hw2_5 thru 8 answers.txt - 6(a...

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6: (a): x<-c(2.4,1.6,2.0,2.6,1.4,1.6,2.0,2.2) y<-c(225,184,220,240,180,184,186,215) cor(x,y) [1] 0.9129053 //sqrt root of R^2 (b): summary(lm(y~x)) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -19.5182 -1.2267 -0.7368 4.2632 14.4818 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 104.06 18.65 5.581 0.00141 ** x 50.73 9.26 5.478 0.00155 ** --- Signif. codes: 0 ¡®***¡¯ 0.001 ¡®**¡¯ 0.01 ¡®*¡¯ 0.05 ¡®.¡¯ 0.1 ¡® ¡¯ 1 Residual standard error: 10.29 on 6 degrees of freedom Multiple R-squared: 0.8334, Adjusted R-squared: 0.8056 F-statistic: 30.01 on 1 and 6 DF, p-value: 0.001546 LS regression equation:104.06+ 50.73 x cat("the ls regression equation is y=",lm(y~x)$coefficients[1],"+x*",lm(y~x) $coefficients[2]) (c): According to (b): Multiple R-squared: 0.8334 summary(lm(y~x))$r.squared [1] 0.8333961 (d): > anova(lm(y~x)) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) x 1 3178.2 3178.2 30.014 0.001546 ** Residuals 6 635.3 105.9 --- Signif. codes: 0 ¡®***¡¯ 0.001 ¡®**¡¯ 0.01 ¡®*¡¯ 0.05 ¡®.¡¯ 0.1 ¡® ¡¯ 1 p-value = 0.001546, so there is significant linear relationship. anova(lm(y~x))$Pr[1] Another way: p-value of slope is the same as p-value in anova summary(lm(y~x))$coeff[2,4] (e):
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predict(lm(y~x),data.frame(x=1.8)) 1 195.3725 lm(y~x)$coefficients[1]+lm(y~x)$coefficients[2]*1.8
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