Week 6 Worked Examples
Jeremy Foote
April 11, 2019
Programming Questions
PC0. First we import the data.
raw_df = read.csv("~/Desktop/DeleteMe/Teaching/owan03.csv") # Note that I saved the file as a CSV for importing to R
head(raw_df)## X1 X2 X3 X4
## 1 70 49 30 34
## 2 77 60 37 36
## 3 83 63 56 48
## 4 87 67 65 48
## 5 92 70 76 65
## 6 93 74 83 91PC1. Let’s reshape the data
library(tidyverse)## ── Attaching packages ────────────────────────────────── tidyverse 1.2.1 ──## ✔ ggplot2 3.1.0 ✔ purrr 0.2.5
## ✔ tibble 2.1.1 ✔ dplyr 0.7.7
## ✔ tidyr 0.8.2 ✔ stringr 1.3.1
## ✔ readr 1.1.1 ✔ forcats 0.3.0## ── Conflicts ───────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()# Rename the columns
colnames(raw_df) <- c("None", "Low", "Medium","High")
# Gather gets all of the columns from the dataframe and
# use them as keys, with the value as the value from that cell.
# This gives us a "long" dataframe
df <- gather(raw_df, dose, weeks_alive)
# We want to treat dose as a factor, so we can group by it easier
df$dose <- as.factor(df$dose)
# Finally, when we look at it, we see some missing data (NA); this is because not all group sizes were the same.
# We can safely remove these.
df <- df[complete.cases(df),]PC2: Now we’re goint to get statistics and create some visualizations
tapply(df$weeks_alive, df$dose, summary)## $High
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 34.00 45.00 56.50 65.25 92.75 102.00
##
## $Low
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 49.00 63.00 70.00 69.89 77.00 89.00
##
## $Medium
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 30.00 58.25 79.50 71.50 89.25 97.00
##
## $None
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 70.00 85.00 93.00 91.36 101.00 103.00# Alternative way to do this using tidyverse
df %>% group_by(dose) %>% summarize_all(c('min','max','mean', 'IQR'))## # A tibble: 4 x 5
## dose min max mean IQR
## <fct> <dbl> <dbl> <dbl> <dbl>
## 1 High 34 102 65.2 47.8
## 2 Low 49 89 69.9 14
## 3 Medium 30 97 71.5 31
## 4 None 70 103 91.4 16When it comes to visualizations, we definitely want to use ggplot. We have lots of options for what to do.
# Histograms
h_plot = df %>% ggplot(aes(x=weeks_alive, # What to summarize
fill = dose # How to group by color
)) + geom_histogram(position = 'dodge', bins = 5)
h_plot
# In this case, faceted histograms is probably better
h_facet = df %>% ggplot(aes(x=weeks_alive, # What to summarize
)) + geom_histogram(bins = 5) + facet_grid(~dose)
h_facet
# Density plots
d_plot = df %>% ggplot(aes(x=weeks_alive, # What to summarize
fill = dose # How to group by color
)) + geom_density(alpha = .2)
d_plot
# Boxplots
box_plot = df %>% ggplot(aes(y=weeks_alive,
x = dose
)) + geom_boxplot()
box_plot
# My favorite - ridgeline plots
# install.packages('ggridges')
library(ggridges) ##
## Attaching package: 'ggridges'## The following object is masked from 'package:ggplot2':
##
## scale_discrete_manualridge_plot = df %>% ggplot(aes(x=weeks_alive, y = dose)) +
geom_density_ridges(jittered_points = T)
ridge_plot## Picking joint bandwidth of 10.5
It’s a bit tough to tell, but the overall assumptions of normality and equal variance seem reasonable.
The global mean is
mean(df$weeks_alive)## [1] 75.55263PC3. T-test between None and Any, and between None and High.
