Source: Mid-Term Formula Sheet, STAT 601 (UCF)
Tags: sampling distribution, central limit theorem, CLT, standard error, confidence interval, margin of error, sample size, point estimate, proportion, finite population correction, STAT 601
Difficulty: Intermediate Prerequisites: Chapters 1–3 (mean, standard deviation, z-scores), Chapter 6 (normal distribution).
Up to now, you have computed a sample mean or proportion and treated it as a single number. Chapters 7 and 8 ask a deeper question: if you drew a different sample, how different would that number be? Chapter 7 introduces the sampling distribution, the distribution of a statistic (like x̄ or p) across all possible samples of the same size. Chapter 8 then uses that distribution to build confidence intervals, ranges that are likely to contain the true population parameter. This is the bridge between describing your data and making inferences about the population it came from. The Central Limit Theorem is the key result that makes most of inferential statistics work.
Sample means and proportions vary from sample to sample, and that variation follows predictable patterns. The Central Limit Theorem says that for large enough samples, the distribution of the sample mean is approximately normal regardless of the population shape. Confidence intervals use this fact to estimate the true population parameter with a stated level of certainty.
Sampling distribution
The probability distribution of a sample statistic (such as x̄ or p) calculated from all possible samples of a given size from the population. In simple terms, it describes the "scatter" you would see if you repeated the same sampling process many times.
Standard error of the mean
σ_x̄ = σ / √n. The standard deviation of the sampling distribution of x̄. As sample size grows, the standard error shrinks, meaning your sample mean becomes a more precise estimate of μ.
Central Limit Theorem (CLT)
If the sample size is large enough (conventionally n ≥ 30), the sampling distribution of x̄ is approximately normal with mean μ and standard deviation σ / √n, regardless of the shape of the population distribution. This is the single most important theorem in introductory statistics.
Finite Population Correction Factor (FPCF)
When sampling more than 5% of a finite population, the standard error is multiplied by √[(N − n) / (N − 1)]. This factor shrinks the standard error because drawing a large fraction of the population reduces the remaining uncertainty.
Sample proportion
p = x / n, where x is the number of successes. The sample proportion is binomially distributed.
Normal approximation to the binomial
When nπ ≥ 10, n(1 − π) ≥ 10, and n ≥ 30, the sample proportion p is approximately normal with mean μₚ = π and standard deviation σₚ = √[π(1 − π) / n]. Think of it as the CLT applied specifically to proportions.
Point estimate
A single-value best guess for a population parameter. x̄ estimates μ, p estimates π, s estimates σ.
Confidence interval (CI)
An interval constructed so that, across many repeated samples, a stated percentage of such intervals would contain the true parameter. The general form: point estimate ± (critical value) × (standard error).
Confidence level
(1 − α) × 100%. If α = 0.05, you have a 95% confidence level. This means 95% of intervals built this way, across all possible samples, contain the true parameter.
Margin of error (e)
The half-width of the confidence interval. For a mean with σ known: e = z_(α/2) · σ / √n. With σ unknown: e = t_(α/2) · s / √n.
Mean of sample means: μ_x̄ = μ (unbiased)
Standard error: σ_x̄ = σ / √n
If the population is normal, x̄ is exactly normal for any n
If the population is non-normal, the CLT says x̄ is approximately normal when n ≥ 30
Standardisation (σ known): z = (x̄ − μ) / (σ / √n)
Standardisation with FPCF: z = (x̄ − μ) / [(σ / √n) · √((N − n) / (N − 1))]
μₚ = π
σₚ = √[π(1 − π) / n]
For finite populations (sample > 5% of N): σₚ = √[π(1 − π) / n] · √[(N − n) / (N − 1)]
Normal approximation conditions: nπ ≥ 10, n(1 − π) ≥ 10, and n ≥ 30
Standardisation: z = (p − π) / σₚ
Parameter | Population symbol | Point estimate |
|---|---|---|
Mean | μ | x̄ |
Proportion | π | p |
Standard deviation | σ | s |
General form: CI = point estimate ± (critical value) × (standard error of the point estimate)
Use z_(α/2) when σ is known (or when n is large by the CLT for proportions)
Use t_(α/2) with d.f. = n − 1 when σ is unknown
Specific intervals:
Mean, σ known: x̄ ± z_(α/2) · σ / √n
Mean, σ unknown: x̄ ± t_(α/2) · s / √n
Proportion (large sample): p ± z_(α/2) · √(pq / n)
To achieve a desired margin of error e:
σ known: n = (zσ / e)²
σ unknown (pilot study available): n = (t · s / e)²
For proportions: n = z² · π(1 − π) / e², using π from a pilot study or π = 0.5 for the "worst case" (maximises the required n)
Always round sample size up to the next whole number.
