The following table presents data on the selling price (y) and annual taxes (x)

Question

The following table presents data on the selling price (y) and annual taxes (x) for 24 houses. a) Assuming that a simple linear regression model is appropriate, obtain the least squares fit relating selling price to taxes paid. b) Plot the residuals against the input, and comment on the results. c) Test the significance of regression at 1% level of significance. d) Is the slope significantly different than 3.100 at 5% level of significance? e) Find the 95% confidence interval for the slope of the regression line. f) Find the 95% confidence interval for the intercept of the regression line. g) Find the 95% confidence interval about the mean response for taxes paid x_0 = 7.500. h) Find the 95% prediction interval for a single new observation at taxes x_0 = 7.500.

Explanation / Answer

Price <- c(25.9,29.5,27.9,25.9,29.9,29.9,30.9,28.9,35.9,31.5,31,30.9,30,36.9,41.9,
+ 40.5,43.9,37.5,37.9,44.5,37.9,38.9,36.9,45.8)
> Taxes <- c(4.9176,5.0208,4.5429,4.5573,5.0597,3.891,5.898,5.6039,5.8282,
+ 5.3003,6.2712,5.9592,5.05,8.2464,6.6969,7.7841,9.0384,5.9898,7.5422,8.7951,6.0831,
+ 8.3607,8.14,9.1416)
> relation <- lm(Price~Taxes)
> print(summary(relation))

Call:
lm(formula = Price ~ Taxes)

Residuals:
Min 1Q Median 3Q Max
-3.8343 -2.3156 -0.3668 1.9786 6.3169

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 13.3199 2.5717 5.179 3.42e-05 ***
Taxes 3.3244 0.3903 8.518 2.05e-08 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.961 on 22 degrees of freedom
Multiple R-squared: 0.7673, Adjusted R-squared: 0.7568
F-statistic: 72.56 on 1 and 22 DF, p-value: 2.05e-08

a) price ^ = 13.3199 + 3.3244 *Taxes

b)

c) over-all model has p-value = 2.05e-8 <<0.01 ,hence it is significant

d) TS = (3.3244 - 3.1 ) /0.3903 = 0.574942

t-critical = 2.074

since TS < t-critical

we fail to reject the null

Please post rest parts again . You can copy Data and paste in next question,

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