> modlyq <- lm(ly poly(x,2,raw=true)) > summary(modlyq) Call: lm(formula = ly poly(x, 2, raw = TRUE))

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1 School of Mathematical Sciences MTH5120 Statistical Modelling I Tutorial 4 Solutions The first two models were looked at last week and both had flaws. The output for the third model with log y and a quadratic model is given below. > modlyq <- lm(ly poly(x,2,raw=true)) > summary(modlyq) lm(formula = ly poly(x, 2, raw = TRUE)) Estimate Std. Error t value (Intercept) poly(x, 2, raw = TRUE) poly(x, 2, raw = TRUE) Pr(> t ) (Intercept) < 2e-16 *** poly(x, 2, raw = TRUE)1 8.04e-10 *** poly(x, 2, raw = TRUE)2 2e-04 *** Residual standard error: on 33 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 2 and 33 DF, p-value: < 2.2e-16 > anova(modlyq) 1

2 Analysis of Variance Table Response: ly Df Sum Sq Mean Sq F value Pr(>F) poly(x, 2, raw = TRUE) < 2.2e-16 Residuals poly(x, 2, raw = TRUE) *** Residuals The fitted model is ly = x x 2. All the parameters are highly significant. The value of R 2 is 96.99% so most of the variation is explained by the model. > stdres3 <-rstandard(modlyq) > fits3<-fitted(modlyq) > plot(fits3,stdres3, main="std res vs fits, Janka3") > qqnorm(stdres3, main="q-q Plot Janka 3") > qqline(stdres3) > shapiro.test(stdres3) Shapiro-Wilk normality test data: stdres3 W = , p-value = > hat <- hatvalues(modlyq) > cook<-cooks.distance(modlyq) > i<-1:36 > plot(i,hat, main="leverage values") > plot(i,cook, main="cooks distance values") > qf(0.5, 3, 33) [1] The plot of standardised residuals versus fitted values looks random. The QQ plot has points close to the line and the Shapiro-Wilk test confirms there is no reason to doubt normality. 2

3 The largest leverage values are above 6/n but not 9/n. The largest Cook s distance is for observation 35 but at about 0.35 it is well below the value of which would indicate a highly influential observation. This model fits very well. I next tried a cubic model. > modlyc<- lm(ly poly(x,3,raw=true)) > summary(modlyc) lm(formula = ly poly(x, 3, raw = TRUE)) Estimate Std. Error t value (Intercept) 2.786e e poly(x, 3, raw = TRUE) e e poly(x, 3, raw = TRUE) e e poly(x, 3, raw = TRUE) e e Pr(> t ) (Intercept) ** poly(x, 3, raw = TRUE) ** poly(x, 3, raw = TRUE) * poly(x, 3, raw = TRUE) Residual standard error: on 32 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 3 and 32 DF, p-value: < 2.2e-16 The R 2 = 97.32%, slightly higher but the cubic term is not significant at the 5% significance level (p=0.0554). By the principle of parsimony a simpler model is to be preferred. And so I would choose the quadratic model over this. Next I tried transforming by taking the square root of y. > sy <- (yˆ0.5) 3

4 > modsy <- lm(sy x) > summary(modsy) lm(formula = sy x) Estimate Std. Error t value Pr(> t ) (Intercept) x <2e-16 *** Residual standard error: 2.06 on 34 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 1 and 34 DF, p-value: < 2.2e-16 > anova(modsy) Analysis of Variance Table Response: sy Df Sum Sq Mean Sq F value Pr(>F) x < 2.2e-16 *** Residuals > plot(x,sy, main="sqrt Y versus X") > (abline (1.045, , lty=3)) The fitted model was y 0.5 = x. The intercept wasn t significant but the slope was. R 2 = 96.54%. The one slight problem with this model was a mild funnel shape in the standardised residual versus fitted value plot, suggesting the variance is not constant. The Shapiro Wilk test showed no problem with normality.no values values had very high leverage. The Cook s distance of observation 35 was about 0.50, less than the cutoff of but higher than for the log y quadratic model. 4

5 Overall, mainly because of the funnel shape I prefer the log quadratic model to this square root linear model but there is not much in it as it is a simpler model and I could understand it if you decided this was a better model. I also tried a reciprocal model. lm(formula = ry x) e e e e e-04 Estimate Std. Error t value Pr(> t ) (Intercept) 2.725e e < 2e-16 *** x e e e-13 *** Residual standard error: on 34 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: 138 on 1 and 34 DF, p-value: 1.618e-13 > anova(modry) Analysis of Variance Table Response: ry Df Sum Sq Mean Sq F value Pr(>F) x e e e-13 *** Residuals e e-08 The fitted model was 1/y = x. The parameters were highly significant. R 2 = 80.23%. It was clear from the scatterplot and residual plots that a quadratic model needed to be fitted. lm(formula = ry poly(x, 2, raw = TRUE)) 5

6 -2.558e e e e e-04 Estimate Std. Error t value (Intercept) 5.350e e poly(x, 2, raw = TRUE) e e poly(x, 2, raw = TRUE) e e Pr(> t ) (Intercept) < 2e-16 *** poly(x, 2, raw = TRUE)1 9.06e-14 *** poly(x, 2, raw = TRUE)2 9.45e-11 *** Residual standard error: on 33 degrees of freedom Multiple R-squared: ,Adjusted R-squared: F-statistic: on 2 and 33 DF, p-value: < 2.2e-16 > anova(modry) Analysis of Variance Table Response: ry Df Sum Sq Mean Sq F value poly(x, 2, raw = TRUE) e e Residuals e e-08 Pr(>F) poly(x, 2, raw = TRUE) < 2.2e-16 *** Residuals I fitted this model which was 1/y = x x 2. R 2 = 94.55%. There was now one clear influential outlier and as a result the assumption of normality was rejected. As there were these problems and the R 2 value is higher I prefer the model with a log transformation of y and a quadratic model. So I decided the best model of those I have looked at is ly = x x 2. (But I would also accept the simple linear regression model with a square root transformation.) 6