Question

4. (20 points) Enterprise Industries produces Fresh, a brand of liquid laundry detergent. In order to study the relationship between price and demand for the large bottle of Fresh, the company has gathered data concerning demand for Fresh over the last 30 sales periods. The response variable, demand, is the demand for the large bottle of Fresh (in hundreds of thousands of bottles) in the sales period. The explanatory variable, pricedif, is the average industry price of competitors detergents in the sales period minus the price of Fresh as offered in the sales period. Use the R output given below to answer the following questions. [Hint: By default, R states C.1, and P.1. as 95% intervals] summary(model) Call: lm(formula - demand pricedif) Coefficients Estimate Std. Error t value Pr(>ltI) Intercept) 7.8141 0.0799 97.82 0.000 pricedif 2.6652 0.2585 10.31 0.000 Residual standard error: 0.316561 on 28 degrees of freedom Multiple R-squared: 0.792,Adjusted R-squared: 0.784 F-statistic: 106.3 on 1 and 28 DF p-value: 0.000 > anova(model) Analysis of Variance Table Response: demand Df Sum Sq Mean Sq F value Pr(>F) pricedif 1 10.653 10.653 106.30 0.000 Residuals 28 2.806 0.100 > newdata -data.frame (pricedif - .1) > predict(model, newdata, interval - "confidence") fit lwr upr 8.2133 8.0806 7.9479 > predict(model, newdata, interval - "predict") fit upr 1 8.0806 7.4187 8.7425 (a) Report a point estimate of and a 95% confidence interval for the mean demand for Fresh in all sales periods when the price difference is 0.1. (b) Report a point prediction of and a 95% prediction interval for the actual demand for Fresh in an individual sales period when the price difference is 0.1 (c) Find 99% confidence and prediction intervals for the mean and individual period de- mands referred to in parts (a) and (b). Hint: Use the formulas given in class to solve for the distance value.

Answer 1

a)
point estimate is fit value from output
= 8.0806

b)
point prediction remains same for prediction interval
= 8.0806

95% prediction interval is (7.4187,8.7425)

c)
margin of error for 95% prediction interval is (8.7425 - 7.4187)/2
= 0.6619

df = 28
t for 95% = t.inv.2t(0.05,28) = 2.0484

t for 99% = t.inv.2t(0.01,28) = 2.7633

hence margin of error for 99%
= 0.6619/2.0484*2.7633
= 0.8929
hence
99% PI =
(8.7425 - 0.8929 , 8.7425 + 0.8929)
= ( 7.8496 , 9.6354)

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