t.test(df[df$dose == 'None', 'weeks_alive'], # Samples with no dose
df[df$dose != 'None','weeks_alive'] # Samples with any dose
)##
## Welch Two Sample t-test
##
## data: df[df$dose == "None", "weeks_alive"] and df[df$dose != "None", "weeks_alive"]
## t = 4.2065, df = 33.732, p-value = 0.0001806
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 11.49879 33.00626
## sample estimates:
## mean of x mean of y
## 91.36364 69.11111# Or, using formula notation
t.test(weeks_alive ~ dose == 'None', data = df)##
## Welch Two Sample t-test
##
## data: weeks_alive by dose == "None"
## t = -4.2065, df = 33.732, p-value = 0.0001806
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -33.00626 -11.49879
## sample estimates:
## mean in group FALSE mean in group TRUE
## 69.11111 91.36364# T-test between None and High
t.test(df[df$dose == 'None', 'weeks_alive'], # Samples with no dose
df[df$dose == 'High','weeks_alive'] # Samples with high dose
)##
## Welch Two Sample t-test
##
## data: df[df$dose == "None", "weeks_alive"] and df[df$dose == "High", "weeks_alive"]
## t = 2.4958, df = 8.5799, p-value = 0.0353
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 2.266399 49.960874
## sample estimates:
## mean of x mean of y
## 91.36364 65.25000# Formula notation is a bit tricker. I would probably create a temprorary dataframe
tmp = df %>% filter(dose %in% c('None', 'High'))
t.test(weeks_alive ~ dose, data = tmp)##
## Welch Two Sample t-test
##
## data: weeks_alive by dose
## t = -2.4958, df = 8.5799, p-value = 0.0353
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -49.960874 -2.266399
## sample estimates:
## mean in group High mean in group None
## 65.25000 91.36364The t-test supports the idea that receiving a dose of RD40 reduces lifespan
PC4. Anova
summary(aov(weeks_alive ~ dose, data = df))## Df Sum Sq Mean Sq F value Pr(>F)
## dose 3 4052 1350.7 3.55 0.0245 *
## Residuals 34 12937 380.5
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1This provides evidence that the group means are different.
Statistical Questions
Q1. a) It is a sample statistic, because it comes from a sample. b) Confidence intervals for proportions are equal to
\[p \pm z * \sqrt{ \frac{p*(1-p)}{n}}\]
For a 95% confidence interval, \(z = 1.96\), so we can calculate it like this:
lower = .48 - 1.96 * sqrt(.48 * .52 / 1259)
upper = .48 + 1.96 * sqrt(.48 * .52 / 1259)
ci = c(lower, upper)
print(ci)## [1] 0.4524028 0.5075972This means that we are 95% confident that the true proportion of Americans who support legalizing marijuana is between ~45% and ~51%.
- We have a large enough sample, which is collected randomly, to assume that the distribution is normal.
- The statement isn’t justified, since our confidence interval include 50%
6.20 We can use the point estimate of the poll to estimate how large a sample we would need to have a confidence interval of a given width.
Basically, we want each half of the confidence interval to be 1%, i.e., \(1.96 * \sqrt{\frac{.48 * .52}{n}} = .01\)
We can solve for \(n\):
\[\sqrt{\frac{.48 * .52}{n}} = .01/1.96\] \[\frac{.48 * .52}{n} = (.01/1.96)^2\] \[n = (.48 * .52)/(.01/1.96)^2\]
(.48 * .52)/(.01/1.96)^2## [1] 9588.634So, we need a sample of approximately 9,589
6.38
The question is whether there has been a change in the proportion over time. While the tools we have learned could allow you to answer that question, they assume that responses are independent. In this case, they are obviously not independent as they come from the same students.
6.50
- We can test this with a \(\chi^2\) test.
chisq.test(x = c(83,121,193,103), p = c(.18,.22,.37,.23))##
## Chi-squared test for given probabilities
##
## data: c(83, 121, 193, 103)
## X-squared = 3.2426, df = 3, p-value = 0.3557# Or manually
# Calculate expected values
500 * c(.18,.22,.37,.23)## [1] 90 110 185 115# Use the formula for chi-squared
chisq = (83-90)^2/90 + (121-110)^2/110 + (193-185)^2/185 + (103-115)^2/115The p-value for this is large, meaning that we don’t have evidence that the sample differs from the census distribution.
- Opinion is the response and location is the explanatory variable, since it’s unlikely that people move to a region based on their opinion.
- One hypothesis is that opinions differ by region. The null hypothesis is that opinion is independent of region, while the alternative hypothesis is that there is a relationship.
- We can again use a \(\chi^2\) test.
x <- matrix(c(29,54,44,77,62,131,36,67), nrow = 4, # this makes a matrix with 4 rows
byrow=T) # And this says that we've entered it row by row
chisq.test(x)##
## Pearson's Chi-squared test
##
## data: x
## X-squared = 0.66724, df = 3, p-value = 0.8809