Problem 4: Consider the data shown below: Please use R, and show the codes, outp
ID: 3066684 • Letter: P
Question
Problem 4: Consider the data shown below:
Please use R, and show the codes, outputs, formulas, and brief explanation.
x
y
x
y
4
24.6
6.5
67.11
4
24.71
6.5
67.24
4
23.9
6.75
67.15
5
39.5
7
77.87
5
39.6
7.1
80.11
6
57.12
7.3
84.67
a.- Fit a second-order polynomial model to these data.
b.- Test for significance of regression.
c.- Test for lack of fit and comment on the adequacy of the second-order model.
d.- Test the hypothesis H0: B2=0. Can the quadratic term be deleted from this equation?
x
y
x
y
4
24.6
6.5
67.11
4
24.71
6.5
67.24
4
23.9
6.75
67.15
5
39.5
7
77.87
5
39.6
7.1
80.11
6
57.12
7.3
84.67
Explanation / Answer
x=c(4,4,4,5,5,6,6.5,6.5,6.75,7,7.1,7.3)
> y=c(24.6,24.71,23.9,39.5,39.6,57.12,67.11,67.24,67.15,77.87,80.11,84.61)
> ##a)
> z=lm(y~x+I(x^2))
> z
Call:
lm(formula = y ~ x + I(x^2))
Coefficients:
(Intercept) x I(x^2)
-4.666 1.466 1.459
> ##b)
## testing:
H0: beta is equal to zero.
Vs
H1: beta is not equal to zero.
> fit=lm(y~x)
> fit
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
-47.40 17.68
> summary(fit)
Call:
lm(formula = y ~ x)
Residuals:
Min 1Q Median 3Q Max
-4.7648 -1.4072 0.1688 1.4367 2.9735
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -47.3966 3.0389 -15.60 2.40e-08 ***
x 17.6758 0.5157 34.28 1.06e-11 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 2.205 on 10 degrees of freedom
Multiple R-squared: 0.9916, Adjusted R-squared: 0.9907
F-statistic: 1175 on 1 and 10 DF, p-value: 1.057e-11
> ##INTERPRETATION: Here p_value is 1.057e-11 which is much less than 0.05.therefore we reject the null hypothesis that beta=0.Hence there is significant relation between the variables in the linear regression model.
> ##c)
## testing:
H0: there is no lack of fit in the given data.
Vs
H1: there is lack of fit in the given data.
> anova(lm(y~x+I(x^2)),lm(y~factor(x)*factor(I(x^2))))
Analysis of Variance Table
Model 1: y ~ x + I(x^2)
Model 2: y ~ factor(x) * factor(I(x^2))
Res.Df RSS Df Sum of Sq F Pr(>F)
1 9 24.6204
2 4 0.3995 5 24.221 48.5 0.001133 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
>
> ##INTERPRETATION: Here p_value is 0.001133 which is less than 0.05.therefore we reject the null hypothesis that there is no lack of fit. Hence the second order polynomial model is inadequate.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.