8.99 118 1.37 126 125 138 26.58 178 156 184 24.52 176 12 01 1692 592 Ex SSE 2448

Question

8.99 118 1.37 126 125 138 26.58 178 156 184 24.52 176 12 01 1692 592 Ex SSE 244886 44 15.65 10 The standard error of the estimate is 15.65 FTEs. An examination of for this reveals that 8 of 12 (67%) within t1s, are vithin s this of error acceptable? Hospital administrators probably that question. EMS 12.24 Determine the sum of squares of error (SSE) and the standard error of the estaa Problem 12.6. Determine how many of the residuals computed in 12.14 (for Problem 12.6) are within one standard error of the estimate the terms are normally distributed, approximately how many of these residuals be within tls? 12.25 Determine the SSE and the s for Problem 12.7. Use the residuals computed in Problem 12.15 (for Problem 12.7) and determine how many of them are tlsandt2s. How do these numbers compare with what the empirical rule sam should occur if the error terms are normally distributed? 12.26 Determine the SSE and the s, for Problem 12.8. Think about the variables being analyzed by regression in this problem and comm on the value of s, 12.27 Determine the SSE and s for Problem 12.9. Examine the variables being analyed by regression in this problem and comment on the value of s, 12.28 Problem 12.10, you were asked to develop the equation of a regression mode predict the number of business by the number of frm births bankruptcies regression model, solve for the standard error of the estimate and comment ont 12.29 Use the data from Problem 12.19 and determine the s 12.30 Determine the SSE and the s for Problem 12.20. Comment on the size ot regression model, which is used to predict the cost of milk. 12.31 Determine the equation of the regression line to predict annual sales of a o from the yearly stock market volume of shares sold in a recent year Comput standard error of the estimate for this model. Does volume of shares so be a good of a sales? Why or why not? 12

Explanation / Answer

12.26. We got regression output from excel :

Here we have intercept = a= -46.2918 and slope = b = 15.2398

Hence teh regression equation will be

Y = a + b *x

Y = -46.2918 + 15.2398 * x

The SSE = sum of square of error = Sum of square of residual = 70969.2

Se = Standard error = 108.7575

SUMMARY OUTPUT Regression Statistics Multiple R 0.947663 R Square 0.898064 Adjusted R Square 0.881075 Standard Error 108.7575 Observations 8 ANOVA df SS MS F Significance F Regression 1 625246.3 625246.3 52.86065 0.000344 Residual 6 70969.2 11828.2 Total 7 696215.5 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept -46.2918 64.89096 -0.71338 0.502402 -205.074 112.4907 Advertising 15.23977 2.096101 7.270533 0.000344 10.1108 20.36875

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