Note that in this question, the empirical distribution of the sample mean is made up from the means of samples drawn from the population (not from resamples from a single sample).
Assume that you have a certain population of interest whose histogram is below.
from datascience import *
import numpy as np
%matplotlib inline
pops = Table().with_column(
'Population', [1] * 40 + [2] * 25 + [3] * 15 + [4] * 10 + [5] * 5 + [6] * 2 + [7] * 1
)
pops.hist("Population", bins = np.arange(1.5, step = 1, stop = 8.5))
Cyrus takes many large random samples with replacement from the population with the goal of generating an empirical distribution of the sample mean. What shape do you expect this distribution to have? Which value will it be centered around?
The distribution will look like a bell curve (normally distributed) centered around the population mean, by the Central Limit Theorem. We are satisfying the conditions of the Central Limit Theorem as we are drawing large random samples with replacement, and that it is a distribution of the sum or mean.
sample_means = make_array()
for i in np.arange(1000):
sample_means = np.append(sample_means, np.mean(pops.sample().column(0)))
Table().with_columns("Sample Means", sample_means).hist("Sample Means")
Why are we able to use the CLT to reason about the empirical distribution of the sample mean’s shape if the population data is skewed?
To use the CLT all we need is a large random sample with replacement from the population and a statistic that is a sum or mean. The CLT applies regardless of the distribution of the population.
Suppose that Cyrus creates two empirical distributions of sample means, with different sample sizes. Which distribution corresponds to a larger sample size? Why?
The distribution to the right corresponds to a larger sample size compared to the one to the left. We know based on the spread of the two distributions. The larger the sample size you take, the less variable the distribution of the sample mean becomes. You can see that increasing the sample size is increasing the denominator in calculating the SD of sample means, which decreases the standard deviation.
sample_means = make_array()
for i in np.arange(10):
sample_means = np.append(sample_means, np.mean(pops.sample().column(0)))
Table().with_columns("Sample Means", sample_means).hist("Sample Means")
sample_means = make_array()
for i in np.arange(1000):
sample_means = np.append(sample_means, np.mean(pops.sample().column(0)))
Table().with_columns("Sample Means", sample_means).hist("Sample Means")
Based solely on the information in the histogram, what is an estimate for the standard deviation of the sample mean on the left? How did you determine this?
Suppose you were told that the distribution on the right was generated based on a sample size of 100 and has a standard deviation of 0.2. How big of a sample size would you need if you wanted the standard deviation of the distribution of sample means to be 10 times smaller?
First, find the Population SD:
\(0.2 = \frac{Population\;SD}{\sqrt{100}}\)
\(Population\;SD = 0.2 * \sqrt{100} = 2\)
Then, calculate the new sample size needed:
\(\frac{0.2}{10} = \frac{2}{\sqrt{new\;sample\;size}}\)
\(0.2 * \sqrt{new\;sample\;size} = 2 * 10\)
\(\sqrt{new\;sample\;size} = 100\)
\(new\;sample\;size = 10000\)
Confidence Interval Visualizer!
UC Berkeley students love studying in Main Stacks! You are working with Aayan on constructing a confidence interval for the mean number of hours students spend in Main Stacks each year. To do this, you take a random sample of 400 UC Berkeley students and record how many hours each student spent studying in Main Stacks over the past year. Then, you compute the mean number of hours for your sample; it is 170 hours. You also calculate the standard deviation of your sample to be 10 hours.
Aayan claims that the distribution of all possible sample means is normal with an SD of 0.5 hours. Use this information to construct an approximate 68% confidence interval for the mean hours spent studying in Main Stacks for all UC Berkeley students.
We know that the distribution is approximately normal and hence we know that roughly 68% of the area under the normal curve (or 68% of the data) is contained within 1 SD of the mean. Therefore, our confidence interval range will be [170 - 1*0.5, 170 + 1*0.5] = [169.5 hours, 170.5 hours].
Suppose Richard and Bing took a different random sample of 400 UC Berkeley students and found a sample mean of 172 hours, with a sample standard deviation of 10 hours. Would it be surprising to observe this sample mean if the true average number of hours all UC Berkeley students spend in Main Stacks is 170?
If the true population mean is 170 hours and the standard deviation of the sample means is 0.5 (as we saw before), then we can compute how many standard deviations away 172 is from the mean using a z-score: \(z = \frac{172 - 170}{0.5} = 4\). A z-score of 4 is very far in the tails of the normal distribution — much further than what we would expect by random chance. Therefore, yes, it would be surprising to observe a sample mean of 172 hours if the true mean were 170. This might suggest the true population mean is not 170 hours.
true_mean = 170
true_sd = 10
n = 400
repetitions = 5000
sample_means = []
for _ in range(repetitions):
sample = np.random.normal(true_mean, true_sd, n)
sample_means.append(np.mean(sample))
sim_table = Table().with_column("Sample Means", sample_means)
prop_extreme = np.count_nonzero((sim_table.column("Sample Means") >= 172) |
(sim_table.column("Sample Means") <= 168)) / repetitions
print("Simulated probability of observing 172 or more extreme:", prop_extreme)
sim_table.hist("Sample Means")Simulated probability of observing 172 or more extreme: 0.0002
If Aayan had not told you what the SD of the distribution of sample means was, could you estimate it from the data in the sample? If yes, how?
We know that the sample size is 400 and that the standard deviation of our sample was 10 hours. Since the SD of sample means is equal to \(Population \; SD / \sqrt{Sample \; size}\), we can substitute the sample SD (i.e., 10 hours) in place of the \(Population \; SD\) in this formula (assuming the sample is representative of the population) and calculate the value!\
SD of all possible sample means = 10 hours / \(\sqrt{400}\) = 10 hours / 20 = 0.5 hours.