Page 97 - Basic Statistics

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Computerized solution:

Table 5.4 Splus print out

*** Linear Model ***

Call: lm(formula = gasoline ~ car, data = riset1, na.action
= na.exclude)

Residuals:
Min 1Q Median 3Q Max
-0.2982 -0.1504 0.04567 0.1123 0.3172

Coefficients:
Value Std. Error t value Pr(>|t|)
(Intercept) -0.7124 1.3361 -0.5332 0.6056
Car 0.2205 0.0459 4.8017 0.0007

Residual standard error: 0.196 on 10 degrees of freedom

Multiple R-Squared: 0.6975

F-statistic: 23.06 on 1 and 10 degrees of freedom, the p-value is
0.0007218

Correlation of Coefficients:
(Intercept)
Car -0.9991

Analysis of Variance Table

Response: gasoline

Terms added sequentially (first to last)
Df Sum of Sq Mean Sq F Value Pr(F)
Car 1 0.8854023 0.8854023 23.05599 0.0007218227
Residuals 10 0.3840227 0.0384023

Comment:

Solving of the above computerized presenting:

a. In the coefficients part, the estimations of both the value of the coefficients:

ˆ
ˆ
β = –0.7124 and β = 0.2205
1
0
ˆ
So that the regression equation is Y = -0.7128 + 0.2205 Xj
j

~~* CHAPTER 5 LINEAR REGRESSION MODEL *~~