Below, we have imported the Python libraries needed for this module. Run the code in this cell before running any other code cells, and be careful not to change any of the code. You can run the cell in any of these ways:

Ctrl + Enter: Run the cell and keep the cursor in the same cell.
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Click the Play button: Click the Run (play) button to the left of the cell to execute it.

# Necessary imports for this module
from utils import *

Sampling Distributions & Central Limit Theorem¶

Estimated Time: 30 Minutes
Professor: Alice Martinez
Developers: James Geronimo, Mark Barranda, Rinrada Maneenop

The sampling distribution of a statistic is the distribution of all values of a statistic when all possible samples of the same size n are taken from the same population.

The basic idea is this: If you were to take multiple samples, what values from those samples will give you the best estimates of the population values?

Example 1: Fair Die¶

n_samples = 10000
sample_size = 5
population = np.arange(1, 7)

Means¶

Sampling Procedure:
Roll a fair six-sided die 5 times and record the sample mean, $\bar{x}$ .
Repeat this process 10,000 times to build a distribution of sample means.

Population Mean ( $\mu$ ): 3.5

🟩 The dashed green line represents the mean of all sample means.

fair_die_means(n_samples, sample_size, population)

In this cell block, we show a few indivudal sample means generate from five samples.

for i in range(5):
    rolls = np.random.choice(population, size=sample_size)
    print(f"Sample {i+1} rolls: {rolls}, Sample Mean: {np.mean(rolls):.2f}")

All outcomes are equally likely so the __________ mean is 3.5;

The __________ of the sample means in 10,000 trials is 3.51. If continued indefinitely, the sample mean will be 3.5.

Note:

The mean of the sample means __________ the value of the pouplation mean
The sample means have a __________ distribution

Variances¶

Sampling Procedure:
Roll a fair six-sided die 5 times and record the sample variance, $s^2$ .
Repeat this process 10,000 times to build a distribution of sample variances.

Population Variance ( $\sigma^2$ ): 2.9

🟩 The dashed green line represents the mean of all sample variances.

fair_die_variances(n_samples, sample_size, population)

In this cell block, we show a few individual sample variances generate from five samples.

for i in range(5):
    rolls = np.random.choice(population, size=sample_size)
    print(f"Sample {i+1} rolls: {rolls}, Sample Mean: {np.var(rolls, ddof=1):.2f}")

All outcomes are equally likely so the __________ variance is 2.9.

The __________ of sample variance in the 10,000 trials is 2.92. If continued indefinitely, the sample variance will be 2.9.

Note:

The mean of the sample variances __________ the value of the pouplation variance
The sample variances have a __________ distribution

Proportions¶

Sampling Procedure:
Roll a fair six-sided die 5 times and record the proportion of odd numbers.
Repeat this process 10,000 times to build a distribution of odd number proportions.

Population Proportion of Odd Numbers ( $p$ ): 0.5
$\hat{p}$ represents the sample proportion of odd numbers

🟩 The dashed green line represents the mean of all odd number proportions.

fair_die_proportions(n_samples, sample_size, population)

In this cell block, we show a few indivudal sample proportions generate from five samples.

for i in range(5):
    rolls = np.random.choice(population, size=sample_size)
    print(f"Sample {i+1} rolls: {rolls}, Sample Mean: {np.sum(rolls % 2 == 1) / sample_size:.2f}")

All outcomes are equally likely so the __________ of odd numbers is 0.5.

The mean of the __________ of the 10,000 trials is 0.503. If continued indefinitely, the mean of sample proportions will be 0.5.

Note:

The mean of the sample proportions __________ the value of the pouplation proportions
The sample proportions have a __________ distribution

Estimators¶

Using the above information, we discover biased and unbiased estimators of population parameters.

Unbiased Estimators:¶

The __________ $\bar{x}$ , __________ $s^2$ , and __________ $\hat{p}$ of the samples are unbiased estimators of the corresponding pouplation parameters $mu$ , $\sigma^2$ , and $p$ because they target the value that the population would have.

Biased Estimators:¶

The __________ , __________ , and __________ $s$ do NOT target their corresponding pouplation parameters, so they are generally NOT good estimators. However: often bias in the $s$ is small enough that it is used to estimate $\sigma$ .

Example 2: Assassinated Presidents¶

There are four U.S. presidents who were assassinated in office. Their ages (in years) were Lincoln 56, Garfield 49, McKinley 59, and Kennedy 46.

a) Assuming that 2 of the ages are randomly selected with replacement from [56, 49, 59, 46], list the 16 different possible samples by replacing the ellipses with appropriate values. We’ve filled out a few as a hint:

[56, 56] [49, 56] [59, ...] [46, ...]
[56, 49] [49, 49] [..., ...] [..., ...]
[56, 59] [49, ...] [..., ...] [..., ...]
[56, 46] [49, ...] [..., ...] [..., ...]

b) Find the sample mean and range of possible samples by completing the functions calculate_mean, calculate_range, and calculate_probability. Then, run the cell to create tables that represent the probability distribution of each statistic.

def calculate_mean(number1, number2):
    """Fill in the ... to calculate the mean of two numbers."""
    return ...
    
def calculate_range(number1, number2):
    """Fill in the ... to calculate the range of two numbers."""
    return ...

def calculate_probability(frequency, total_samples):
    """Fill in the ... to calculate the probability given frequency and total_samples."""
    return ...

mean_range_tables(calculate_mean, calculate_range, calculate_probability)

c) Calculate the population_mean and population_range. Then, for each statistic, compare the mean of sample statistics to the population statistic. Which sampling distributions target the population parameter?

population_mean = ...
population_range = ...

print(f"Population Mean: {population_mean}")
print(f"Population Range: {population_range}")

Mean of sample means __________ the population mean which makes this a __________ estimator

Mean of sample ranges __________ the population range which makes this a __________ estimator

Standard Deviation and Variance¶

Variance measures how spread out the data are around the mean.
- For a population with values $x_1, x_2, \dots, x_N$ and mean $\mu$ :
  $\sigma^2 = \frac{1}{N} \sum_{i=1}^N (x_i - \mu)^2$
  (1)
- For a sample with values $x_1, x_2, \dots, x_n$ and sample mean $\bar{x}$ , the sample variance is:
  $s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2$
  (2)
Standard deviation is the square root of the variance.
- For a population: $\sigma = \sqrt{\sigma^2}$
- For a sample: $s = \sqrt{s^2}$

It is in the same units as the original data and tells you a typical distance of values from the mean.

d) Find the sample standard deviation and variance of possible samples by completing the functions standard_deviation and variance. Then, run the cell to create tables that represent the probability distribution of each statistic.

def standard_deviation(number1, number2):
    """Fill in the ... to calculate the standard deviation of two numbers."""
    return ...

def variance(number1, number2):
    """Fill in the ... to calculate the variance of two numbers."""
    return ...

Complete each of the following expressions:

Mean of sample medians __________ the population median which makes this a __________ estimator

Mean of the sample proportions __________

Mean of the __________ variances __________

__________ of the __________ standard deviations __________

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