HW4.Rmd

---
title: "STAT645 - Homework 4"
author: "Salih Kilicli"
date: "10/1/2019"
output: html_document
editor_options: 
  chunk_output_type: console
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```



**Problem 1: Carry out a simulation to investigate the sampling distribution of a difference in two sample medians and compare it with that of two sample means. Let $n = 100$ (the sample size), and carry out $B = 1000$ simulations. Simulate samples from:**

**1) Gamma with shape 2 and scale 0.5**

**2) Gamma with shape 2.5 and scale 0.75; each sample should be of size n = 100. **

**Simulate the difference in medians (first group minus the second). Report a histogram of the simulated median differences. Overlay a density curve from the normal distribution with mean and variance same as that of the sampling distribution. Do a similar sampling distribution plot for the difference of the sample means, and comment on the differences.**

```{r}
B = 1000
n = 100
dmedian = 0*1:B
dmean = 0*1:B

for (i in 1:B)
{
set.seed(i)
s1 = rgamma(n, shape=2, scale=0.5)
s2 = rgamma(n, shape=2.5, scale=0.75)
dmedian[i] = median(s1)-median(s2)
dmean[i] = mean(s1-s2)
}

par(mfrow = c(1, 2))

x1 = seq(min(dmedian), max(dmedian), by = 0.01)
y1 = dnorm(x1, mean=mean(dmedian), sd=sd(dmedian))
hist(dmedian, main = "Median Difference", xlab = "n=100, B=1000", probability= TRUE)
lines(y1 ~ x1, col="blue", lwd=2)

x2 = seq(min(dmean),max(dmean), by=0.01)
y2 = dnorm(x2, mean=mean(dmean), sd=sd(dmean))
hist(dmean, main = "Mean Difference", xlab = "n=100, B=1000", probability= TRUE)
lines(y2 ~ x2, col="blue", lwd=2)
```

The plot of median difference looks skewed as compared to plot of mean difference.


**Problem 2: Carry out a simulation to compare the two sample t-test (with equal variance) to the paired t-test under a variety of correlation values. Specifically, for pairwise correlations, $\sigma$, of $-0.5, -0.45, \dots , 0.0, 0.05, \dots, 0.95,$ use simulation (number of simulations, B, equal to $5000$) to approximate the power to reject the null hypothesis of equal means against the two-sided alternative hypothesis. Let the sample size $n$ be $30$; so, there will be $30$ pairs. For the population variance, use $\sigma = 1$ for both groups; i.e.,**

