Sampling Distributions

Population vs. Sample

• Population: All elements/members (individuals, items, objects) whose characteristics are studied.
• Sample: A portion/subset of the population selected for study.

Reasons for Sampling

• Gathering information on the entire population is too expensive.
• It is often impossible to gather information on the entire population.

Representative Sample

• A sample representing the population's characteristics as closely as possible.
• Also called random sample.

Types of Samples

• Random/Representative Samples
  • Simple Random Sample
  • Systematic Sample
  • Stratified Sample
  • Cluster Sample
• Non-random/Non-representative
  • Voluntary Response Sample / Convenience Sample

Sampling Methods

• Simple Random Sample: Each element has the same chance of being included.
• Systematic Sample: Select a starting point, then select every nth element (e.g., 10th or 50th).
• Cluster Sample: Divide the population into groups (clusters), randomly select clusters, and then include all members from selected clusters.
• Stratified Sample: Subdivide the population into subgroups (strata) with shared characteristics (e.g., gender, age), then draw a sample from each stratum.
• Voluntary Response/Convenience Sample: Respondents decide whether to be included.

Inferential Statistics

• Making statements about a population by examining sample results.
• Using sample statistics to infer population parameters.
• Inference: Estimating unknown population parameters from sample evidence.
• Estimation: Estimating population mean weight using the sample mean weight.
• Hypothesis Testing: Using sample evidence to test a claim about a population parameter (e.g., population mean weight is 120 pounds).
• Drawing conclusions and/or making decisions about a population based on sample results.

Population and Sampling Distribution

• Population Distribution: Probability distribution of the population data.
• Sampling Distribution: Probability distribution of a sample statistic.
• Sample information helps infer population parameters ($\mu$, p, and $\sigma$) based on sample statistics ($\bar{x}$, $\hat{p}$, and S).

Sampling Distributions

• Sampling Distributions of Sample Means
• Sampling Distributions of Sample Proportions

Sampling Distribution of the Mean ($\sigma$ known)

• Central Limit Theorem:
  • Let $X<em>1, X</em>2, …, X_n$ be a sequence of independent and identically distributed random variables each having a mean $\mu$ and a standard deviation $\sigma$.
  • For a large n (by the law of large numbers), the Central Limit Theorem asserts that: $E(X) = \mu_x = \mu$
  • Standard error of the mean: \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}}
  • For samples from finite populations: \sigma_\bar{x} = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{N-n}{N-1}}
  • Finite population correction factor: $\sqrt{\frac{N-n}{N-1}}$
  • If $\frac{n}{N} \leq 0.05$, then the correction factor can be ignored.

Central Limit Theorem Visualization

• Sampling distribution of $\bar{X}$ becomes normal as n increases, regardless of population distribution (shown with uniform and exponential distributions).
• As sample size (n) increases, the sampling distribution becomes approximately normal.

Standardized Sample Mean

• $Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}$
• When n is large and the population is infinite or finite but $\frac{n}{N} \leq 0.05$
• If random samples are from a normal distribution, the sampling distribution of the mean is normal regardless of sample size.
• In practice, the normal distribution provides an excellent approximation to the sample distribution of the mean for n as small as 30.

Sampling Distribution of the Mean ($\sigma$ unknown)

• If $\bar{X}$ is the mean of a random sample of size n from a normal population with mean $\mu$ and standard deviation $\sigma$:
  • $t = \frac{\bar{X} - \mu}{\frac{S}{\sqrt{n}}}$, where $\frac{S}{\sqrt{n}} = S_x$
  • S is a random variable having a t-distribution with df = n-1 (degrees of freedom).
  • $S = \sqrt{\frac{\sum<em>{i=1}^{n} (X</em>i - \bar{X})^2}{n-1}}$

Student’s t-Distribution

• t-distributions are bell-shaped and symmetric, but have ‘fatter’ tails than the normal distribution.
• Standard Normal distribution is equivalent to a t-distribution with df = ∞.
• t approaches Z as n increases.
• The t-distribution is similar to the standard normal distribution and both are bell shaped and symmetrical about the mean.
• The standard deviation of the t-distribution exceeds 1, but approaches 1 as n approaches ∞.
• The standard normal distribution provides a good approximation to the t-distribution for a sample size of 30 or more.
• The assumption that the sample must come from a normal distribution is not so severe restriction as it may seem. Studies have shown that the distribution of random variable is fairly close to a t-distribution even for a sample from certain non-normal populations. In practice, it is necessary to make sure that the population from which we draw the sample is approximately bell shaped and not too skewed.

