| Name | Wesley Zheng |
| Pronouns | He/him/his |
| wzheng0302@berkeley.edu | |
| Discussion | Wednesdays, 12–2 PM @ Etcheverry 3105 |
| Office Hours | Tuesdays/Thursdays, 2–3 PM @ Warren Hall 101 |
Contact me by email at ease — I typically respond within a day or so!
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?
By the Central Limit Theorem, the distribution will be normal (bell curve), centered around the population 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?
We are able to use the CLT, as we are drawing large random samples, with replacement, and the statistic is sum or mean. If these conditions are applied, the CLT can be used regardless of the population’s distribution.
Note, given a large enough population, and small relative sample size, we do not have to sample with replacement, as the behavior approaches that of drawing with replacement. This is out of scope for Data 8.
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?
Approximately 0.3.
This is because the point of inflection on a normal distribution reflects 1 standard deviation away from the mean. Looking at the histogram, we can see that the inflection points occur at around 1.7 and 2.3, with the mean of the histogram being 2.0.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:
\(0.2 = \frac{population\;SD}{\sqrt{100}}\)
\(\Rightarrow population\;SD = 0.2 * \sqrt{100} = 2\)
Then, calculate the new sample size needed:
\(\frac{0.2}{10} = \frac{2}{\sqrt{new\;sample\;size}}\)
\(\Rightarrow 0.2 * \sqrt{new\;sample\;size} = 2 * 10\)
\(\Rightarrow \sqrt{new\;sample\;size} = 100\)
\(new\;sample\;size = 10000\)
Confidence Interval Visualizer!
UC Berkeley students love studying in Main Stacks! You are working with Lena 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.
Lena 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].
If Lena 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? If no, why not?
We know that the sample size is 400 and that the standard deviation of our sample was 10 hours. Recall the equation for the SD of sample means is
\(standard\;deviation\;of\;sample\;means = \frac{population\;standard\;deviation}{\sqrt{Sample \; size}}\)
We can substitute the sample SD (10 hours) in place of the population SD in this formula (assuming the sample is representative of the population) and calculate the value!
\(standard\;deviation\;of\;sample\;means = \frac{10\;hours}{\sqrt{400}} = \frac{10\;hours}{20} = 0.5\;hours\)
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: 0.0
You are a super fan of the girl group TLC and are interested in estimating the average amount of plays their songs have online. You generate an 80% confidence interval for this parameter to be [700000, 1200000] based on a random sample of 50 songs using the Central Limit Theorem. Are each of the following statements true or false?
Note: Generally, \(n \geq 30\) is considered the minimum sample size for CLT to take effect!
plays = np.random.normal(loc=950000, scale=150000, size=50)
plays = np.clip(plays, 500000, 1500000)
tlc_songs = Table().with_columns(
"Plays", plays
)The value of our population parameter changes depending on our sampling process.
The empirical distribution of any statistic we choose will be roughly normal based on the Central Limit Theorem, but it requires our population to have a normal distribution to begin with.
If we generate a 95% confidence interval using the same sample, the interval will be narrower than the original confidence interval because we are more certain of our results.
Using a 95% confidence level would result in a wider interval than an 80% confidence level. In fact, the 95% confidence interval will envelop the original 80% confidence interval. The “increased confidence” comes from having a wider interval.
arr = make_array()
for i in np.arange(1000):
arr = np.append(arr, np.mean(tlc_songs.sample().column(0)))
print(f'95% Confidence Interval for average plays: [{percentile(2.5, arr)}, {percentile(97.5, arr)}]')
print(f'80% Confidence Interval for average plays: [{percentile(10, arr)}, {percentile(90, arr)}]')95% Confidence Interval for average plays: [927380.9355463004, 1002753.505085332]
80% Confidence Interval for average plays: [942785.6492003608, 990490.8098108827]Using the same process to generate another 80% confidence interval, there is an 80% chance that the next confidence interval you generate will contain the true average number of plays for TLC songs.
There is an 80% chance that a confidence interval we generated contains the true average number of plays for TLC songs.
Once we have calculated a confidence interval, it is fixed - it either contains the parameter or it does not. There is no chance involved.
80% of TLC’s songs have between 700,000 and 1,200,000 plays.
Our confidence interval is estimating the average number of plays but makes absolutely NO claim about what percentage of songs have between 700,000 and 1,200,000 plays. A \(n\)% confidence interval contains \(n\)% of the statistics you simulated, but it does not suggest that \(n\)% of the population is in that interval!
tlc_songs.where(0, are.between(70000, 1200000)).num_rows / tlc_songs.num_rows0.96The original sample mean you obtained was approximately 950,000 plays.