Homework 3 Solution (Due Oct. 1, Monday)

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1 Homework 3 Solution (Due Oct. 1, Monday) Chapter 2 Problems: [1] Text Problems 1.27, 2.27 and 2.28 (for 2.28, a) and b) only). [SAS output] The REG Procedure Model: MODEL1 Dependent Variable: muscle Number of Observations Read 61 Number of Observations Used 60 Number of Observations with Missing Values 1 Model Error Corrected Total Root MSE R-Square Dependent Adj R-Sq Coeff Var Variable DF Estimate Error t Value Pr > t 95% Confidence Limits Intercept age Output Statistics Dependent Predicted Std Error Obs Variable Value Predict 95% CL 95% CL Predict Residual < outputs for case 10 - case 57 were omitted >

2 1.27 a) The estimated regression function is Ŷ = X. The linear regression looks like a good fit in terms of picking up the general trend in how muscle mass changes with age, and supports the idea that muscle mass is decreasing with age. b) (1) is simply asking for an estimate of β 1, which is (2) Ŷh = = (3) e 8 = Y 8 Ŷ8 = 112( (41)) = (4) MSE = a) This is a one-sided t-test for β 1. We are testing H 0 : β 1 =0vsH a : β 1 < 0. For α =0.05, we would reject H 0 if the observed value of t = b1 s(b 1) is less than t(0.95, 58) = (an approximation from table B.2 with degrees of freedom = 60). Since t obs = 13.19, reject H 0 and conclude β 1 < 0. As stated on the assignment, you do not need to calculate an exact p-value. b) No. Doing so would be extrapolation since we are fitting the model for ages 40 and above, since the X data values start at age 40. There is no reason to believe the model will hold at lower ages. Interpreting β 0 as E(Y ) at X = 0 only makes sense if the linear model works all the way down to 0 (it often doesn t), or when the data includes X values near 0. c) This is just a confidence interval for β 1 which is given in the SAS output as ( , ). The difference [E(Y ) at (X + 1)] - [E(Y ) at X] isβ 1 no matter what X is a) From SAS output, the confidence interval for the expected mass at age 60 is ( , ). We conclude with 95% confidence that the mean muscle mass at age 60 is between and Note that this confidence coefficient are interpreted with respect to repeated sampling where the values of a predictor are kept at the same level as in the observed sample. b) From SAS output, the prediction interval for the mass of a randomly selected woman from those age 60 is ( , ). Whether the prediction interval is precise or not depends on i) if the assumptions in a simple linear regression model are satisfied and ii) how important a range of 68 to 105 is. But, since that is close to the full range of mass values over the data, it looks very imprecise to me. 5

3 [2] a) The estimate of β 0, β 1, σ 2 and σ are b 0 = 168.6, b 1 = , s 2 = MSE = and s = s 2 = b) Using table B.2 in the book, t(.975, 14) = The 95% confidence interval for β 0 is ± 2.145( ) = ( , ) and The 95% confidence interval for β 1 is ±2.145(.09039) = ( , ). c) The t-statistic of is t obs =2.0343/ which is the estimate/ standard error. The p-value is the probability (before the data is collected) when H 0 is true (that is H 0 : β 0 = 0) that we will get a t statistic t such that t > This probability is less than d) The value Ŷh at X h = 24 is = For the confidence interval of E(Y ) at X h = 24, we need to calculate s 2 (Ŷh) =MSE above, MSE = and s 2 (b 1 )= 1 n + (X h X) 2 MSE (Xi X) 2 = MSE n + MSE(X h X) 2 (Xi X) 2. From outputs (Xi X) 2 =( ) 2. Since (X h X) 2 = (24 28) 2 = 16, s 2 (Ŷ ) = and s(ŷ ) = Using a critical value of t(0.975, 14) = 2.145, we have the confidence interval for E(Y ), ± = ( , ) (subject to rounding error). e) We are now trying to predict the response from a single observation that will be taken at X = 24. The predicted value is the same as the estimate of the mean at 24, namely s 2 (pred) = MSE + s 2 (Ŷh) = = and s(pred) = The prediction interval is ± = ( , ) b) The following is the ANOVA table from SAS Model Error Corrected Total c) Here we are testing H 0 : β 1 =0vsH a : β 1 = 0. Reject H 0 if Fobs >F(0.95; 1, 58) = Here, Fobs = from the SAS output, so reject H 0. Also, reject H 0 if p-value <α=0.05. From the SAS output, p-value <.0001 which is less than So reject H 0. Conclude that there is significant linear relationship between muscle mass and age. Variable DF Estimate Error t Value Pr > t Intercept age d) 1 R 2 = = is amount unexplained, which is relatively small. Root MSE R-Square Dependent Adj R-Sq Coeff Var e) R 2 = SSR/SST O =0.7501(or from SAS output). r = R 2 = r is negative since the slope of the regression line is negative from SAS output. 6

4 Pearson Correlation Coefficients, N = 60 Prob > r under H0: Rho=0 muscle age muscle See the following SAS outputs age Variable DF Estimate Error t Value Pr > t 99% Confidence Limits Intercept population a) Here we are testing H 0 : β 1 =0vsH a : β 1 = 0. Reject H 0 if t obs >t(0.995; 82) = Here, t obs = 4.1 = 4.1 from the SAS output. Since t obs =4.1 > 2.637, we reject H 0. The p- value<0.0001, from the SAS output. We conclude that there is a significant linear association between crime rate and percentage of high school graduates. c) The 99% CI for β 1 is b 1 ± t(0.995,n 2)s(b 1 )= ± 2.637(41.57) = ( 280.2, 60.94) a) See the following SAS outputs Model Error Corrected Total b) Here we are testing H 0 : β 1 =0vsH a : β 1 = 0. Reject H 0 if Fobs >F(0.99; 1, 82) = Here, Fobs = from the SAS output, so reject H 0. Also, reject H 0 if p-value <α=0.01. From the SAS output, p-value <.0001 which is less than So reject H 0. We conclude that there is a significant linear association between crime rate and percentage of high school graduates. Root MSE R-Square Dependent Adj R-Sq Coeff Var Here t osb 2 = F,since(4.1) 2 = (with some rounding error). Yes. the p-value for the F-test is the same as that for the t-test. c) The percent explained is about 17% (R 2 =0.1703). This seems to be somewhat low. d) r = R 2 = This is negative since the regression slope is negative. Pearson Correlation Coefficients, N = 84 Prob > r under H0: Rho=0 rate population rate population [3] Text Problems

5 a) The 95% confidence interval for β 1 allows us to test H 0 : β 1 =0vs. H a : β 1 = 0 at α =.05 by rejecting H 0 if 0 is not in the interval ; which it is not. Hence the conclusion is warranted with an implied level of significance of.05. This assumes that a plot shows that the simple linear regression model is reasonable. b) First, there is no reason to be sure that the regression relationship used still holds at low values of X. If it does not then we would not interpret β 0 as the expected sales at population 0 (in this case, at zero population you would have zero sales, exactly). 8