# Necessary imports for this module
from utils import *

Sampling Distributions & Central Limit Theorem¶

Estimated Time: 30 Minutes
Developers: James Geronimo, Mark Barranda

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:

1. The mean of the sample means __________ the value of the pouplation mean

2. 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:

1. The mean of the sample variances __________ the value of the pouplation variance

2. 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:

1. The mean of the sample proportions __________ the value of the pouplation proportions

2. 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

d) 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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