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. For this question, refer to the SHAZAM printout at the end of the exam Feel free to separate the SHAZAM page from the rest of the exam; you do not need to hand in the SHAZAM page. (a) Draw a graph that shows how you would predict SALES, for two different levels of TEMP: 70 and 90. (Hint: feel free to round off any numbers you use, so that they are whole numbers, to simplify vour results Represent the results from both OLS commands on the same graph, labeling your axes, and identify the numerical values for predicted SALES for both temperatures from both models.Question 4, continued (b) Suppose that the observed value of SALES is 300 for a particular week i, in which the average temperature is 70 degrees. Give the two predicted values for SALES, and the two corresponding values for ê for this week, using the two sets of OLS results shown i.e., one prediction and one ê from each model (c) The printout contains a test of the hypothesis that the true coeffi- cient for TEMP, in the second model estimated, is equal to zero. Is this hypothesis rejected? (Yes or No, and justify your answer.)

Question 4, continued (d) The printout contains a test of the hypothesis that the true coeffi cient for TEMP, in the second model estimated, is equal to 0.25 Is this hypot hesis rejected? (Yes or No, and justify your answer.)

SHAZAM Results to Accompany Midterm Exam 1: 2010 stat NAME N MEAN 25 13.000 25 75.810 25 327.34 ST. DEV 7.3598 8.7926 26.388 VARIANCE 54.167 77.310 696.34 MINIMUM 1.0000 60.607 282.64 MAXIMUM 25.000 88.730 365.99 TEMP SALES |_*ols command number 1 lols sales / nogf noanova REQUIRED MEMORY IS PAR 2 CURRENT PAR 781 OLS ESTIMATION 25 OBSERVATIONS DEPENDENT VARIABLE= SALES ..NOTE..SAMPLE RANGE SET TO 25 R-SQUARE- 0 . 0000 VARIANCE OF THE ESTIMATE-SIGMA*2696.34 STANDARD ERROR OF THE ESTIMATE-SIGMA 26.388 SUM OF SQUARED ERRORS-SSE= 16712 MEAN OF DEPENDENT VARIABLE327.34 LOG OF THE LIKELIHOOD FUNCTION116.786 R-SQUARE ADJUSTED 0.0000 VARIABLE ESTIMATED STANDARD T-RATIO PARTIAL STANDARDIZED ELASTICITY NAME COEFFICIENT ERROR CONSTANT 327.34 24 DF P-VALUE CORR. COEFFICIENT AT MEANS 62.02 5.278 0.000 0.997 0.0000 1.0000 |_confid constant USING 95% AND 90% CONFIDENCE INTERVALS CONFIDENCE INTERVALS BASED ON T-DISTRIBUTION WITH 24 D.F. T CRITICAL VALUES2.064 AND 1.711 LOWER 2.5% LOWER 5% 318.3 UPPER 5% 336.4 NAME COEFFICIENT UPPER 2.5% STD. ERROR 5.278 CONSTANT 316.4 |_*ols command number 2 lols sales temp / nogf noanova REQUIRED MEMORY IS PAR 327.34 338.2 2 CURRENT PAR 781 OLS ESTIMATION 25 OBSERVATIONS DEPENDENT VARIABLE-SALES. . .NOTE..SAMPLE RANGE SET TO: 25 R-SQUARE= 0.9984 VARIANCE OF THE ESTIMATE-SIGMA *+2 = 1.1667 STANDARD ERROR OF THE ESTIMATE-SIGMA = 1.0801 SUM OF SQUARED ERRORS-SSE= 26.834 MEAN OF DEPENDENT VARIABLE 327.34 LOG OF THE LIKELIHOOD FUNCTION -36.3584 R-SQUARE ADJUSTED0.9983 VARIABLE ESTIMATED STANDARD T-RATIO COEFFICIENT ERROR PARTIAL STANDARDIZED ELASTICITY NAME TEMP 2.9988 CONSTANT 100.00 0.2508E-01 119.6 52.27 23 DF P-VALUE CORR. COEFFICIENT AT MEANS 0.6945 0.3055 0.000 0.999 0.000 0.996 0.9992 0.0000 1.913 I_confid constant USING 95% AND 90% CONFIDENCE INTERVALS CONFIDENCE INTERVALS BASED ON T-DISTRIBUTION WITH 23 D.F T CRITICAL VALUES = 2.069 AND 1.714 NAME LOWER 2.5% CONSTANT 96.04 LOWER 5% 96.72 UPPER 5% 103.3 COEFFICIENT UPPER 2.5% STD. ERROR 1.913 100.00 104.0 I_confid temp USING 95% AND 90% CONFIDENCE INTERVALS CONFIDENCE INTERVALS BASED ON T-DISTRIBUTION WITH 23 D.F T CRITICAL VALUES = LOWER 2.5% 2.069 AND 1.714 LOWER 5% 2.956 UPPER 5% 3.042 UPPER 2.5% NAME TEMP COEFFICIENT STD. ERROR 0.025 2.947 2.9988 3.051 test temp 0.25 TEST VALUE 2 , 7488 T STATISTIC 109.61855 F STATISTIC 12016.226 WALD CHI-SQUARE STATISTIC 12016.226 UPPER BOUND ON P-VALUE BY CHEBYCHEV INEQUALITY = 0.00008 STD. ERROR OF TEST VALUE 0.25076E-01 WITH 23 D.F. WITH 1 AND P-VALUE-0.00000 23 D. F. P-VALUE= 0.00000 WITH 1 D . F. P-VALUE= 0.00000 -stop

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