1 The two variable regression. For the regression model y = a + Bx+ɛ(a) Show that the least squares normal equations imply Ee = 0 and E;x;e¡ = 0.(b) Show that the solution for the constant term is a = ỹ - bx.(c) Show that the solution for b is b = [E (x – x)(y - y)]/[E* (x; – x)²].2 Change in the sum of squares. Suppose that b is the least squares coefficient vector inthe regression of y on X and that c is any other K x 1 vector. Prove that thedifference in the two sums of squared residuals is(у — Хc)' (у — Хс) - (у-Xb) (у - Xb) 3D (с - b)'X'X(с — b)Prove that this difference is positive.3(Difficult) Adding an observation. A data set consists of n observations on X, and yn.The least squares estimator based on these n observations is b, = (X,Xn) X,'yn-%3DAnother observation, x, and y,, becomes available. Prove that the least squares estimatorcomputed using this additional observation isbns = bn +-(X„Xn)¯1xg(Ys – x,bn)._("x"x);x+14 Prove the theorem of Change in the Sum of Squares When a Variable is Added to aRegression.

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MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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In a comprehensive road test on new car models, one variable measured is the time it takes a car to accelerate from 0 to 60 miles per hour. To model acceleration time, a regression analysis is conducted on a random sample of 129 new cars. TIME60: y = Elapsed time (in seconds) from 0 mph to 60 mph MAX: x = Maximum speed attained (miles per hour) The simple linear model E(y) = Bo + B1x was fit to the data. Computer printouts for the analysis are given below: NWEIGHTED LEAST SQUARES LINEAR REGRESSION OF TIME60 PREDICTOR VARIABLES COEFFICIENT STD ERROR STUDENT'S T CONSTANT 187171 0.63708 29.38 0.0000 0.0000 MAX -0.08365 0.00491 -17.05 0.6960 0.6937 R-SQUARED RESID. MEAN SQUARE (MSE) 1.28695 ADJUSTED R-SQUARED STAND ARD DEVIATION 113444 SOURCE DF MS F REGRESSION 374.285 0.0000 374.285 1.28695 290.83 RESIDUAL 127 163.443 TOTAL 128 537.728 CASES INCLUDED 129 MISSING CASES 0 Fill in the blank: "At a =.05, there is between maximum speed and acceleration time." O sufficient evidence of a…
M5
A researcher wants to know if there is a significant correlation between hours spent studying for an exam (X) and exam performance (Y). Level of significance is 0.05, degrees of freedom is 3, and the critical value is 0.878. Interpret the strength and direction of the correlation, calculate and interpret the coefficient of determination, then develop the simple regression equation for the two variables. X (Time Spent Studying in Hours) Y (Exam Performance, 0 to 100) 3 56 6 77 7 79 8 70 11 96 Mean 7.00 75.60 SD 2.92 14.54 For this question, on your hand calculation document clearly state: a) the null hypothesis, b) the alternative hypothesis, c) the alpha level you are using, d) the critical value, e) process for calculating r, f) your decision about the null hypothesis, g) the coefficient of determination, h) and regression equation. This is what I will be marking. You may also enter your responses here, but it is…
Consider the data. X;| 1 2 3 4 5 y, 4 8 11 12 The estimated regression equation for these data is ý = 2.30 + 1.90x. (a) Compute SSE, SST, and SSR using equations SSE = E(y; - 9)?, SST = E(y; - y)?, and SSR = E(ŷ; - y)?. SSE = SST = SSR = (b) Compute the coefficient of determination r2. r2 Comment on the goodness of fit. (For purposes of this exercise, consider a proportion large if it is at least 0.55.) O The least squares line did not provide a good fit as a small proportion of the variability in y has been explained by the least squares line. O The least squares line did not provide a good fit as a large proportion of the variability in y has been explained by the least squares line. O The least squares line provided a good fit as a small proportion of the variability in y has been explained by the least squares line. O The least squares line provided a good fit as a large proportion of the variability in y has been explained by the least squares line. (c) Compute the sample…
12
To examine whether there is any association between the performances on the two tests, T1 and T2 (test T1 was conducted before T2), and whether T2 scores could be predicted in terms of T1 scores. A random sample of 150 students in Statistics major was selected. Let the scores be denoted S1 and S2 for tests T1 and T2, respectively. The tests had different maximum points. The following outputs using R codes were obtained. fit[t]) (Intercept) 0.10152 2.51003 0.04 0.968 S1 0.26723 0.01921 13.91 <2e-16 *** Signif. codes: O ***' 0.001 **' 0.01 (*' 0.05 '.' 0.1 '? 1 Residual standard error: 4.472 on 148 degrees of freedom Multiple R-squared: F-statistic: 193.6 on 1 and 148 DF, 0.5667, Adjusted R-squared: 0.5638 p-value: < 2.2e-16 plot (fit$fitted,fit$residuals, xlab="fitted values", ylab="residuals") qqnorm (fit$residuals) Normal Q-Q Plot oo 00 00 00 O Go a god 00 00 0. of 25 30 35 40 45 -2 2 fitted values Theoretical Quantiles esiduals 15 Score: Sample Quantiles 091
The regression model Yi=−303.8+1.5949X1i−0.0699X2i predicts standby hours based on total staff present, X1i, and remote hours, X2i, for week i. The data from which the model was constructed are provided. b. If appropriate, perform the Durbin-Watson test, using α=0.05. Determine the Durbin-Watson statistic.

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1 The two variable regression. For the regression model y = a + Bx +ɛ (a) Show that the least squares normal equations imply Eiei = 0 and Exej = 0. (b) Show that the solution for the constant term is a = ỹ - bx. (c) Show that the solution for b is b [E (x - x)(yi – y)]/[E# (x¡ – x)²]. %3D