From: aaronshaw <aaron.d.shaw@gmail.com>
Date: Thu, 2 May 2019 02:35:15 +0000 (-0500)
Subject: updating to w06 content outline. still no substance
X-Git-Url: https://code.communitydata.science/stats_class_2019.git/commitdiff_plain/6b6a21a00c0fc0eff5645b65b080c5cae393fe25?ds=inline;hp=d1fbf48eefd7acc23c5d311c256662f830a51a63

updating to w06 content outline. still no substance
---

diff --git a/r_lectures/w06-R_lecture.Rmd b/r_lectures/w06-R_lecture.Rmd
index 1861cb6..56c75cf 100644
--- a/r_lectures/w06-R_lecture.Rmd
+++ b/r_lectures/w06-R_lecture.Rmd
@@ -1,7 +1,8 @@
 ---
-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
 ---
 
@@ -9,97 +10,17 @@ 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.
\ No newline at end of file