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Lesson 102 — Confidence interval for the mean

Construction and interpretation of confidence intervals for the population mean. z-cases (known sigma) and Student's t-cases (unknown sigma). Margin of error and sample size.

Used in: 3.º ano do EM (17-18 anos) · Equiv. Stochastik LK alemão · Equiv. Math B japonês · H2 Statistics singapurense

\bar X \pm z_{\alpha/2}\,\frac{\sigma}{\sqrt{n}} \quad \text{ou} \quad \bar X \pm t_{\alpha/2,\,n-1}\,\frac{s}{\sqrt{n}}

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Rigorous notation, full derivation, hypotheses

Rigorous definition

Pivotal statistics and the CI for the mean

"A 95% confidence interval means that if we construct many confidence intervals from many different samples, we expect 95% of these intervals to contain the true population parameter." — OpenStax Statistics, §8.1

Case 1: $\sigma$ known (z-pivot)

Definition· CI with known sigma

By the CLT, $Z = (\bar X - \mu)/(\sigma/\sqrt{n}) \overset{\text{approx.}}{\sim} \mathcal{N}(0,1)$ . Therefore:

P\!\left(-z_{\alpha/2} \leq \frac{\bar X - \mu}{\sigma/\sqrt{n}} \leq z_{\alpha/2}\right) = 1 - \alpha

what this means · Two-sided 1-alpha level CI for mu when sigma is known.

Rearranging for $\mu$ :

\mathrm{CI}_{1-\alpha}(\mu) = \left[\bar X - z_{\alpha/2}\frac{\sigma}{\sqrt{n}},\ \bar X + z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\right]

what this means · The z-confidence interval for mu. Extremes are sample mean plus and minus the margin of error.

Case 2: $\sigma$ unknown (t-pivot)

Definition· CI with unknown sigma — Student's t

Replacing $\sigma$ with $s = \sqrt{\frac{1}{n-1}\sum(X_i-\bar X)^2}$ , the pivotal statistic is:

T = \frac{\bar X - \mu}{s/\sqrt{n}} \sim t_{n-1}

what this means · Student's t-statistic: follows the t-distribution with n-1 degrees of freedom when the population is normal.

The resulting CI is:

\mathrm{CI}_{1-\alpha}(\mu) = \left[\bar X - t_{\alpha/2,\,n-1}\frac{s}{\sqrt{n}},\ \bar X + t_{\alpha/2,\,n-1}\frac{s}{\sqrt{n}}\right]

what this means · Student's t CI for mu. Uses the t-quantile with n-1 degrees of freedom.

The $t_{n-1}$ distribution has heavier tails than $\mathcal{N}(0,1)$ ; for $n \geq 30$ , the differences are small.

"When the population is not normal but $n$ is large, the t-distribution still approximates the behavior of the pivot well due to the robustness of the CLT." — OpenIntro Statistics, §4.2

Reference quantiles

Level $(1-\alpha)$	$z_{\alpha/2}$	$t_{\alpha/2,\,29}$	$t_{\alpha/2,\,9}$
90%	1.645	1.699	1.833
95%	1.960	2.045	2.262
99%	2.576	2.756	3.250

Margin of error and sample size

Solved examples

Example— 102.1· CI with known sigma (basic)

Problem. A laboratory measures glucose concentration in blood samples. The equipment has $\sigma = 5$ mg/dL (known from calibration). A sample of $n = 25$ patients gives $\bar X = 98$ mg/dL. Construct the 95% CI for the true mean concentration $\mu$ .

Strategy. Since $\sigma$ is known, use the z-pivot. The two-sided 95% quantile is $z_{0.025} = 1.960$ .

Resolution.

$E = z_{0.025} \cdot \frac{\sigma}{\sqrt{n}} = 1.960 \cdot \frac{5}{\sqrt{25}} = 1.960 \cdot 1 = 1.96$

$\mathrm{CI}_{95\%}(\mu) = [98 - 1.96;\; 98 + 1.96] = [96.04;\; 99.96] \text{ mg/dL}$

Verification. The CI width is $2 \times 1.96 = 3.92$ mg/dL. For $n = 100$ (four times larger), $E = 1.96 \cdot 5/10 = 0.98$ — half the width, consistent with $\mathrm{SE} \propto 1/\sqrt{n}$ .