$$\Sigma = \begin{bmatrix}
1&\sigma\\
\sigma&1\\
\end{bmatrix}.$$

**You can simulate correlated data using *mvrnorm* in the *MASS* package of R.**

**For each of $\mu = [0, 0]$,  $\mu = [0, 0.25]$,  $\mu= [0, 0.5]$, and  $\mu= [0, 0.75]$, do the above and report a plot of $\sigma$ versus the rejection probabilities. Each plot should have 2 connected line segments, one for the two-sample t-test and one for the paired t-test; use different colors to distinguish between the two. You may use the t.test function of R. What do the plots tell you? **

```{r}
library(MASS)

rho = seq(-0.5, 0.95, by=0.05)
m=rbind(c(0,0),c(0,0.25),c(0,0.5),c(0,0.75))
p1=NULL; rej1=NULL;
p2=NULL; rej2=NULL;

  for (i in 1:30){
    sigma=matrix(c(1,rho[i],rho[i],1), ncol=2, byrow=T)
  
      for (j in 1:5000){
        data=mvrnorm(30, m[1,], sigma)
        p1[j]=t.test(data[,1], data[,2], mu=0, alternative="two.sided", var.equal=T)$p.value
        p2[j]=t.test(data[,1], data[,2], mu=0, alternative="two.sided", paired=T)$p.value
        
      }

  rej1[i]=sum(p1<0.05)/length(p1)
  rej2[i]=sum(p2<0.05)/length(p1)
  }

plot(rho, rej1, ylim=c(0,1), pch=20, ylab="Rejection Probability", xlab="Rho",
     main="Power comparison of t.tests - mu'=[0,0]", col="blue", las=1)
points(rho, rej2, ylim=c(0,1), pch=20, col="red")
legend("topleft",  c("2-Sample", "Paired"), fill=c("blue","red"))

p1=NULL; rej1=NULL;
p2=NULL; rej2=NULL;

  for (i in 1:30){
    sigma=matrix(c(1,rho[i],rho[i],1), ncol=2, byrow=T)
  
      for (j in 1:5000){
        data=mvrnorm(30, m[2,], sigma)
        p1[j]=t.test(data[,1], data[,2], mu=0, alternative="two.sided", var.equal=T)$p.value
        p2[j]=t.test(data[,1], data[,2], mu=0, alternative="two.sided", paired=T)$p.value
      }

  rej1[i]=sum(p1<0.05)/length(p1)
  rej2[i]=sum(p2<0.05)/length(p2)
  }

plot(rho, rej1, ylim=c(0,1), pch=20, ylab="Rejection Probability", xlab="Rho",
     main="Power comparison of t.tests - mu'=[0,0.25]", col="blue", las=1)
points(rho, rej2, ylim=c(0,1), pch=20, col="red")
legend("topleft",  c("2-Sample", "Paired"), fill=c("blue","red"))

p1=NULL; rej1=NULL;
p2=NULL; rej2=NULL;

  for (i in 1:30){
    sigma=matrix(c(1,rho[i],rho[i],1), ncol=2, byrow=T)
  
      for (j in 1:5000){
        data=mvrnorm(30, m[3,], sigma)
        p1[j]=t.test(data[,1], data[,2], mu=0, alternative="two.sided", var.equal=T)$p.value
        p2[j]=t.test(data[,1], data[,2], mu=0, alternative="two.sided", paired=T)$p.value
      }

  rej1[i]=sum(p1<0.05)/length(p1)
  rej2[i]=sum(p2<0.05)/length(p2)
  }

plot(rho, rej1, ylim=c(0,1), pch=20, ylab="Rejection Probability", xlab="Rho",
     main="Power comparison of t.tests - mu'=[0,0.5]", col="blue", las=1)
points(rho, rej2, ylim=c(0,1), pch=20, col="red")
legend("topleft",  c("2-Sample", "Paired"), fill=c("blue","red"))

p1=NULL; rej1=NULL;
p2=NULL; rej2=NULL;

  for (i in 1:30){
    sigma=matrix(c(1,rho[i],rho[i],1), ncol=2, byrow=T)
  
      for (j in 1:5000){
        data=mvrnorm(30, m[4,], sigma)
        p1[j]=t.test(data[,1], data[,2], mu=0, alternative="two.sided", var.equal=T)$p.value
        p2[j]=t.test(data[,1], data[,2], mu=0, alternative="two.sided", paired=T)$p.value
      }

  rej1[i]=sum(p1<0.05)/length(p1)
  rej2[i]=sum(p2<0.05)/length(p2)
  }

plot(rho, rej1, ylim=c(0,1), pch=20, ylab="Rejection Probability", xlab="Rho",
     main="Power comparison of t.tests - mu'=[0,0.75]", col="blue", las=1)
points(rho, rej2, ylim=c(0,1), pch=20, col="red")
legend("topleft",  c("2-Sample", "Paired"), fill=c("blue","red"))

```
 
 It is apparent from the plots that the power of paired t-test is higher than 2 sample t-test since it takes higher probabilities in similar cases as $rho$ increases.


**Problem 3: Consider the calories data in “calories.txt.” These contain data on the difference between the true and reported calorie contents (we will use the differences per item)**

**(a) Construct side by side box plots of the calorie differences by region. Comment.**

```{r}
library(nlme)
setwd("~/Desktop/STAT645/Data")
cal = read.delim("calories.txt")
calories=cal[,3]
region = factor(c(rep(1,20),rep(2,12),rep(3,8)))
boxplot(calories ~ region, names=c("National","Regional","Local"), col=rainbow(3), las=1)
```

Both variability and the median in the calorie information increases from National to Local. Also, there is an outlier for the National region with a minimal value.

**(b) Fit an ordinary least squares model using lm. Report the coefficient estimates, standard errors, and p-values.**

```{r}
model1 = lm(calories ~ region)
summary(model1)
```

**(c) What assumption of the ordinary least squares model appears to be violated?**

```{r}
par(mfrow=c(2,2))
plot(model1)

par(mfrow=c(1,1))
plot(model1$fitted, model1$residuals,  main="Plot of residuals for OLS")
abline(h=0, col="red")
```

Clearly, from "sqrt(standard residuals) vs fitted values" (bottom left) or "fitted values vs residuals" plots we can see that the error variance is increasing. Therefore, **constant error variance** assumption is violated.

