---
title: "Week 7 R Lecture"
author: "Jeremy Foote"
date: "April 4, 2019"
output: html_document
---
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```
## Categorical Data
The goal of this script is to help you think about analyzing categorical data, including proportions, tables, chi-squared tests, and simulation.
### Estimating proportions
If a survey of 50 randomly sampled Chicagoans found that 45% of them thought that Giordano's made the best deep dish pizza, what would be the 95% confidence interval for the true proportion of Chicagoans who prefer Giordano's?
Can we reject the hypothesis that 50% of Chicagoans prefer Giordano's?
```{r}
est = .45
sample_size = 50
SE = sqrt(est*(1-est)/sample_size)
conf_int = c(est - 1.96 * SE, est + 1.96 * SE)
conf_int
```
What if we had the same result but had sampled 500 people?
```{r}
est = .45
sample_size = 500
SE = sqrt(est*(1-est)/sample_size)
conf_int = c(est - 1.96 * SE, est + 1.96 * SE)
conf_int
```
### Tabular Data
The Iris dataset is composed of measurements of flower dimensions. It comes packaged with R and is often used in examples. Here we make a table of how often each species in the dataset has a sepal width greater than 3.
```{r}
table(iris$Species, iris$Sepal.Width > 3)
```
The chi-squared test is a test of how much the frequencies we see in a table differ from what we would expect if there was no difference between the groups.
```{r}
chisq.test(table(iris$Species, iris$Sepal.Width > 3))
```
The incredibly low p-value means that it is very unlikely that these came from the same distribution and that sepal width differs by species.
## BONUS: Using simulation to test hypotheses and calculate "exact" p-values
When the assumptions of $\chi^2$ tests aren't met, we can use simulation to approximate how likely a given result is. The material here comes from the final two sections of Chapter 6 of the *OpenIntro* textbook. The book uses the example of a medical practitioner who has 3 complications out of 62 procedures, while the typical rate is 10%.
The null hypothesis is that this practitioner's true rate is also 10%, so we're trying to figure out how rare it would be to have 3 or fewer complications, if the true rate is 10%.
```{r}
# We write a function that we are going to replicate
simulation <- function(rate = .1, n = 62){
# Draw n random numbers from a uniform distribution from 0 to 1
draws = runif(n)
# If rate = .4, on average, .4 of the draws will be less than .4
# So, we consider those draws where the value is less than `rate` as complications
complication_count = sum(draws < rate)
# Then, we return the total count
return(complication_count)
}
# The replicate function runs a function many times
simulated_complications <- replicate(5000, simulation())
```
We can look at our simulated complications
```{r}
hist(simulated_complications)
```
And determine how many of them are as extreme or more extreme than the value we saw. This is the "exact" p-value.
```{r}
sum(simulated_complications <= 3)/length(simulated_complications)
```