Sum2 2011 NCSU ST512 QUIZ 1 NAME We are interested in studying the relationship, if any, between yield (Y, in bushels/acre), and rainfall (X, inches/yr) on corn raised in farms of similar acreage and management located in the midwest, during the period from 1890 to 1927. Our main goal is to find out an equation that will help in the prediction of corn yield for a given rainfall. We decided to run a multiple regression analysis after looking at a representation of the relationship between Year, Rainfall and Yield Model YIELD = RAINFALL YEAR YEAR*RAIN yi o 1 x1i 2 x2i 3 x3i ei X 1 Rainfall Regression model fitted X 2 Year from 1890 X 3 X 1 *X2 Y= Yield e ~ N 0, 2 i Next we present the Analysis of variance table and parameter estimates. 1. Please fill in the missing entries. Analysis of Variance 2. Source DF Sum of Squares Mean Square F Value Pr > F Model 33 340.60856 113.53619 10.61 <.0001 Error 34 363.94197 10.70418 Corrected Total 37 704.55053 Give the null (and alternative) hypothesis tested by the overall F-test for the model . 1 July 07, 2011 Sum2 2011 NCSU ST512 QUIZ 1 Next we have the table of Parameter estimates 3. Please fill in the missing entries. Parameter Estimates Variable DF Parameter Estimate Standard Error t Value Pr > |t| Type I SS Intercept 1 4.48573 5.48728 0.82 0.4193 38707 X1 1 2.31128 0.50479 4.58 <.0001 114.21474 X2 1 1.08545 0.27376 3.96 0.0004 95.99361 -0.08781 0.02516 -3.49 0.0014 130.40020 X3 4. Write the estimated regression equation. yi 4.49 2.31x1i 1.08x2i 0.09 x3i yi 4.49 2.31x1i 1.08 x2i 0.09 x3i o 4.49 1 2.31 2 1.08 3 0.09 yi 1 x1i x2i 4.49 2.31 x3i 1.08 0.09 5. Calculate an estimate for the conditional mean yield of farms raising corn in year 1900 receiving a rainfall of 9 inch/year. rainfall 9 year 1900 x1new 9 x2 new 1900 1890 10 x3new x1new x2 new 9 10 90 yi 4.49 2.31x1i 1.08 x2i 0.09 x3i ynew 4.49 2.31 9 1.08 10 0.09 90 27.98 2 July 07, 2011 Sum2 2011 NCSU ST512 QUIZ 1 The variance-covariance matrix for the estimates of the regression coefficients is presented next. Covariance of Estimates Variable Intercept rainfall Intercept 30.110197368 -2.719640859 rainfall -2.719640859 0.2548146526 0.1197135611 -0.01120618 YEAR1 -1.320670004 0.1197135611 0.0749470666 -0.006778894 YR_RAIN 0.1195693604 -0.01120618 YEAR1 YR_RAIN -1.320670004 0.1195693604 -0.006778894 0.0006329419 6. Show how you would calculate the standard error of the conditional mean in 5. You do not need to do the actual calculations y aβ var y var aβ aΣa Σ var β 4.49 2.31 yi 1 x1i x2i x3i 1.08 0.09 a 1 9 10 90 30.1102 2.7196 1.3207 0.1196 1 2.7196 0.2548 0.1197 0.0112 9 var ynew 1 10 9 90 1.3207 0.1197 0.0749 0.0068 10 0.1196 0.0112 0.0068 0.0006 90 s.e ynew var ynew 7. A researcher is analyzing the effect of the year by rainfall interaction, and ask your help. She needs to draw the regression line between Yield and Rainfall at each of the following years: 1890, 1915 and 1935. a. Compute the regression equation for each of these three years 3 July 07, 2011 Sum2 2011 x2i 0 NCSU ST512 QUIZ 1 x1i = x1i x2i x1i 0 0 yi 4.49 2.31x1i 1.08 0 0.09 0 yi 4.49 2.31x1i x2i 25 x3i x1i x2i x1i 25 25 x1i yi 4.49 2.31x1i 1.08 25 0.09 25 x1i yi 4.49 2.31x1i 27 2.25 x1i yi 31.49 0.06 x1i x2i 35 x3i x1i x2i x1i 35 35 x1i yi 4.49 2.31x1i 1.08 35 0.09 35 x1i yi 4.49 2.31x1i 37.8 3.15 x1i yi 42.29 0.84 x1i Year X2 = (Year-1890) Intercept Regression coefficient (Slope for rainfall) 1890 0 4.49 +2.31 1915 25 31.49 +0.06 1925 35 42.29 -0.84 b. Explain changes in these three slopes? Changes in intercept and slope for regression equation of Y on X1 shows the effect of a negative regression coefficient for the interaction between X1 and X2. Slope for rainfall is 0 in year 1890, our baseline year, and 0.06 in year 1915 while in year 1925 is negative, -0.84, which indicates that while in year 1915, for one inch/yr increase in rainfall, yield increases in 0.06units; for year 1925, for each inch/yr increase in rainfall, yield decreases in 0.84 units 4 July 07, 2011 Sum2 2011 NCSU ST512 QUIZ 1 8. I regressed Y on X1 and X2 getting these Type I and Type II sums of squares from PROC REG. What would have happened if I had run the regression in the opposite order (PROC REG; MODEL Y = X2 X1; ) Variable DF Type I SS INTERCEP 1 90000 18000 X1 1 800 900 X2 1 700 700 Model SS Type II SS 1500 Fill missing entries Variable DF Type I SS INTERCEP 1 90000 18000 X2 1 __600_ __700_ X1 1 __900_ __900_ Model SS 1500 R(X1|X2)=900=1500-R(X2| o ) 5 Type II SS R(X2| o ) = 1500 – 900 = 600 July 07, 2011
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