , where h i is the ( i, i ) diagonal element of H . The diagonal elements of Cov ( ˆ y ) = σ 2 H are the variances of ˆ y 1 , . . . , ˆ y n . These diagonal elements are σ 2 h 1 , . . . , σ 2 h n . 5

(d) (3 pts) (i) Give the numerical values for the degrees of freedom of the F-test in the table. (ii) Is the null hypothesis rejected at level α = 0 . 05 ? How do you know? (i) Degrees of freedom (numerator, denominator) = (3, 44). (ii) H 0 is not rejected at level 0.05 because p = 0 . 06315 . 7

Problem 5. (4 parts) Data on fuel consumption were collected for each of the 50 states and Washington D.C. for a total sample size of n = 51 . The variables are gasoline Tax (cents/gallon), Fuel consumption per 1000 pop. over 16, Dlic (Licensed Drivers per 1000 population over 16), and logMiles (log 10 miles of highway in the state). The following linear model was fit to the data. Fuel = β 0 + β 1 Tax + β 2 Dlic + β 3 logMiles + error Below is a plot of standardized residuals versus diagonals of the “hat” matrix after fitting the above model by ordinary least squares. The points are labeled with the two letter state abbreviations. 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 -3 -2 -1 0 1 2 3 Leverage (h_i) Standardized Residual AL AK AZ AR CA CO CT DE DC FL GA HI ID IL IN IA KS KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC ND OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY (a) (3 pts) We know that in general QQQQQQQ n i =1 h i = p , where p is the number of columns of the design matrix X including the constant column for the intercept. Based on the information given, (i) find the average (sample mean) of the leverages in the data, and (ii) identify the states that have leverage more than twice the average value (identify by their labels). (i) Average leverage = 4/51 = 0.0784 (ii) Twice the average = 8/51 = 0.157. According to the graph, the states above this level are VT, GA, HI, RI, AK, and DC. 8

(d) (2 pts) Consider Box-Cox models of the form g λ ( Fuel ) = β 0 + β 1 Tax + β 2 Dlic + β 3 logMiles + error where g λ ( y ) = 999 ===== ;;;; y λ − 1 λ , if λ ̸ = 0 log e ( y ) , if λ = 0 . The plot below shows log-likelihood versus the value of λ , where for each λ we obtain a set of coefficient estimates for the regression parameters specific to that value of λ by the method of maximum likelihood. For each λ the result is the same as if we performed ordinary least squares regression of g λ ( Fuel ) on the predictor variables. -1 0 1 2 3 4 2 4 6 8 10 12 λ log-Likelihood 95% Consider the null hypothesis that λ = 1 . Based on the results given, should we reject or fail to reject this hypothesis at the at level α = 0 . 05 ? Why or why not? The 95% confidence interval for λ in the graph includes the value λ = 1 so we fail to reject the null hypothesis. 10