just updating the name here from wk 6 to wk 7

r_lectures/w07-R_lecture.Rmd

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+---
+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.
+
+
+
+## Using Simulation
+
+When the assumptions of Chi-squared tests aren't met, we can use simulation to approximate how likely a given result is.
+
+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 p-value.
+
+```{r}
+
+sum(simulated_complications <= 3)/length(simulated_complications)
+```
+