In roulette, a bet on a single number has a 1/38 probability of success and pays 35-to-1. That is, if you bet 1 dollar, your net winnings are -1 with probability 37/38 and +35 with probability 1/38. Consider betting on a single number on each of \(n\) spins of a roulette wheel. Let \(\bar{X}_n\) be your average net winnings per bet.
For each of the values \(n = 10\), \(n = 100\), \(n = 1000\):
Let \(X\) be the winnings on a single bet. The population mean is \[ \mu = \text{E}(X) = (35)(1/38) + (-1)(37/38) = -2/38 \approx -0.05 \] The population variance is \[ \sigma^2 = \text{Var}(X) = \text{E}(X^2) - (\text{E}(X))^2 = (35)^2(1/38) + (-1)^2(37/38) - (-2/38)^2 = 33.2 \] The population SD is \(\sigma = 5.76\) dollars.
n = 10
N_samples = 10000
sample_means = vector(length = N_samples)
for (i in 1:N_samples) {
x = sample(c(-1, 35), size = n, prob = c(37, 1) / 38, replace = TRUE)
sample_means[i] = mean(x)
}
hist(sample_means)
mean(sample_means)[1] -0.04348sd(sample_means)[1] 1.834692sum(sample_means > 0) / N_samples[1] 0.2357n = 100
N_samples = 10000
sample_means = vector(length = N_samples)
for (i in 1:N_samples) {
x = sample(c(-1, 35), size = n, prob = c(37, 1) / 38, replace = TRUE)
sample_means[i] = mean(x)
}
hist(sample_means)
mean(sample_means)[1] -0.0559sd(sample_means)[1] 0.5723487sum(sample_means > 0) / N_samples[1] 0.4938n = 1000
N_samples = 10000
sample_means = vector(length = N_samples)
for (i in 1:N_samples) {
x = sample(c(-1, 35), size = n, prob = c(37, 1) / 38, replace = TRUE)
sample_means[i] = mean(x)
}
hist(sample_means)
mean(sample_means)[1] -0.0518968sd(sample_means)[1] 0.1815168sum(sample_means > 0) / N_samples[1] 0.39631 - pnorm(0, -2 / 38, 5.76 / sqrt(1000))[1] 0.3863095qnorm(0.99)[1] 2.326348pnorm(0, - 2 / 38, 5.76 / sqrt(64878))[1] 0.9900282n = 64878
N_samples = 10000
sample_means = vector(length = N_samples)
for (i in 1:N_samples) {
x = sample(c(-1, 35), size = n, prob = c(37, 1) / 38, replace = TRUE)
sample_means[i] = mean(x)
}
hist(sample_means)
mean(sample_means)[1] -0.05282828sd(sample_means)[1] 0.02280221sum(sample_means < 0) / N_samples[1] 0.9892The standard measurement of the alcohol content of drinks is alcohol by volume (ABV), which is given as the volume of ethanol as a percent of the total volume of the drink. A report states that on average the ABV for beer is 4.5%. Is this true? In a sample of 67 brands of beer, the mean ABV is 4.61 percent and the standard deviation of ABV is 0.754 percent. Note: the data set and some R summaries are provided in Canvas so you can check your answers. But you should answer these questions by hand, using only the information provided here.
data = read.csv("beer.csv")head(data) Brand Brewery ABV Calories Carbohydrates
1 Anheuser Busch Natural Light Anheuser Busch 4.2 95 3.2
2 Aspen Edge Adolph Coors 4.1 94 2.6
3 Blatz Beer Pabst 4.8 153 12.5
4 Blue Moon Adolph Coors 5.4 171 13.7
5 Bud Dry Anheuser Busch 5.0 130 7.8
6 Bud Ice Anheuser Busch 5.5 148 8.9summary(data$ABV) Min. 1st Qu. Median Mean 3rd Qu. Max.
0.40 4.20 4.60 4.61 5.00 6.50 sd(data$ABV)[1] 0.7540063library(tidyverse)── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
✔ dplyr 1.1.2 ✔ readr 2.1.4
✔ forcats 1.0.0 ✔ stringr 1.5.0
✔ lubridate 1.9.2 ✔ tibble 3.2.1
✔ purrr 1.0.1 ✔ tidyr 1.3.0
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
✖ dplyr::group_rows() masks kableExtra::group_rows()
✖ dplyr::lag() masks stats::lag()
ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorsdata |>
summarize(n = n(),
mean = mean(ABV),
sd = sd(ABV),
min = fivenum(ABV)[1],
Q1 = fivenum(ABV)[2],
median = fivenum(ABV)[3],
Q3 = fivenum(ABV)[4],
max = fivenum(ABV)[5]) n mean sd min Q1 median Q3 max
1 67 4.610448 0.7540063 0.4 4.2 4.6 5 6.5hist(data$ABV, freq = FALSE)
lines(density(data$ABV))
boxplot(data$ABV, horizontal = TRUE)
ggplot(data,
aes(x = ABV)) +
geom_histogram(aes(y = ..density..)) +
geom_density()Warning: The dot-dot notation (`..density..`) was deprecated in ggplot2 3.4.0.
ℹ Please use `after_stat(density)` instead.`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
t.test(data$ABV, mu = 4.5)
One Sample t-test
data: data$ABV
t = 1.199, df = 66, p-value = 0.2348
alternative hypothesis: true mean is not equal to 4.5
95 percent confidence interval:
4.426531 4.794364
sample estimates:
mean of x
4.610448 data = data |>
filter(!(Brand == "O'Doul's"))
summary(data$ABV) Min. 1st Qu. Median Mean 3rd Qu. Max.
3.800 4.200 4.600 4.674 5.000 6.500 sd(data$ABV)[1] 0.5480906data |>
summarize(n = n(),
mean = mean(ABV),
sd = sd(ABV),
min = fivenum(ABV)[1],
Q1 = fivenum(ABV)[2],
median = fivenum(ABV)[3],
Q3 = fivenum(ABV)[4],
max = fivenum(ABV)[5]) n mean sd min Q1 median Q3 max
1 66 4.674242 0.5480906 3.8 4.2 4.6 5 6.5hist(data$ABV, freq = FALSE)
lines(density(data$ABV))
boxplot(data$ABV, horizontal = TRUE)
ggplot(data,
aes(x = ABV)) +
geom_histogram(aes(y = ..density..)) +
geom_density()`stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
t.test(data$ABV, mu = 4.5)
One Sample t-test
data: data$ABV
t = 2.5827, df = 65, p-value = 0.01206
alternative hypothesis: true mean is not equal to 4.5
95 percent confidence interval:
4.539505 4.808980
sample estimates:
mean of x
4.674242 (I had much longer instructions written, but I think they weren’t helpful. If you have questions about how the applet is working, don’t hesitate to ask.)
You are going to use a simulation applet that randomly generates confidence interals for a population mean to help you understand some ideas. The applet calculates a confidence interval for each set of randomly generated data.
Experiment with different simulations; be sure to change
Then write a paragraph or two or some bullet points of your main observations. Based on this applet, what can you say about how confidence intervals work? How does changing the inputs affect the confidence intervals and the Results?