Problems and Solutions: Sample Mean

• Problem 1:

  • SAT test mean score: 904, standard deviation: 152. Assume normal distribution.
  • Calculate mean and standard deviation of $\bar{X}$ for:
    • a) n = 16
      • $E(\bar{X}) = \mu_{\bar{x}} = \mu = 904$
      • $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{152}{\sqrt{16}} = 38$
    • b) n = 100
      • $E(\bar{X}) = \mu_{\bar{x}} = \mu = 904$
      • $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{152}{\sqrt{100}} = 15.2$

• Problem 2:

  • Cookie weight mean: 8 ounces, standard deviation: 3 ounces.
  • Find P(7.8 ≤ $\bar{X}$ ≤ 8.2) for n = 36.
  • Solution:
    • $\mu_{\bar{x}} = 8$
    • $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = 0.5$
    • $P(7.8 \leq \bar{X} \leq 8.2) = P(\frac{7.8 - 8}{0.5} \leq Z \leq \frac{8.2 - 8}{0.5}) = P(-0.4 \leq Z \leq 0.4) = 0.3108$

Sampling Distribution of the Proportion

• The information that is usually available for the estimation of a proportion is the number of times, X, that an appropriate event occurs in n trials, occasions, or observations.
• The point estimator of the population proportion, itself, is usually the sample proportion: $\hat{p} = \frac{X}{n}$
  • $\hat{p}$ - proportion of the time that the event actually occurs or sample proportion.
• If the n trials satisfy the assumptions underlying the binomial distribution, we know the mean and standard deviation of the number of successes.
• If we divide both of these quantities by n, we will find the mean and the standard deviation of the proportions of successes:
  • $\mu_{\hat{p}} = p$
  • $\sigma_{\hat{p}} = \sqrt{\frac{pq}{n}} = \sqrt{\frac{p(1-p)}{n}}$

Standardized Sample Mean for Proportion

• $Z = \frac{\hat{p} - p}{\sqrt{\frac{pq}{n}}}$
• When n is large, we can construct a Z-value for a binomial parameter p, by using the normal approximation to the binomial distribution.
• Usually, the normal distribution is used as an approximation to the binomial distribution when np ≥ 5 and nq ≥ 5

Proportion Problems and Solutions

• Problem 1:
  • 44% of teenage boys are bad drivers. Sample of 100 male teenager drivers.
  • Find mean and standard deviation of $\hat{p}$.
  • $\mu_{\hat{p}} = p = 0.44$
  • $\sigma_{\hat{p}} = \sqrt{\frac{(0.44)(0.56)}{100}} = 0.0496$
• Problem 2:
  • True proportion of voters supporting Proposition A is 0.4. Sample size 200.
  • Find P(0.40 ≤ $\hat{p}$ ≤ 0.45).
  • $\sigma_{\hat{p}} = \sqrt{\frac{0.4(1-0.4)}{200}} = 0.03464$
  • $P(0.40 \leq \hat{p} \leq 0.45) = P(\frac{0.40 - 0.40}{0.03464} \leq Z \leq \frac{0.45 - 0.40}{0.03464}) = P(0 \leq Z \leq 1.44) = 0.4251$

Additional Problems

• Problem 1:
  • Maureen Webster claims 53% favor in a large city. Sample 400 voters. Find probability less than 49% favor her.
  • P(\hat{p} < 0.49) = P(Z < \frac{0.49 - 0.53}{\sqrt{\frac{(0.53)(0.47)}{400}}}) = P(Z < -1.6) = 0.0548
• Problem 2:
  • Machine produces 5% defective parts. Sample of 400 parts every week. Stop and readjust if 8% or more defective.
  • Find probability process will be stopped.
  • P(\hat{p} > 0.08) = P(Z > \frac{0.08 - 0.05}{\sqrt{\frac{(0.05)(0.95)}{400}}}) = P(Z > 2.75) = 0.003