Source. OpenStax Statistics, §8.1, Example 8.1 — CC-BY.

Example— 102.2· CI with unknown sigma — Student's t (intermediate)

Problem. A survey of $n = 16$ students on weekly study hours produced $\bar X = 18$ h and $s = 4$ h. Construct the 95% CI for the true mean.

Strategy. $\sigma$ unknown, small $n$ : use $t_{n-1} = t_{15}$ . Quantile: $t_{0.025,\,15} = 2.131$ .

Resolution.

$E = t_{0.025,\,15} \cdot \frac{s}{\sqrt{n}} = 2.131 \cdot \frac{4}{\sqrt{16}} = 2.131 \cdot 1 = 2.131$

$\mathrm{CI}_{95\%}(\mu) = [18 - 2.13;\; 18 + 2.13] = [15.87;\; 20.13] \text{ hours}$

Verification. If we had erroneously used $z_{0.025} = 1.960$ , $E = 1.96$ — underestimating the uncertainty of estimating $\sigma$ . The difference $2.131 - 1.960 = 0.171$ is the "cost" of not knowing $\sigma$ .

Source. OpenIntro Statistics, §4.2, Example 4.4 — CC-BY-SA.

Example— 102.3· Sample size for prescribed margin of error (intermediate)

Problem. A tax audit wants to estimate the average value of electronic invoices with a maximum margin of error of 50 at 99% confidence. Previous studies indicate $\sigma \approx 300$ . What is the minimum sample size?

Strategy. Use the formula $n \geq (z_{\alpha/2}\,\sigma/E)^2$ with $z_{0.005} = 2.576$ .

Resolution.

$n \geq \left(\frac{z_{\alpha/2}\,\sigma}{E}\right)^2 = \left(\frac{2.576 \times 300}{50}\right)^2 = (15.456)^2 = 238.9$

Rounding up: $n = 239$ invoices.

Verification. With $n = 239$ : $E = 2.576 \times 300/\sqrt{239} = 2.576 \times 19.41 = 50.0$ — exactly at the prescribed limit. Correct.

Source. OpenStax Statistics, §8.1, Example 8.4 — CC-BY.

Example— 102.4· Correct interpretation of CI — conceptual trap (advanced)

Problem. An engineer calculates the 95% CI for the average strength of steel bars: $[425;\; 435]$ MPa. Their colleague claims: "There is a 95% probability that $\mu$ is between 425 and 435 MPa." Is this statement correct?

Strategy. Identify what is random and what is fixed in the frequentist interpretation.

Resolution.

The statement is incorrect in the frequentist view. Once calculated, the interval $[425;\; 435]$ is deterministic — it either contains $\mu$ or it does not. The parameter $\mu$ is fixed; it does not have a probability distribution.

The correct statement is: the procedure that generated this interval has a 95% coverage property. If we repeated the experiment many times and calculated a CI each time, we would expect 95% of them to contain $\mu$ .

A valid version: "We are 95% confident that $\mu \in [425;\; 435]$ " — provided "confident" is understood as referring to the procedure, not the probability of the specific parameter.

Verification. In the Bayesian approach, with a non-informative Jeffreys prior, the 95% credible interval would have the same width and be interpretable as "95% probability that $\mu$ is in this interval given the data" — but this requires specification of a prior.

Source. OpenIntro Statistics, §4.2, Section "Common misconceptions" — CC-BY-SA.

Example— 102.5· CI by t-distribution with level comparison (advanced)

Problem. Monthly income of $n = 10$ families in a low-income neighborhood: 1200, 850, 1500, 970, 1100, 800, 1350, 1050, 920, 1080. Construct the 90%, 95%, and 99% CIs for the average income.

Strategy. Calculate $\bar X$ and $s$ of the sample; apply the t-formula with 9 degrees of freedom.

