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Grade 11-12 (age 16-18)

high school Statistical Estimation

When a whole population cannot be surveyed, statistical estimation uses the values from a single sample to judge the range where the population mean is likely to be. The sample mean wobbles around the population mean with standard deviation σ/√n, so a confidence interval is built by putting the sample mean in the middle and attaching a margin of error z × σ/√n on each side. Raising the confidence level makes z larger and the interval wider, while a larger sample makes σ/√n smaller and the interval narrower, so there is a trade-off. Move the sample-size and confidence sliders and the interval really widens and narrows on a fixed scale; below it you also see how many of eight intervals, each from a fresh sample, contain the population mean.

Population and Sample

🔬 What statistical estimation is
①When surveying the whole population is hard, a sample drawn from part of it is used to judge the population
②The sample mean computed from the sample estimates the population mean
③The key question is how close the sample mean is to the population mean and how large an error to allow

The Distribution of the Sample Mean

25
10
Distribution of the sample mean
X̄ ∼ N(μ, σ2n)
The distribution of the mean X̄ of a sample of size n when the population is N(μ, σ²)
💡 The effect of sample size
①As n grows, the distribution of X̄ narrows
②Its standard deviation σ/√n is called the standard error, and multiplying n by 4 halves it
③A larger σ widens both the population distribution and the distribution of the sample mean
④The law of large numbers says that the sample mean approaches the population mean as n grows

Confidence Intervals

36
95%
Confidence interval for the population mean
x̄ - z × σ√n ≤ μ ≤ x̄ + z × σ√n
The confidence interval for the population mean μ when the population SD σ is known
📐 Confidence level and z
①At 90% confidence, z = 1.645
②At 95% confidence, z = 1.96
③At 99% confidence, z = 2.576
④Raising the confidence level widens the interval, and increasing n narrows it
⑤Among the eight intervals under the figure, more contain μ as the confidence level goes up

What a Confidence Interval Means

Margin of error
E = z × σ√n
Half the width of the confidence interval
Required sample size
n ≥ (zσE)2
The smallest sample size that brings the margin of error down to E or less
⚠️ Clearing up a common misunderstanding
①A "95% confidence interval" does not mean "there is a 95% probability that μ lies in this interval"
②It means that if intervals are made the same way many times, about 95% of them contain μ
③μ is a fixed value; what changes from sample to sample is the interval

Worked Examples

Example 1
A sample of size n = 25 is drawn from a population with mean μ = 100 and standard deviation σ = 20. Find the mean and the standard deviation of the sample mean X̄.
1
The mean of the sample mean equals the population mean, and its standard deviation is σ/√n.
E(X̄) = μ, σ(X̄) = σ/√n
2
Put in μ = 100, σ = 20 and n = 25.
E(X̄) = 100, σ(X̄) = 20/√25 = 4
▸ E(X̄) = 100, σ(X̄) = 4
The standard deviation of the sample mean (the standard error) gets smaller as the sample gets larger.
Example 2
A sample of size n = 25 from a population with σ = 10 has sample mean 50. Find the 95% confidence interval for the population mean μ, given that P(|Z| ≤ 1.96) = 0.95.
1
The standard error is σ/√n = 10/5 = 2.
X̄ − 1.96·σ/√n ≤ μ ≤ X̄ + 1.96·σ/√n
2
Subtract the margin of error 1.96 × 2 = 3.92 from the sample mean and add it.
50 − 3.92 ≤ μ ≤ 50 + 3.92, i.e. 46.08 ≤ μ ≤ 53.92
▸ 46.08 ≤ μ ≤ 53.92
A confidence interval is the sample mean ± (the z for the confidence level × the standard error).

Summary

Key result
x̄ - z σ√n ≤ μ ≤ x̄ + z σ√n
The confidence interval for the population mean when σ is known
CSAT Math (Probability and Statistics) type
A sample of size n = 64 from a population with standard deviation σ gives a 95% confidence interval for the mean of length 4.9. Find σ, using the confidence coefficient 1.96.
①8
②9
③10
④11
⑤12
▸ ③ 10
1
The length of the interval is twice the margin of error (√64 = 8).
L = 2 × 1.96 × σ/√n
2
Substitute the values and solve for σ.
4.9 = 2 × 1.96 × σ/8 = 0.49σ ⇒ σ = 10
🎯 Exam Points
①X̄ follows N(μ, σ²/n)
②The standard error is σ/√n
③95% confidence uses z = 1.96 and 99% uses z = 2.576
④The margin of error is E = zσ/√n
⑤Multiplying n by 4 halves the margin of error
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Normal Distribution
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Distribution of Sample Mean

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