35. Use the computer printout below to answer the following questions. Coefficie

Question

35. Use the computer printout below to answer the following questions. Coefficients Std. Error P-value t-Stat 169.25751 4.3121659 Intercept 729,8665 0.0010099 -10.887 Price 3.4952397 3.1148078 0.0089406 Advertising 0.0465 0.0176228 2.6386297 0.0216284 ANOVA df SS MS Significance F 37.56127994 Regression 12442.8 6221.4. 0.00000683 Residual 1987.6 12 165.63333 14430.4 Tota 14 Se -12.86986 R-sq 0.862263 R-sq (ad 5% 0.8393068 a) Write and interpret the multiple regression cquation b) Does the model with Price and Advertising contribute to the prediction of Y? c) Which independent variablc appcars to bc thc best d) What is thc numbcr of obscrvations uscd in this study? predictor of sales? Explain. c)Find SS l' Assuming that thc cocfficient on Advertising has What is thc standard crior of cstimatc? Can you usc this statistic to assess the model's fit? If so, how? Ha: b. 0, what's S alistical decision g) What is the coefficient of determination, and what does it h) What is the coefficient of determination, adjusted for tell you about the regression model? degrees of freedom? What do this statistic and the statistic referred to in part (g) tell you about how well this model fits that data. Test the overall utility of the model. What does the p-value of the test statistic tell you?

Explanation / Answer

Part-a

Multiple regression equation is

Sales=729.8665 - 10.887 Price + 0..0465 Advertising

Corresponding to unit increase in price there is on an average a decrease of 10.887 units in Sales, holding advertising fixed.

Corresponding to unit increase in advertising there is on an average an increase of 0.0465 units in Sales, holding price fixed.

Part-b

Yes model contribute to predicting Sales as both price(p<0.05) and advertising (p<0.05) are significant predictors of Y

Part-c

Price appears to be the best predictor as is has lower p-value than advertising.

Part-d

Number of observations =14+1=15

Part-e

SST=Total Sum of square=14430.4

One tailed p-value=0.0216284/2 =0.010814

As p-value<0.05 we reject the null hypothesis and conclude that there is significant positive impact of advertising on sales.

Part-f

Standard error of estimate=12.86986.

This is used to compare two models. The model with lower standard error is the better.

Part-g

Coefficient of determination-R-square=0.862263

This means that price and advertising explained 86.23% of the variations in sales

Part-h

Adjusted coefficient of determination =0.8393068

As R2 and adjusted R2 are close so the model is valid

Part-i

The overall utility of model is significant with F(2, 12)=37.56127994, p-value=0.00000683<0.05.

Hence this indicates that at least one of the predicotrs is significant in explaining the sales.

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