Last modified: September 21, 2024

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Central Limit Theorem (CLT)

The Central Limit Theorem (CLT) is a fundamental concept in statistics, explaining why the distribution of sample means approximates a normal distribution, often known as the bell curve, as the sample size becomes larger, irrespective of the population's original distribution.

Mathematical Background

• The CLT states that if you have a population with a mean $\mu$ and a standard deviation $\sigma$, and you take sufficiently large random samples from this population (with replacement), then the distribution of the sample means will approximate a normal distribution.
• The mean of the sample means ($\bar{X}$) will be approximately equal to the population mean ($\mu$).
• The standard deviation of the sample means, also known as the standard error, will be $\frac{\sigma}{\sqrt{n}}$, where $n$ is the sample size.

Formal Description

Let $X_1, X_2, \ldots, X_n$ be a sequence of independent and identically distributed random variables, each with a mean $\mu$ and a variance $\sigma^2$.

As $n$ (the sample size) tends to infinity, the distribution of the standardized sum

$$\frac{X_1 + X_2 + \ldots + X_n - n\mu}{\sigma\sqrt{n}}$$

converges in distribution to a standard normal distribution. Mathematically, this is expressed as:

$$P\left(\frac{X_1 + X_2 + \ldots + X_n - n\mu}{\sigma\sqrt{n}} \leq a\right) \to \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{a} e^{-x^2/2} \, dx$$

Key Points:

1. "Large" is typically considered to be a sample size of 30 or more, though this can vary based on the distribution's initial shape.
2. Samples should be independent of each other.
3. It's the distribution of the means (and other statistics like sum and percentage) of these samples that becomes normal, not the distribution of the individual data points (it can still be skewed).

Implications and Applications

The Central Limit Theorem (CLT) enables us to approximate probabilities and percentages for large samples using the normal distribution. This has wide-ranging implications, particularly in statistical analysis and data inference. For example:

• In an online game with a 20% chance of winning a small prize, the distribution of wins in a large number of attempts (n) follows a binomial distribution. However, when the sample size (n) is large, the binomial distribution can be approximated by a normal distribution, simplifying the analysis.

Key points regarding the CLT:

• The CLT applies to almost any population distribution, regardless of whether the distribution is normal, as long as the sample size is sufficiently large.
• The CLT allows statisticians to make inferences about population parameters using sample statistics. This is especially useful when analyzing large datasets where accessing the entire population is impractical or impossible.
• Many statistical procedures, experiments, and tests are grounded in the CLT, making it a fundamental concept in statistical analysis. Its ability to approximate complex distributions with a normal one is crucial for practical applications in various fields.

Limitations

• The theorem applies only when the sample size is large enough. "Large enough" varies depending on the underlying distribution, but a common rule of thumb is a sample size greater than 30.
• The sampled values must be independent of each other. This assumption can be violated in many real-world scenarios.

Visualization

A histogram of the sample means will tend to form a bell-shaped curve as the number of samples increases, reflecting the normal distribution predicted by the CLT.

Data Generation

• Population data generated from a non-normal (exponential) distribution.
• Distribution used is an exponential distribution with a scale parameter of 1.0.
• The population size consists of 10,000 data points.
• Sample details: 1000 samples, each with a sample size of 50.

The plot below shows the non-normal exponential distribution, which is right-skewed:

Distribution of Sample Means

• For each of the 1000 samples, a sample mean is calculated.
• The formula used is: $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$, where $\bar{x}$ represents the sample mean, $n$ is the sample size, and $x_i$ are the data points in the sample.

The second plot demonstrates the distribution of sample means, where the sample means form a bell-shaped (approximately normal) distribution, even though the original population is non-normal. This illustrates the Central Limit Theorem in action.

• The mean of the population data is approximately 0.9775.
• The mean of the sample means is approximately 0.9843, indicating that the average of the sample means closely aligns with the population mean, consistent with the prediction of the Central Limit Theorem.
• The standard error was calculated using the formula $\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. The resulting standard error was approximately 0.1378.
• The histogram of the sample means formed a bell-shaped curve, indicating a normal distribution as predicted by the Central Limit Theorem.

Standardizing Using CLT

When standardizing a sample statistic, we use the following formula to calculate the z-score:

$$ z = \frac{{\text{statistic} - \text{expected value}}}{{\text{Standard Error (SE) of the statistic}}} $$

Step-by-Step Example

Let’s assume we are sampling incomes with the following population parameters:

• Population mean: $\mu = 67,000$
• Population standard deviation: $\sigma = 38,000$
• Sample size: $n$

Step 1: Calculate the Standard Error of the Sample Mean

The standard error of the sample mean ($SE(\bar{x}_n)$) is calculated using the formula:

$$ SE(\bar{x}_n) = \frac{\sigma}{\sqrt{n}} $$

Where:

• $\sigma = 38,000$ is the population standard deviation.
• $n$ is the sample size.

For example, if we take a sample size of $n = 100$, we can calculate the standard error as follows:

$$ SE(\bar{x}_{100}) = \frac{38,000}{\sqrt{100}} = \frac{38,000}{10} = 3,800 $$

Step 2: Calculate the z-score

Once we have the standard error, we can calculate the z-score, which measures how far the sample statistic (e.g., sample mean) is from the expected value (population mean), in terms of standard errors. The z-score formula is:

$$ z = \frac{{\bar{x} - \mu}}{{SE(\bar{x}_n)}} $$

Where:

• $\bar{x}$ is the sample mean.
• $\mu = 67,000$ is the population mean.
• $SE(\bar{x}_n)$ is the standard error of the sample mean.