Standard error of the mean: σ_x̄ = σ / √n
Standard error of the proportion: σₚ = √[π(1 − π) / n]
Confidence interval (general): CI = point estimate ± (critical value) × (standard error)
Margin of error: e = z_(α/2) · σ / √n (σ known) e = t_(α/2) · s / √n (σ unknown)
Required sample size for a mean: n = (zσ / e)²
Required sample size for a proportion: n = z² · π(1 − π) / e²
Political polls report a "margin of error of ±3 percentage points." That margin of error comes directly from the confidence interval formula for a proportion. When a poll says a candidate has 52% support with ±3%, the underlying 95% CI runs from 49% to 55%.
Clinical trials use sample-size formulas to determine how many patients are needed before the trial begins. Getting this wrong means either wasting resources (too many patients) or failing to detect a real treatment effect (too few).
Students often interpret a 95% confidence interval as "there is a 95% probability the true parameter is in this interval." Strictly, the parameter is fixed; it is the interval that is random. 95% refers to the long-run proportion of intervals that would capture the parameter across repeated sampling.
Increasing the confidence level (say from 95% to 99%) makes the interval wider, not narrower. Higher confidence requires more "room" to be sure you have captured the parameter.
The CLT does not say the data becomes normal. It says the distribution of the sample mean becomes approximately normal. The raw data can remain skewed; it is x̄ that normalises.
When computing required sample size, always round up. Rounding down gives you a smaller sample than needed, which means your actual margin of error will exceed the target.
⚠️ Know the conditions for applying the CLT: n ≥ 30 for means, and nπ ≥ 10 plus n(1 − π) ≥ 10 for proportions.
⚠️ z vs. t: use z when σ is known; use t (with d.f. = n − 1) when σ is unknown and you substitute s.
⚠️ Finite Population Correction: apply it when n / N > 0.05.
⚠️ Confidence interval interpretation: practice the precise wording. Exams test whether you say "95% of intervals" rather than "95% probability the parameter is in this interval."
⚠️ Sample size formulas: expect a question asking you to solve for n given a desired margin of error.
True or False: The standard error of the mean decreases as the sample size increases.
Fill in the blank: The Central Limit Theorem requires n ≥ ________ for non-normal populations.
True or False: A 99% confidence interval is narrower than a 95% confidence interval for the same data.
Fill in the blank: When σ is unknown, you use the ________ distribution to build confidence intervals for the mean.
True or False: To find the required sample size for a proportion when no pilot data exists, use π = 0.5.
Answers: 1. True. 2. 30. 3. False (it is wider). 4. t (Student's t). 5. True.
Q: A population has σ = 20. You want a 95% CI for the mean with a margin of error no larger than 4. What sample size do you need?
A: n = (zσ / e)² = (1.96 × 20 / 4)² = (9.8)² = 96.04. Round up to n = 97.
Q: A sample of n = 64 has x̄ = 50 and σ = 16. Construct a 95% confidence interval for μ.
A: Standard error = 16 / √64 = 2. CI = 50 ± 1.96 × 2 = 50 ± 3.92 = (46.08, 53.92).
Q: In a sample of 200, 60 are successes. Construct a 90% confidence interval for the population proportion.
A: p = 60/200 = 0.30, q = 0.70. Standard error = √(0.30 × 0.70 / 200) = √(0.00105) ≈ 0.0324. z_(0.05) = 1.645. CI = 0.30 ± 1.645 × 0.0324 = 0.30 ± 0.0533 = (0.247, 0.353).
Q: Why do we apply the finite population correction factor?
A: When the sample is more than 5% of the population, the standard error is overstated without the correction. The FPCF = √[(N − n)/(N − 1)] reduces the standard error to reflect the fact that a large fraction of the population has already been observed.
The confidence interval framework here is the mirror image of hypothesis testing in Chapter 9: a 95% CI for μ contains all values of μ₀ that a two-tailed test at α = 0.05 would fail to reject. The sample-size formulas connect to study design and power analysis. The standard error concept extends directly to the two-sample problems in Chapter 10, where you estimate the difference between two means or two proportions.
sampling distribution, central limit theorem, CLT, standard error, confidence interval, margin of error, point estimate, interval estimate, z-interval, t-interval, sample size determination, proportion, finite population correction, FPCF, alpha, confidence level, STAT 601, UCF