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Module 6 test2 - Regression Analysis Step 1 2 ANOVA table...

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Regression Analysis Step 1 2 ANOVA table Source SS df MS F p-value Regression 48.0000 1 48.0000 Residual 54.0000 9 6.0000 Total 102.0000 10 3 4 5 6 7 8 9
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SS Total = SS Regression + SS Residual SS Total = 48.0000 54.0000 SS Total = 102.0000 SS Regression = SS Total - SS Residual SS Regression = 102.0000 54.0000 SS Regression = 48.0000 SS Regression / SS Total 48.0000 102.0000 0.471 Delta Check 0.0000 n = 11 Formula for Standard Error S = Square Root of SS Residual / n - 2 S= SS Residual n - 2 S= 54.0000 9 S= 2.449 df Regression = df Total - df Residual 10 9 F-Test Formula F = MS Regression / MS Residual F = (SS Regression / df Regression) / SS Residual / df Residual 48.0000 1 54.0000 9 F = MS Regression / MS Residual 48.000 6.0000 F = 8.00 Find the value of r r = 0.686 Is r Positive or Negative ( + / - )? It is positive (+) because the slope of the line goes upward and right on the plot. r 2 = r 2 = r 2 = r = square root of r 2
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Regression Analysis r² 0.366 n 24 r 0.605 k 1 Std. Error 109.162 Dep. Var. Sales ANOVA table Source SS df MS F p-value Regression 151,642.2653 1 151,642.2653 12.73 .0017 Residual 262,159.4365 22 11,916.3380 Total 413,801.7018 23 Regression output confidence interval variables coefficients std. error t (df=22) p-value 95% lower 95% upper Intercept 203.8348 43.5917 4.676 .0001 113.4312 294.2384 Advertising 30.0232 8.4162 3.567 .0017 12.5690 47.4774 Predicted values for: Sales 95% Confidence Intervals 95% Prediction Intervals Advertising Predicted lower upper lower upper Leverage 2 263.881 200.900 326.862 28.896 498.867 0.077 7 413.997 349.857 478.136 178.698 649.296 0.080 Regression Equation: 203.8348 tells us how much the natural gas is estimeated to be consumed when 30.0232 For each additional Advertising 30.0232 Expect an x 0 should be substituted with the independent variable. 233.858 x Find x To find the x-intercept assumes x = 0 The Simple Coefficient of Determination Explained Variation / Total Variation the simple linear regression model, which includes the independent variable explains  36.6% of the variability in  Sales r = r = the simple coefficient of r = 0.605  relationship between  0 AND 0 Y^ = b 0 + b 1 x b 0 b 1 Y^ = b 0 + b 1 r 2  =  r 2  =  r 2  =  r 2  =  Our interpretation is that there is
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for each additional independent variable we can expect an increase/decrease of the dependent vari Dependent Sales Independen Advertising IS 30.0232 < OR > THAN 1? Increase, Increase or Decrease? n the temperature = 0 Increase, of the Sales 203.8348 Advertising expenditures 0 s a fairly strong  positive  or  negative  linear
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iable.
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Regression Analysis r² 0.366 n 24 r 0.605 k 1 Std. Error 109.162 Dep. Var. Sales ANOVA table Source SS df MS F p-value Regression 151,642.2653 1 151,642.2653 12.73 .0017 Residual 262,159.4365 22 11,916.3380 Total 413,801.7018 23 Regression output confidence variables coefficients std. error t (df=22) p-value 95% lower Intercept 203.8348 43.5917 4.676 1.16E-04 113.4312 Advertising 30.0232 8.4162 3.567 .0017 12.5690 Predicted values for: Sales 95% Confidence Intervals 95% Prediction Intervals Advertising Predicted lower upper lower upper 2 263.881 200.900 326.862 28.896 498.867 7 413.997 349.857 478.136 178.698 649.296 Regression Equation: 203.8348 tells us how much the natural gas is estimeated to be consumed w 30.0232 For each additional Advertising 30.0232 x 0 should be substituted with the independent variabl 233.858 x Find x To find the x-intercept assumes x = 0 Y^ = b 0 + b 1 x b 0 b 1 Y^ = b 0 + b 1
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Dependent Sales
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