Sampling Distribution of Statistics and the Central Limit Theorem

Conceptual Foundation: Every random sample taken from a population involves inherent randomness or uncertainty. This uncertainty is transmitted to the statistics calculated from that sample (such as the mean).
Importance of Distribution: To perform accurate statistical inference (point estimation, confidence intervals, and hypothesis testing), researchers must understand the sampling distribution of the specific statistics being used.

Definitions and Notation:
- Population Parameters: Defined by mean mu and standard deviation sigma .
- Random Sample: Represented as X_1, X_2, dots, X_n.
- Sample Mean (bar{x}): Defined as the average of random variables: bar{x} = frac{sum_{i=1}^{n} X_i}{n}.
- Sample Total ( $T$ ): Defined as the summation of random variable values: T = sum_{i=1}^{n} X_i.
Expected Values:
- Mean of Sample Mean: The expected value of the sample mean is always the population mean: E[bar{X}] = mu.
- Mean of Sample Total: The expected value of the total is the sample size times the population mean: E[T] = n times mu.
Variance and Standard Deviation:
- Variance of Sample Mean: Var(bar{X}) = frac{sigma^2}{n}.
- Variance of Sample Total: Var(T) = n times sigma^2.
- Standard Deviation of Sample Mean: sigma_{bar{x}} = frac{sigma}{sqrt{n}}.
- Standard Deviation of Sample Total: sigma_T = sqrt{n} times sigma.
Standard Error of the Mean (SEM):
- Term Definition: The standard deviation of the sample mean (frac{sigma}{sqrt{n}}) is frequently referred to as the Standard Error of the Mean.
- Function: It describes the magnitude of a typical deviation of the sample mean from the population mean. It measures how spread out or variable the sample mean is relative to the center (mu).

Scenario: A notch tensile fatigue test is performed on titanium specimens.
Given Population Parameters:
- Expected cycles to first acoustic emission (mu): $28,000$ .
- Standard deviation (sigma): $5,000$ .
Sample Details: A random sample of $n = 25$ observations (X_1, dots, X_{25}) are taken from an Independently and Identically Distributed (IID) sample.
Calculations for Sample Mean (bar{X}):
- Expected Value: E[bar{X}] = mu = 28,000.
- Standard Deviation: sigma_{bar{x}} = frac{5,000}{sqrt{25}} = frac{5,000}{5} = 1,000.
Calculations for Sample Total ( $T$ ):
- Expected Total: E[T] = n times mu = 25 times 28,000 = 700,000.
- Standard Deviation: sigma_T = sqrt{25} times 5,000 = 5 times 5,000 = 25,000.

Principle: If the population distribution is Normal (N(mu, sigma^2)), then a random sample taken from it will result in sample mean and sample total distributions that are also Normal.
Distribution Parameters:
- Sample Mean: bar{X} sim N(mu, frac{sigma^2}{n}).
- Sample Total: T sim N(nmu, nsigma^2).
Sample Size Impact:
- Smaller sample size results in a more spread-out distribution (larger standard deviation).
- Larger sample size results in a distribution more concentrated around the mean (smaller standard deviation).

Scenario Context: Data from the article "Evaluation of egg quality traits of chicken reared under backyard system in Western Arta Pradesh."
Population Data:
- Distribution: Normal.
- Mean (mu): 53text{ grams}.
- Standard Deviation (sigma): 0.3text{ grams}.
Problem Part A: Total Weight of 12 Eggs:
- Sample Size ( $n$ ): $12$ .
- E[T] = 12 times 53 = 636.
- Var(T) = 12 times 0.3^2 = 1.08 (Note: The speaker corrected a typo in the notes where $3^2$ was written instead of $0.3^2$ ).
- Probability Calculation: Find P(635 < T < 640).
- Standardization ( $z$ -score): Z_1 = frac{635 - 636}{sqrt{1.08}} approx -0.96; Z_2 = frac{640 - 636}{sqrt{1.08}} approx 3.85.
- Standard Normal Result: P(-0.96 < Z < 3.85) = P(Z < 3.85) - P(Z < -0.96).
- Result: Yields a probability of approximately $0.832$ .
Problem Part B: Mean Weight of 4 Eggs:
- Sample Size ( $n$ ): $4$ .
- E[bar{X}] = 53.
- sigma_{bar{x}} = frac{0.3}{sqrt{4}} = 0.15.
- Probability Calculation: Find P(bar{X} > 53.5).
- Standardization: Z = frac{53.5 - 53}{0.15} = 3.33.
- Result: Yields a probability of approximately $0.0004$ .

General Application: Applied when the underlying population distribution is unknown (could be discrete, continuous, skewed, etc.).
Central Limit Theorem (CLT) Definition: If the sample size ( $n$ ) is sufficiently large, the sample mean (bar{X}) has approximately a Normal distribution, regardless of the shape of the population distribution.
Appropriate Normal Parameters Under CLT:
- Mean: mu_{bar{x}} = mu.
- Variance: sigma^2_{bar{x}} = frac{sigma^2}{n}.
Rule of Thumb: In this course, a sample size is considered "sufficiently large" if n > 30.
Conceptual Demonstrations:
- Coffee Simulation: Population mean = $2.5$ cups per week. Heavily right-skewed distribution. With $n = 10$ or $n = 48$ , the distribution of sample averages becomes bell-shaped and symmetric around $2.5$ .
- Spanish Flu Simulation: Ages of death distribution was non-normal. As $n$ increased from $10$ to $40$ , the sampling distribution of the mean centered at $43$ and became increasingly normal.
- Visual Resource: The simulation tool used was created by a geology professor at the University of British Columbia (UBC) Canada.

Scenario: Batch of chemical products.
Given Population Parameters:
- Mean (mu): 4text{ grams}.
- Standard Deviation (sigma): 1.5text{ grams}.
Sample Size: $n = 50$ independently prepared batches.
Task: Find the approximate probability that sample average bar{x} is between $3.5$ and $3.8$ .
Applying CLT:
- Since n = 50 > 30, we assume bar{X} sim N(4, frac{1.5^2}{50}) .
- sigma_{bar{x}} = frac{1.5}{sqrt{50}} approx 0.2121.
- Standardization:
  - Lower: frac{3.5 - 4}{0.2121} = -2.36 .
  - Upper: frac{3.8 - 4}{0.2121} = -0.94.
- Result: P(-2.36 < Z < -0.94) approx 0.164.

Context: Outcomes are either $1$ (success) or $0$ (failure).
Binary Parameters:
- $P(X_i = 1) = p$ .
- $P(X_i = 0) = 1 - p$ .
Criteria for Normal Approximation: For binary data, the rule of n > 30 is replaced by checking success/failure counts:
- Condition 1: (n times p) geq 10.
- Condition 2: (n times (1-p)) geq 10.
Distribution Parameters:
- Sample Proportion (hat{p} or bar{X}): mu = p; Var = frac{p(1-p)}{n}.
- Sum of Ones ( $X$ ): mu = np; $Var = np(1-p)$ .

General Properties (Any Distribution):
- E[bar{X}] = mu
- Var(bar{X}) = frac{sigma^2}{n}
- E[T] = nmu
- Var(T) = nsigma^2
Normal Population Requirement: If population is N(mu, sigma^2), then bar{X} and $T$ are exactly Normal.
Central Limit Theorem Requirement: If population is not Normal but n > 30, then bar{X} is approximately Normal.
Binary Requirement: If population is Binary, approximation is valid if np geq 10 and n(1-p) geq 10.