---
title: "STAT 513 fa 2018 hw XX"
author: "YOUR NAME HERE"
date: "THE DATE HERE"
output:
  pdf_document: default
  html_document: default
---

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

# 1. 

  a) $H_0$: $p = 1/6$ versus $H_1$: $p \neq 1/6$
  b) $H_0$: $p \leq 1/6$ versus $H_1$: $p > 1/6$
  c) $H_0$: $p \geq 1/6$ versus $H_1$: $p <  1/6$
  d)  i) Flip the coin very many times; if "heads"" turns up close to half the time, it seems to support the claim; however, someone could always say, "well, maybe the probability of heads is $0.5001$."
    
      ii) Even if from a very large number of tosses "heads" comes up close to half the time, one can never defend a claim that the coin is perfectly balanced: an objector can always posit that the true probability of "heads" is $0.50001$ or $0.5000001$, and then one has to do more tosses \emph{ad infinitum}.  Thus it is easier to collect evidence against the claim of balancedness than in favor of it. 
      
# 2.

  a) $H_0$: $p=1/2$ versus $H_0$: $p \neq 1/2$.
  b)  i) `sum(dbinom(8:10,10,.6),dbinom(0:2,10,.6))=0.1795843`
    ii) `1-sum(dbinom(8:10,10,.3),dbinom(0:2,10,.3))}= 0.6156268`
    iii) $\gamma(p) = \sum_{y=0}^2{ 10 \choose y }p^y(1-p)^{10-y} + \sum_{y=8}^{10}{ 10 \choose y }p^y(1-p)^{10-y}$ 
    iv) The size is 
        $$
        \begin{aligned}\gamma(1/2) &= \sum_{y=0}^2{ 10 \choose y }(1/2)^y(1-1/2)^{10-y} + \sum_{y=8}^{10}{ 10 \choose y }(1/2)^y(1-1/2)^{10-y}\\
                &=\texttt{sum(dbinom(0:2,10,.5),dbinom(8:10,10,.5))}\\
                &= 0.109375.    
        \end{aligned}
        $$
      v) The code below plots the power curve.
      vi) The power is equal to the size at $p=1/2$, which is the value of $p$ specified in the null hypothesis.
      
```{r}
p.seq <- seq(.01,.99,length=99)
power <- numeric()
for(j in 1:99)
{
    power[j] <- sum(dbinom(0:2,10,p.seq[j]),dbinom(8:10,10,p.seq[j]))
}
plot(p.seq,power,type="l",ylim=c(0,1),xlab="p")
abline(v=0.5,lty=3)		# vert line at null value
abline(h=0.109375,lty=3)# horiz line at size
```
  
  c) Consider tests of the form
    $$
    \text{Reject $H_0$ iff $X_1+\dots+X_{20} \in \mathcal{R}$}
    $$
    for different rejection regions $\mathcal{R}$.  One option is to choose 
    $$
    \mathcal{R} = \{0,1,2,3,4,5\}\cup\{15,16,17,18,19,20\},
    $$ 
    with which the test has size
    $$
    \texttt{sum(dbinom(0:5,20,.5),dbinom(15:20,20,.5))} =  0.04138947 < 0.05
    $$
  d) The following code plots the power curve.
```{r}
p.seq <- seq(.01,.99,length=99)
power <- numeric()
for(j in 1:99)
{
    power[j] <- sum(dbinom(0:5,20,p.seq[j]),dbinom(15:20,20,p.seq[j]))
}
plot(p.seq,power,type="l",ylim=c(0,1),xlab="p")
abline(v=0.5,lty=3)  # vert line at null value
abline(h=0.05,lty=3) # horiz line at max. allowed size
```