For instance, if the sample mean $\bar{x}$ is 70,000, the z-score would be:

$$ z = \frac{{70,000 - 67,000}}{{3,800}} = \frac{3,000}{3,800} = 0.79 $$

This means the sample mean is 0.79 standard errors above the population mean.

Example: Applying CLT

Consider a scenario where the heights of a certain plant species are normally distributed with a population mean $\mu = 15$ cm and a population standard deviation $\sigma = 3$ cm. We will analyze random samples of different sizes and calculate the probability that the sample mean falls between 14 cm and 16 cm.

Definitions:

• $\mu = 15$: Population mean height.
• $\sigma = 3$: Population standard deviation of heights.
• $n$: Sample size.
• Standard Error of the Mean (SEM): $\text{SEM} = \frac{\sigma}{\sqrt{n}}$.
• Z-score formula: $Z = \frac{X - \mu_{\text{sample mean}}}{\text{SEM}}$.

We will calculate the probability that the sample mean is between 14 cm and 16 cm for various sample sizes.

Step 1: Sample Size of 16 Plants

Estimate the probability that the sample mean height of 16 plants lies between 14 cm and 16 cm.

I. Calculate the Standard Error of the Mean (SEM):

$$ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{16}} = \frac{3}{4} = 0.75 $$

II. Calculate the Z-scores for 14 cm and 16 cm:

• For $X = 14$:

$$ Z_{14} = \frac{14 - 15}{0.75} = \frac{-1}{0.75} = -1.33 $$

• For $X = 16$:

$$ Z_{16} = \frac{16 - 15}{0.75} = \frac{1}{0.75} = 1.33 $$

III. Find the probabilities associated with the Z-scores:

Using standard normal distribution tables (or a calculator):

• $P(Z \leq -1.33) \approx 0.0918$
• $P(Z \leq 1.33) \approx 0.9082$

IV. Calculate the probability that the sample mean lies between 14 and 16 cm:

$$ P(14 \leq \bar{X} \leq 16) = P(Z \leq 1.33) - P(Z \leq -1.33) $$

$$ P(14 \leq \bar{X} \leq 16) = 0.9082 - 0.0918 = 0.8164 $$

Thus, the probability is approximately 81.64%.

Step 2: Sample Size of 64 Plants

Estimate the probability that the sample mean height of 64 plants lies between 14 cm and 16 cm.

I. Calculate the Standard Error of the Mean (SEM):

$$ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{64}} = \frac{3}{8} = 0.375 $$

II. Calculate the Z-scores for 14 cm and 16 cm:

• For $X = 14$:

$$ Z_{14} = \frac{14 - 15}{0.375} = \frac{-1}{0.375} = -2.67 $$

• For $X = 16$:

$$ Z_{16} = \frac{16 - 15}{0.375} = \frac{1}{0.375} = 2.67 $$

III. Find the probabilities associated with the Z-scores:

Using standard normal distribution tables (or a calculator):

• $P(Z \leq -2.67) \approx 0.0038$
• $P(Z \leq 2.67) \approx 0.9962$

IV. Calculate the probability that the sample mean lies between 14 and 16 cm:

$$ P(14 \leq \bar{X} \leq 16) = P(Z \leq 2.67) - P(Z \leq -2.67) $$

$$ P(14 \leq \bar{X} \leq 16) = 0.9962 - 0.0038 = 0.9924 $$

Thus, the probability is approximately 99.24%.

Step 3: Sample Size of 144 Plants

Estimate the probability that the sample mean height of 144 plants lies between 14 cm and 16 cm.

I. Calculate the Standard Error of the Mean (SEM):

$$ \text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{144}} = \frac{3}{12} = 0.25 $$

II. Calculate the Z-scores for 14 cm and 16 cm:

• For $X = 14$:

$$ Z_{14} = \frac{14 - 15}{0.25} = \frac{-1}{0.25} = -4 $$

• For $X = 16$:

$$ Z_{16} = \frac{16 - 15}{0.25} = \frac{1}{0.25} = 4 $$

III. Find the probabilities associated with the Z-scores:

Using standard normal distribution tables (or a calculator):

• $P(Z \leq -4) \approx 0.00003$
• $P(Z \leq 4) \approx 0.99997$

IV. Calculate the probability that the sample mean lies between 14 and 16 cm:

$$ P(14 \leq \bar{X} \leq 16) = P(Z \leq 4) - P(Z \leq -4) $$

$$ P(14 \leq \bar{X} \leq 16) = 0.99997 - 0.00003 = 0.99994 $$

Thus, the probability is approximately 99.99%.

Step 4: Unknown Population Distribution

Estimate the probability for the same range if the population distribution is unknown.

If the distribution of plant heights is unknown, the Central Limit Theorem assures us that the sampling distribution of the sample mean will still approximate normality as long as the sample size is sufficiently large (usually $n \geq 30$).

CLT Estimation:

• For a sample size $n = 64$ (as in Step 2), the sampling distribution of the sample mean will still be approximately normal, even if the population distribution is not normal.
• Therefore, the same calculations as in Step 2 can be applied, and the probability remains approximately 99.24%.