Question

For expert using R , I solve it but i need to figure out what I got is correct or wrong. Thank you

(a) What is the second degree polynomial model in terms of mnemonic variable names and its estimated regression equation? Test hypotheses for slope parameters and write a short paragraph stating your findings. Use a significance level of α-0.05. (b) (c) Complete the following analysis of variance table for this model Sum SqMean S Source df F valuePrF) Total Corrected) (d) Perform regression diagnostics and comment on the validity of each assumption. This includes four diagnostic plots and the Shapiro-Wilk test. For Shapiro-Wilk, give hypotheses, result fron use a-0.01 to make a decision about the null hypothesis, and give a conclusion (e) Make a scatterplot of the sample data and include the estimated regression curve for the second degree polynomial Comment on this gr () Compare the regression models for simple linear regression and the second degree polynomial model. Use Adjusted R and Mean Square Error (or Mean Square Residuals). Comment on the results. Which model do you recommend?

# Simple Linear Regression and Polynomial Regression

# HW 2

#

# Read data from csv file

data <- read.csv("C:\data\SweetPotatoFirmness.csv",header=TRUE, sep=",")

head(data)

str(data)

# scatterplot of independent and dependent variables

plot(data$pectin,data$firmness,xlab="Pectin, %",ylab="Firmness")

par(mfrow = c(2, 2)) # Split the plotting panel into a 2 x 2 grid

model <- lm(firmness ~ pectin , data=data)

summary(model)

anova(model)

plot(model)

shapiro.test(resid(model))

# examine histogram and boxplot of residuals

par(mfrow = c(1, 1))

hist(resid(model))

boxplot(resid(model))

# predict dependent variable for specified value of independent variable

predict(model, data.frame(pectin = 1.5))

# Estimated regression line and scatterplot of data

plot(data$pectin,data$firmness,xlab="Pectin, %", ylab="Firmness",

ylim=c(45,75),xlim=c(0,3),main="Simple Linear Regression",

pch=19,cex=1.5)

lines(sort(data$pectin),fitted(model)[order(data$pectin)], col="blue", type="l")

par(mfrow = c(2, 2))

# fit a second degree polynomial

# create quadratic term for pectin

data$pectinSq <- data$pectin^2

model2 <- lm(firmness ~ pectin + pectinSq , data=data)

summary(model2)

anova(model2)

plot(model2)

shapiro.test(resid(model2))

par(mfrow = c(1, 1))

hist(resid(model2))

boxplot(resid(model2))

# predict dependent variable for specified value of independent variable

predict(model2, data.frame(pectin = 1.5, pectinSq=2.25))

# Estimated regression line and scatterplot of data

plot(data$pectin,data$firmness,xlab="Pectin, %", ylab="Firmness",

ylim=c(45,75),xlim=c(0,3),main="Simple Linear Regression",

pch=19,cex=1.5)

lines(sort(data$pectin),fitted(model2)[order(data$pectin)], col="blue", type="l")

Answer 1

The code is perfectly fine here. But the conclusions are needed in each part.

Both for linear model and second degree polynomial model fitting, we should comment on our findings in the following way :

(C) The ANOVA table give the conclusions in such a way that :

If the p value <= .05, we should reject the null hypothesis at level 0.05.

And p value > .05 implies we should accept the null hypothesis at level 0.05.

(d) In Shapiro Wilks normality test, if the p value comes out to be less than or equal to 0.05 , we conclude that the errors are not from Normal distribution.

If p value > 0.05, we conclude that the errors are Normally distributed.

(e) From the scatterplot, if the data goes more or less closely to the estimated regression curve for second degree polynomial , then fitting second degree polynomial to the data will be the best option.

(f) on fitting both linear regression and second degree polynomial regression we get adjusted R square values.

The one with the greater adjusted R square value will be better to fit to the given data..