Question

Instructions: Questions labeled with (R) require use of R. Please provide appropriate R commands and their output, along with sufficient explanation and interpretation of the output to demonstrate your understanding. All other questions should be completed without reference to any R commands or output. Make sure you give enough explanation so your tutor can follow your reasoning if you happen to make a mistake. Please also try to be as succinct as possible. Each assignment will include marks tor good presentation and for attempting all problems

NOTE: PLEASE INCLUDE ALL THE R CODES ALONG WITH THE PLOTS

Answer 1

(a)

R code:

x=c(28.8,24.4,30.1,25.6,26.4,23.9,22.1,22.5,27.6,28.1,20.8,27.7,24.4,25.1,24.6,
26.3,28.2,22.2,26.3,24.4)
summary(x)
boxplot(x)

Output:

Min. 1st Qu. Median Mean 3rd Qu. Max.
20.80 24.28 25.35 25.48 27.63 30.10

26 30 24 26 2

Center=25.35, Inter quartile range=27.63-24.28=3.35. Since 3rd quartile-2nd quartile>2nd quartile-1st quartile so the distribution is positively skewed.

$(b)~Suppose~X_1,X_2,...,X_n~are~random~sample~from~N(\mu,\sigma^2).\\ The~likelihood~function=L(\mu,\sigma^2)=\prod_{i=1}^n\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{1}{2\sigma^2}(x_i-\mu)^2 \right )\\ =\frac{1}{(2\pi\sigma^2)^{\frac{n}{2}}}\exp\left(-\frac{1}{2\sigma^2}\sum_{i=1}^n(x_i-\mu)^2 \right )\\ l(\mu,\sigma^2)=\log L(\mu,\sigma^2)=-\frac{n}{2}\log 2\pi-\frac{n}{2}\log \sigma^2-\frac{1}{2\sigma^2}\sum_{i=1}^n(x_i-\mu)^2\\ \frac{\partial l(\mu,\sigma^2)}{\partial \mu}=\frac{1}{\sigma^2}\sum_{i=1}^n(x_i-\mu)=0\Rightarrow \hat{\mu}=\frac{1}{n}\sum_{i=1}^nx_i=\bar{x}\\ \frac{\partial l(\mu,\sigma^2)}{\partial \sigma^2}=-\frac{n}{2\sigma^2}+\frac{1}{2\sigma^4}\sum_{i=1}^n(x_i-\mu)^2=0\Rightarrow \hat{\sigma}^2=\frac{1}{n}\sum_{i=1}^n(x_i-\hat{\mu})^2\\ =\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2\\$

$\left.\begin{matrix} \frac{\partial^2 l(\mu,\sigma^2)}{\partial \mu^2} \end{matrix}\right|_{\mu=\hat{\mu},\sigma^2=\hat{\sigma}^2}=-\frac{n}{\hat{\sigma}^2},~ \left.\begin{matrix} \frac{\partial^2 l(\mu,\sigma^2)}{\partial \mu\partial \sigma} \end{matrix}\right|_{\mu=\hat{\mu},\sigma^2=\hat{\sigma}^2}=0,\\ ~\left.\begin{matrix} \frac{\partial^2 l(\mu,\sigma^2)}{\partial \sigma^2} \end{matrix}\right|_{\mu=\hat{\mu},\sigma^2=\hat{\sigma}^2}=-\frac{n}{2\hat{\sigma}^4}\\ then~\begin{pmatrix} \left.\begin{matrix} \frac{\partial^2 l(\mu,\sigma^2)}{\partial \mu^2} \end{matrix}\right|_{\mu=\hat{\mu},\sigma^2=\hat{\sigma}^2}&\left.\begin{matrix} \frac{\partial^2 l(\mu,\sigma^2)}{\partial \mu\partial \sigma} \end{matrix}\right|_{\mu=\hat{\mu},\sigma^2=\hat{\sigma}^2} \\ & \left.\begin{matrix} \frac{\partial^2 l(\mu,\sigma^2)}{\partial \sigma^2} \end{matrix}\right|_{\mu=\hat{\mu},\sigma^2=\hat{\sigma}^2} \end{pmatrix} is~negative~definite.\\$

$Hence~MLEs~of~\mu~and~\sigma^2~are~\bar{x}~and~\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2.\\ For~this~data~set,~ML~estimates~are:~\bar{x}=25.475,~\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2=5.9069\\$

(c)

x=c(28.8,24.4,30.1,25.6,26.4,23.9,22.1,22.5,27.6,28.1,20.8,27.7,24.4,25.1,24.6,
26.3,28.2,22.2,26.3,24.4)
h<-hist(x, breaks=6, col="red", xlab="Pull force",
main="Histogram with Normal Curve",ylim=c(0,7))
xfit<-seq(min(x),max(x),length=40)
yfit<-dnorm(xfit,mean=mean(x),sd=sqrt(5.9069))
yfit <- yfit*diff(h$mids[1:2])*length(x)
lines(xfit, yfit, col="blue", lwd=2)

(d)

R code:

x=c(28.8,24.4,30.1,25.6,26.4,23.9,22.1,22.5,27.6,28.1,20.8,27.7,24.4,25.1,24.6,
26.3,28.2,22.2,26.3,24.4)
qqnorm(x,lwd=2);qqline(x, col = 2,lwd=2)

Normal Q-Q Plot 0 CH 丈 CH 2 0 1 Theoretical Quantiles

It is observed that the sample quantiles are closed to reference line hence we can assume data come from normal distribution.