**(d) Fit a weighted least squares model using gls, allowing for different variances for each region. Report the coefficient estimates, standard errors,and p-values. Comment on how these compare to the ordinary least squares results.**

```{r}
model2 = gls(calories ~ 1+region, weights = varIdent(form=~1|region))
summary(model2)
```

Coefficient estimates are exactly same; however, std. errors and t-values therefore p-values are fixed (are different).

**Problem 4: For the “calories.txt” data: Use the bootstrap to test the null hypothesis $H_0$ that the population median per-item calorie difference equals zero.**

**(a) Report a 95% BCa confidence interval for the population median. Comment on whether it supports $H_0$ or not.**

```{r}
library(boot)
setwd("~/Desktop/STAT645/Data")
cal = read.delim("calories.txt")
calories=cal[,3]

boot_median = function(calories, indices){
boot_data = calories[indices]
return(median(boot_data))
}

newdata=calories-median(calories)
model4a=boot(calories, boot_median, 10000)
boot.ci(model4a, type="bca")
```

Since 0 is not in the confidence interval we reject the Null hypothesis. There it **doesn't** support the $H_0$.

**(b) Use the bootstrap to compute a p-value, using the sample median $\bar{y}$ as your test statistic.**

```{r}
x = boot(newdata, boot_median, 10000)
tobs = median(calories)
jj=(1:10000)[!is.na(model4a$t)]
p0=sum(as.numeric(x$t>abs(tobs) | x$t<=-abs(tobs)))/length(jj)
cat("p-value using bootstrap method is", p0)
```

**(c) Using a bootstrap approach estimate the bias of the sample median.**

```{r}
model4a
print("Estimate of the bias of the sample median is 0.211075")
```

**Problem 5: For the “mouse.txt” data: Use the bootstrap to test the null hypothesis $H_0$ that the population means of the treatment and control groups are equal.**

**(a) Report a 95% BCa confidence interval for the difference in population means. Comment on whether it supports $H_0$ or not.**

```{r}
library(boot)
setwd("~/Desktop/STAT645/Data")
mouse = read.delim("mouse.txt",header=F)

boot_diff_means_ci = function(data, ind){
  data = data[ind,]
  if(!all(data[,1]=="C")&!all(data[,1]=="TX")){
  return(mean(data[data[,1]=="TX", 2])- mean(data[data[,1]=="C", 2]))  
  } else {return(NA)}
}

model5a = boot(mouse, boot_diff_means_ci, 10000)
summary(model5a)
boot.ci(model5a, type="bca")
```

Since 0 is within the interval we fail to reject the Null hypothesis. In other words, the CI we have found supports $H_0$.

**(b) Use the bootstrap to compute a p-value, using the two-sample t-statistic (unequal variances) as your test statistic.**

```{r}
diff_pval = function(data, ind){
  tx=data[ind,1]
  y =data[ind,2]
  if(sum(tx=="C")>=2 & sum(tx=="TX")>=2){
    return(t.test(y[tx=="C"], y[tx=="TX"], var.equal=F)$stat)
  } else {return(NA)}
}
data = mouse
y0=y=data[,2]
y[data[, 1]=="TX"]=y[data[,1]=="TX"]-mean(y[data[,1]=="TX"])
y[data[, 1]=="C"]=y[data[,1]=="C"]-mean(y[data[,1]=="C"])

datanew=data
datanew[,2]=y

tobs= t.test(y0[data[,1]=="TX"], y0[data[,1]=="C"], var.equal=F)$stat

model5b = boot(datanew, diff_pval, 10000)


jj=(1:10000)[!is.na(model5b$t)]
p = sum(model5b$t[jj]>abs(tobs) | model5b$t[jj]<= -abs(tobs) )/length(jj)
cat("p-value using bootstrap method is", p)
```

**(c) Compare the bootstrap p-value with the p-value obtained using the standard two-sample t test assuming unequal variances.**

```{r}
p1=t.test(y0[data[,1]=="TX"], y0[data[,1]=="C"], var.equal=F)$p.value
cat("p-value using bootstrap is ", p, "whereas p-value using t-test is",p1)
```

p-values in both methods are pretty close to each other.

**Problem 6: For the pollution data in “pollute data.csv” [Note: There is at least one missing value in these data. Just remove any records with at least one missing value prior to doing the following analyses.]: Consider the model from 1(d) in homework 3. For each of the “bootstrapping pairs” and “bootstrapping residuals” approaches, approximate the sampling distribution of the sum of the coefficients for percent white collar and percent non white, and report the following:**

**(a) A histogram of the estimated sampling distribution.**

**(b) 95% confidence intervals, using the normal-theory approach, the percentile approach, and the BCa approach.**