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Sales $400,000 $1,042,000 $1,129,000 $1,110,000 $1,336,000 $1,363,000 $1,177,000 $603,000 $582,000 $697,000 $586,000 $673,000 $546,000 $334,000 $27,000 $76,000 $298,000 $746,000 $962,000 $907,000 Price $15 $12 $24

Sales $400,000 $1,042,000 $1,129,000 $1,110,000 $1,336,000 $1,363,000 $1,177,000 $603,000 $582,000 $697,000 $586,000 $673,000 $546,000 $334,000 $27,000 $76,000 $298,000 $746,000 $962,000 $907,000 Price $15 $12 $24 $18 $18 $30 $27 $24 $36 $27 $24 $27 $30 $33 $24 $27 $30 $18 $21 $24 Sales $400,000 $1,042,000 $1,129,000 $1,110,000 $1,336,000 $1,363,000 $1,177,000 $603,000 $582,000 $697,000 $586,000 $673,000 $546,000 $334,000 $27,000 $76,000 $298,000 $746,000 $962,000 $907,000 Price $15 $12 $24 $18 $18 $30 $27 $24 $36 $27 $24 $27 $30 $33 $24 $27 $30 $18 $21 $24 Let Yt be the sales during month t (in thousands of dollars) for a photography studio, and let P t This equation indicates that last month's sales and the current month's price are explanatory variables. The la SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.853628 0.728681 0.696761 216073.7 20 ANOVA df SS Regression Residual Total Intercept Price Yt-1 MS 1.07E+12 4.67E+10 F 2 17 19 2.13E+12 7.94E+11 2.93E+12 22.82843 Coefficients 791065.2 -23378 Standard Error 214468.98 8152.7234 t Stat 3.688483 -2.86751 P-value 0.001823 0.010672 0.745564 0.117816 6.328205 7.56E-06 Yt = a + b1Yt1 + b2Pt + et Yt=791065.2+0.745564Yt--23378pt Yt=791065.2+0.745564($962000)-23378(10) =791065.2+0.745564($962000)-233780 =$1274517.768 There is no linear relationship between residuals and the dependent variable. Therefore there is no problem of autocorrelation 1 0.8 0.6 Residuals 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 PT-1 , and let P t be the price charged for portraits during month t. The data are in the attachment below. Use regression to fit lanatory variables. The last term, e t, is an error term. a. If the price of a portrait during month 21 is $10, what would you Significance F 1.53E-05 Lower 95% 338575.2 -40578.7 Upper 95% Lower 95.0% Upper 95.0% 1243555 338575.2 1243555 -6177.25 -40578.7 -6177.25 0.496994 no problem of autocorrelation if residuals. 0.994134 0.496994 0.994134 1 0.8 0.6 0.4 0.2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 PT-1 1 elow. Use regression to fit the following model to these data: Y t = a + b1Yt1 + b2Pt + et 21 is $10, what would you predict for sales in month 21? b. Does there appear to be a problem with autocorrelation of th with autocorrelation of the residuals

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