Resolution.

Sum: $1200+850+1500+970+1100+800+1350+1050+920+1080 = 10,820$ . Mean: $\bar X = 1082$ .

Deviation: $s^2 = \frac{\sum(X_i - \bar X)^2}{n-1}$ . Calculating squared deviations: $(118)^2 + (-232)^2 + (418)^2 + (-112)^2 + (18)^2 + (-282)^2 + (268)^2 + (-32)^2 + (-162)^2 + (-2)^2$ :

$= 13924 + 53824 + 174724 + 12544 + 324 + 79524 + 71824 + 1024 + 26244 + 4 = 433960$

$s^2 = 433960/9 \approx 48218$ , $s \approx 219.6$ .

$\mathrm{SE} = 219.6/\sqrt{10} \approx 69.4$ .

Level	$t_{\alpha/2,\,9}$	Margin	CI
90%	1.833	127.2	$[954.8;\; 1209.2]$
95%	2.262	157.0	$[925.0;\; 1239.0]$
99%	3.250	225.6	$[856.4;\; 1307.6]$

Verification. The three intervals are nested: 99% CI $\supset$ 95% CI $\supset$ 90% CI. More confidence implies a wider interval. Consistent with theory.

Source. OpenIntro Statistics, §4.2, Exercise 4.11 — CC-BY-SA.

Exercise list

30 exercises · 7 with worked solution (25%)

