Bulletin

Today

By the end of today you will…

Getting started

Download this application exercise by pasting the code below into your console

download.file("https://sta101.github.io/static/appex/ae10.Rmd",
destfile = "ae10.rmd")

Load packages

library(tidyverse)
library(tidymodels)
manhattan = read_csv("https://sta101.github.io/static/appex/data/manhattan.csv")

Notes

How do we know when to expect a normal distribution to show up?

Last time we saw an example where the distribution of sample means was normal while the distribution of sample medians was not.

Example:

set.seed(1)
boot_dist = manhattan %>%
  specify(response = rent) %>% 
  generate(reps = 1000, type = "bootstrap")

boot_dist %>%
  calculate(stat = "mean") %>%
  visualize() +
  labs(x = "Sample mean", 
       title = "Simulated distribution of the sample mean")

boot_dist %>%
  calculate(stat = "median") %>%
  visualize() + 
  labs(x = "Sample median",
       title = "Simulated distribution of the sample median")

Are there times when the sample mean will not look normal?

Returning to our binomial example from ae8: you ask numPeople if they prefer dogs or cats. You repeat this 1000 times and plot the distribution of the proportion of who prefer dogs.

Exercise 1

How does the shape of the distribution change as you increase the number of people per sample?

set.seed(714)
numPeople = 1
numDog = rbinom(n = 1000, size = numPeople, prob = 0.85)
sample = data.frame(numDog) # new data frame called sample
sample %>%
mutate(propDog = numDog / numPeople) %>%
ggplot(aes(x = propDog)) +
geom_histogram(binwidth = .01)

What is the central limit theorem?

The central limit theorem is a statement about the distribution of the sample mean, \(\bar{x}\)

The central limit theorem guarantees that, when certain criteria are satisfied, the sample mean (\(\bar{x}\)) is normally distributed.

Specifically,

If

  1. Observations in the sample are independent. Two rules of thumb to check this:

    • completely random sampling
    • if sampling without replacement, sample should be less than 10% of the population size

and

  1. The sample is large enough. The required size varies in different contexts, but some good rules of thumb are:

    • if the population itself is normal, sample size does not matter.
    • if numerical require, >30 observations
    • if binary outcome, at least 10 successes and 10 failures.

then

\(\bar{x} \sim N(\mu, \sigma / \sqrt{n})\)

i.e. \(\bar{x}\) is normally distributed (unimodal and symmetric with bell shape) with mean \(\mu\) and standard deviation \(\sigma / \sqrt{n}\). The standard deviation of the sampling distribution is called the standard error.

Note the standard deviation depends on the number of samples, \(n\).

Practice using CLT & Normal distribution

Suppose the bone density for 65-year-old women is normally distributed with mean \(809 mg/cm^3\) and standard deviation of \(140 mg/cm^3\).

Let \(x\) be the bone density of 65-year-old women. We can write this distribution of \(x\) in mathematical notation as

\[x \sim N(809, 140)\]

Visualize the population distribution

ggplot(data = data.frame(x = c(809 - 140*3, 809 + 140*3)), aes(x = x)) +
  stat_function(fun = dnorm, args = list(mean = 809, sd = 140),
                color = "black") +
  stat_function(fun = dnorm, args = list(mean = 809, sd = 140/sqrt(10)),
                color = "red",lty = 2) + theme_bw() +
  labs(title = "Black solid line = population dist., Red dotted line = sampling dist.")

Exercise 2

Before typing any code, based on what you know about the normal distribution, what do you expect the median bone density to be?

What bone densities correspond to \(Q_1\) (25th percentile), \(Q_2\) (50th percentile), and \(Q_3\) (the 75th percentile) of this distribution? Use the qnorm() function to calculate these values.

Exercise 3

The densities of three woods are below:

  • Plywood: 540 mg/cubic centimeter

  • Pine: 600 mg/cubic centimeter

  • Mahogany: 710 mg/cubic centimeter

  • What is the probability that a randomly selected 65-year-old woman has bones less dense than Pine?

  • Would you be surprised if a randomly selected 65-year-old woman had bone density less than Mahogany? What if she had bone density less than Plywood? Use the respective probabilities to support your response.

Exercise 4

Suppose you want to analyze the mean bone density for a group of 10 randomly selected 65-year-old women.

  • Are the conditions for the Central Limit Theorem met?

    • Independence?
    • Sample size/distribution?

  • What is the shape, center, and spread of the distribution of \(\bar{x}\), the mean bone density for a group of 10 randomly selected 65-year-old women?

  • Write the distribution of \(\bar{x}\) using mathematical notation.

Exercise 5

  • What is the probability that the mean bone density for the group of 10 randomly-selected 65-year-old women is less dense than Pine?

  • Would you be surprised if a group of 10 randomly-selected 65-year old women had a mean bone density less than Mahogany? What the group had a mean bone density less than Plywood? Use the respective probabilities to support your response.

Exercise 6

Explain how your answers differ in Exercises 2 and 4.

On your own

Suppose the distribution of the number of minutes users engage with apps on an iPad has a mean of 8.2 minutes and standard deviation of 1 minute. Let \(x\) be the number of minutes users engage with apps on an iPad, \(\mu\) be the population mean and \(\sigma\) the population standard deviation. Then,

\[x \sim N(8.2, 1)\]

Suppose you take a sample of 60 randomly selected app users and calculate the mean number of minutes they engage with apps on an iPad, \(\bar{x}\). The conditions (independence & sample size/distribution) to apply the Central Limit Theorem are met. Then by the Central Limit Theorem

\[\bar{x} \sim N(8.2, 1/\sqrt{60})\]

    #add code

Data: Pokemon

We will be using the pokemon data set, which contains information about 42 randomly selected Pokemon (from all generations). You may load in the data set with the following code:

pokemon = read_csv("https://sta101.github.io/static/appex/data/pokemon.csv")

In this analysis, we will use CLT-based inference to draw conclusions about the mean height among all Pokemon species.

Exercise 7

Let’s start by looking at the distribution of height_m, the typical height in meters for a Pokemon species, using a visualization and summary statistics.

ggplot(data = pokemon, aes(x = height_m)) +
  geom_histogram(binwidth = 0.25, fill = "steelblue", color = "black") + 
  labs(x = "Height (in meters)", 
       y = "Distributon of Pokemon heights")

pokemon %>%
  summarise(mean_height = mean(height_m), 
            sd_height = sd(height_m), 
            n_pokemon = n())
## # A tibble: 1 × 3
##   mean_height sd_height n_pokemon
##         <dbl>     <dbl>     <int>
## 1       0.929     0.497        42

In the previous lecture (and in the review questions), we were given the mean, \(\mu\), and standard deviation, \(\sigma\), of the population. That is unrealistic in practice (if we knew \(\mu\) and \(\sigma\), we wouldn’t need to do statistical inference!).

Today we will start on using the Central Limit Theorem to draw conclusions about the \(\mu\), the mean height in the population of Pokemon.

  • What is the point estimate for \(\mu\), i.e., the “best guess” for the mean height of all Pokemon?

  • What is the point estimate for \(\sigma\), i.e., the “best guess” for the standard deviation of the distribution of Pokemon heights?

Exercise 8

Before moving forward, let’s check the conditions required to apply the Central Limit Theorem. Are the following conditions met:

  • Independence?
  • Sample size/distribution?

Idea: confidence intervals using CLT