high school Statistical Estimation

Statistical Estimation

When you cannot survey a whole population, statistical estimation uses a single sample to guess the range where the population mean is likely to lie. Since the sample mean wobbles around the population mean with a spread of σ/√n, you attach a margin of error on either side of that sample mean to build a confidence interval. Raising the confidence level to 95% makes the z value bigger and widens the interval, while enlarging the sample narrows it, a genuine trade-off. Move the sample-size n and confidence-level sliders and the interval widens and narrows; 95% confidence is not the probability that the population mean lies in this particular interval.

Population & Sample

🔬 What is Estimation?

①When surveying the whole population is impractical

②Estimate the population from a sample

③Key question: how precisely does the sample mean estimate the population mean?

Distribution of Sample Mean

Sample size n25

Population σ10

Sample Mean Distribution

X̄ ∼ N(μ, σ²n)

For population N(μ,σ²) the sample mean X̄ has this distribution

💡 Magic of Sample Size

①Larger n → narrower distribution of X̄

②Standard error = σ/√n; quadruple n to halve error

③A smaller standard error is a sampling-distribution fact. The LLN says the sample mean converges to the population mean

Confidence Interval

Sample size n36

Confidence (%)95%

CI for Population Mean

x̄ - z × σ√n ≤ μ ≤ x̄ + z × σ√n

Confidence interval for μ when σ is known

📐 Confidence Level and z

①90%: z = 1.645

②95%: z = 1.96

③99%: z = 2.576

④Higher confidence → wider interval. Larger n → narrower

⑤Slider: 90, 95, 99. Example μ is the population mean

Meaning of CI

Margin of Error

E = z × σ√n

Half-width of the interval

Required Sample Size

n ≥ (zσE)²

Minimum n for error within E

⚠️ Common Misconception

①"95% CI" does NOT mean "P(μ in this interval) = 95%"

②Correct: of 100 intervals built this way, about 95 contain μ

③μ is fixed; the interval changes from sample to sample

Work It Out

Example 1

From a population with mean μ = 100 and SD σ = 20, a sample of size n = 25 is drawn. Find the mean and SD of the sample mean X̄.

The mean of the sample mean equals μ, and its SD is σ/√n.

E(X̄) = μ, σ(X̄) = σ/√n

Substitute.

E(X̄) = 100, σ(X̄) = 20/√25 = 4

▸ E(X̄) = 100, σ(X̄) = 4

The SD of the sample mean (standard error) shrinks as the sample grows.

Example 2

From a population with σ = 10, a sample of size n = 25 has sample mean 50. Find the 95% confidence interval for μ. (Given P(|Z| ≤ 1.96) = 0.95)

Substitute into the confidence-interval formula (σ/√n = 10/5 = 2).

X̄ − 1.96·σ/√n ≤ μ ≤ X̄ + 1.96·σ/√n

Add and subtract 1.96 × 2 = 3.92.

50 − 3.92 ≤ μ ≤ 50 + 3.92, i.e. 46.08 ≤ μ ≤ 53.92

▸ 46.08 ≤ μ ≤ 53.92

Confidence interval = sample mean ± (confidence coefficient × standard error).

Wrap-up

Core

x̄ - z σ√n ≤ μ ≤ x̄ + z σ√n

Confidence interval for μ (σ known)

exam-style

From a population with SD σ, a size n = 64 sample gives a 95% confidence interval for the mean of length 4.9. Find σ. (Confidence coefficient 1.96)

①8

②9

③10

④11

⑤12

▸ ③ 10

Write the formula for the length of the confidence interval (√64 = 8).

L = 2 × 1.96 × σ/√n

Substitute and solve for σ.

4.9 = 2 × 1.96 × σ/8 = 0.49σ ⇒ σ = 10

🎯 Exam Points

①Sampling distribution: X̄ ~ N(μ, σ²/n)

②Standard error: σ/√n

③95% → z = 1.96, 99% → z = 2.576

④E = zσ/√n

⑤Quadrupling n halves the error

← Previous

Normal Distribution

Distribution of Sample Mean

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