Application 19Understanding 3Modeling 4Challenge 2Proof 2

Ex. 102.1ApplicationAnswer key
Height of military recruits: $\sigma = 10$ cm, $n = 100$ , $\bar X = 165$ cm. Construct the 95% CI for the average height.
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Ex. 102.2Application
Using the same data from exercise 102.1, construct the 90% and 99% CIs and compare the three levels.
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Ex. 102.3ApplicationAnswer key
Weekly work hours: $n = 25$ , $\bar X = 45$ h, $s = 8$ h. Construct the 95% CI using the t-distribution.
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Ex. 102.4Application
A measuring device has $\sigma = 15$ units (known). What is the minimum $n$ to estimate the mean with a maximum margin of error of 3 units at 95% confidence?
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Ex. 102.5Application
Monthly tuition of $n = 12$ private colleges in a city: $\bar X = 1,500$ , $s = 200$ . 95% CI.
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Ex. 102.6Application
With $\sigma = 10$ and current 95% CI with $n = 16$ (width 9.8), what $n$ is needed to reduce the width to 5?
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Ex. 102.7Application
If you double the sample size, what is the percentage effect on the margin of error? And if you want to reduce the margin by half, by how much should you multiply $n$ ?
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Ex. 102.8ApplicationAnswer key
Notebook battery life: $n = 36$ , $\bar X = 200$ min, $s = 30$ min. 95% CI.
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Ex. 102.9Application
Weight of dogs of a specific breed: $n = 9$ , $\bar X = 15$ kg, $s = 1.2$ kg. 95% CI.
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Ex. 102.10Understanding
The statement "the 95% CI for $\mu$ is [45; 55]" means:
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Ex. 102.11UnderstandingAnswer key
Which of the following actions simultaneously produces a narrower CI AND higher confidence?
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Ex. 102.12ApplicationAnswer key
A tax auditor wants to estimate the average value of invoices with $\sigma = 500$ , margin of error 100, and 99% confidence. What is the minimum $n$ ?
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Ex. 102.13Application
Monthly food expenses of $n = 20$ families: $\bar X = 500$ , $s = 50$ . 95% CI with Student's t.
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Ex. 102.14Application
Compare the t-quantiles for $n = 9$ and $n = 100$ at 95% confidence. Explain why CIs with small samples are much wider.
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Ex. 102.15Application
Daily water consumption of $n = 8$ apartments (in liters): 5.2 | 4.8 | 5.5 | 4.9 | 5.1 | 5.3 | 4.7 | 5.0. Construct the 95% CI.
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Ex. 102.16ApplicationAnswer key
Blood glucose: $\sigma = 25$ mg/dL (from previous studies). What $n$ guarantees a margin of error of 5 mg/dL at 95%?
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Ex. 102.17Modeling
A metalworkers' union collected salaries of $n = 25$ employees: $\bar X = 1,800$ , $s = 200$ . The union claims the real average salary is below 1,883. Does the 95% CI support this claim?
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Ex. 102.18ModelingAnswer key
Body temperature of $n = 30$ healthy adults: $\bar X = 36.8$ °C, $s = 0.5$ °C. 95% CI. Is the classic value of 37°C compatible with these data?
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Ex. 102.19Modeling
An economist wants to estimate average quarterly GDP growth with a margin of error of 0.5 percentage points at 95%. If $\sigma \approx 2$ pp (historical variability), how many quarters of data are needed? Discuss practical feasibility.
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Ex. 102.20Application
A 90% CI has width $A_{90}$ . How many times wider is the 99% CI for the same sample and the same $\sigma$ ?
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Ex. 102.21Application
Weekly screen time of adolescents: $\sigma = 3$ h, $n = 49$ , $\bar X = 12$ h. 95% CI. Is the value 10 h/week plausible?
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Ex. 102.22Application
Cholesterol: $\sigma = 8$ mg/dL. What $n$ for 99% CI with a margin of 2 mg/dL?
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Ex. 102.23Application
Battery duration: $n = 40$ , $\bar X = 520$ h, $s = 60$ h. 95% CI. The manufacturer claims the average duration is 500 hours. Do the data support this claim?
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Ex. 102.24Understanding
What happens to the 95% CI when you increase $n$ from 25 to 100, keeping everything else constant?
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Ex. 102.25Modeling
The agency collected $n = 100$ retirement cases: $\bar X = 37$ days, $s = 15$ days. The legal target is 45 days. Construct the 95% CI and interpret it in relation to the target.
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Ex. 102.26Challenge
For monthly income data of $n = 20$ workers, compare the 95% CI for the mean (using t) with the 95% CI for the median (using order statistics). Which is more suitable for describing "typical" income? Why?
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Ex. 102.27Proof
Formally derive the $(1-\alpha)$ CI for $\mu$ with known $\sigma$ from the symmetry property of the standard normal distribution. Clearly identify what is random and what is fixed in $P(L \leq \mu \leq U) = 1 - \alpha$ .
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Ex. 102.28Proof
Demonstrate that $T = (\bar X - \mu)/(S/\sqrt{n}) \sim t_{n-1}$ when $X_i \overset{\text{iid}}{\sim} \mathcal{N}(\mu, \sigma^2)$ . What are the three properties you need to establish?
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Ex. 102.29Challenge
Show the duality between CI and hypothesis testing: "rejecting $H_0: \mu = \mu_0$ at level $\alpha$ is equivalent to $\mu_0$ being outside the $(1-\alpha)$ CI." Use this duality to explain how a CI can be used as a simultaneous two-sided test for all values of $\mu_0$ .
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Ex. 102.30Application
In $n = 200$ students taking an exam at a public school, 65 scored above 700 on the essay. Construct the 95% CI for the true proportion.
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Sources

OpenIntro Statistics (4th ed.) — Diez, Çetinkaya-Rundel, Barr · CC-BY-SA. Sections §4.2–4.4 (CI for means, duality with tests, sample size).
Statistics (OpenStax) — Illowsky, Dean · CC-BY. Chapter 8 (CI for mean with z and Student's t, margin of error).
Statistical Thinking for the 21st Century — Russell Poldrack · CC-BY-NC. Chapter 9 (frequentist vs. Bayesian interpretation of CI).

Rigorous definition

Pivotal statistics and the CI for the mean

Case 1: σ\sigmaσ known (z-pivot)

Case 2: σ\sigmaσ unknown (t-pivot)

Reference quantiles

Margin of error and sample size

Solved examples

Sources

Case 1: $\sigma$ known (z-pivot)

Case 2: $\sigma$ unknown (t-pivot)