---
-title: "Week 7 R Lecture"
-author: "Jeremy Foote"
-date: "April 4, 2019"
+title: "Week 6 R lecture"
+subtitle: "Statistics and statistical programming \nNorthwestern University \nMTS 525"
+author: "Aaron Shaw"
+date: "May 3, 2019"
output: html_document
---
knitr::opts_chunk$set(echo = TRUE)
```
-## Categorical Data
+## T-tests
+You learned the theory/concepts behind t-tests last week, so here's a brief run-down on how to use built-in functions in R to conduct them and interpret the results.
-The goal of this script is to help you think about analyzing categorical data, including proportions, tables, chi-squared tests, and simulation.
+## ANOVAs
-### Estimating proportions
+Analogous situation with t-tests. Here's a brief introduction to how they work in R.
-If a survey of 50 randomly sampled Chicagoans found that 45% of them thought that Giordano's made the best deep dish pizza, what would be the 95% confidence interval for the true proportion of Chicagoans who prefer Giordano's?
+## Visualizing confidence intervals
-Can we reject the hypothesis that 50% of Chicagoans prefer Giordano's?
+We spent a lot of time on confidence intervals in the past few weeks. Since they can be so useful, surely we should learn some approaches to incorporating them into data visualizations.
+## Date/time arithmetic
-```{r}
-est = .45
-sample_size = 50
-SE = sqrt(est*(1-est)/sample_size)
-
-conf_int = c(est - 1.96 * SE, est + 1.96 * SE)
-conf_int
-```
-
-What if we had the same result but had sampled 500 people?
-
-
-```{r}
-est = .45
-sample_size = 500
-SE = sqrt(est*(1-est)/sample_size)
-
-conf_int = c(est - 1.96 * SE, est + 1.96 * SE)
-conf_int
-```
-
-### Tabular Data
-
-The Iris dataset is composed of measurements of flower dimensions. It comes packaged with R and is often used in examples. Here we make a table of how often each species in the dataset has a sepal width greater than 3.
-
-```{r}
-
-table(iris$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)
-```
-
+Last, but not least, another wrinkle in time...or at least how to manage date-time objects in R.
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