v1 · padrão canônico

Lesson 102 — Confidence Interval for the Mean

Construction and interpretation of confidence intervals for the population mean. Cases z (sigma known) and t-Student (sigma unknown). Margin of error and sample size.

Used in: 3rd year of high school (17-18 years old) · Equiv. Stochastik LK German · Equiv. Math B Japanese · H2 Statistics Singaporean

\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}}

The confidence interval for $\mu$ transforms the point estimate $\bar X$ into a plausible range. Use $z$ when $\sigma$ is known; use $t_{n-1}$ when $\sigma$ is estimated from the sample. Increasing the confidence level widens the interval; increasing $n$ narrows it.

Choose your door

Rigorous notation, full derivation, hypotheses

Rigorous definition

Pivotal statistic and the CI for mean

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

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

Definition· CI with sigma known

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 CI of level 1-alpha for mu when sigma is known.

Rearranging around $\mu$ :

\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. Endpoints are sample mean plus and minus the margin of error.

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

Definition· CI with sigma unknown — Student's t

Substituting $\sigma$ 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 distribution $t_{n-1}$ 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 well the behavior of the pivot via 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 minimum sample size

Worked examples

Example— 102.1· CI with sigma known (basic)

Problem. A laboratory measures blood glucose concentration. The device has $\sigma = 5$ mg/dL (known by 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 95% bilateral quantile is $z_{0.025} = 1.960$ .

Solution.

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

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

Verification. The amplitude of the CI 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 amplitude, consistent with $\mathrm{SE} \propto 1/\sqrt{n}$ .

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

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

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

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

Solution.

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

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

Verification. If we had wrongly 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 mean value of electronic invoices with maximum margin of error of USD 50 at 99% confidence. Previous studies suggest $\sigma \approx 300$ dollars. What is the minimum sample size?

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

Solution.

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

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 pitfall (advanced)

Problem. An engineer calculates the 95% CI for the mean strength of steel bars: $[425;\; 435]$ MPa. His colleague states: "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.

Solution.

The statement is incorrect in the frequentist view. Once calculated, the interval $[425;\; 435]$ is deterministic — it either contains $\mu$ or it doesn't. 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 reference to the procedure, not the probability of the specific parameter.

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

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

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

Problem. Monthly income of $n = 10$ families from a poor community: 1200, 850, 1500, 970, 1100, 800, 1350, 1050, 920, 1080 (in dollars). Construct the 90%, 95%, and 99% CIs for the mean income.

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

Solution.

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

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

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

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

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

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: CI 99% $\supset$ CI 95% $\supset$ CI 90%. More confidence implies wider interval. Consistent with theory.

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

Exercise list

42 exercises · 10 with worked solution (25%)

Application 26Understanding 4Modeling 7Challenge 3Proof 2

Ex. 102.1ApplicationAnswer key
Height of military recruits: $\sigma = 10$ cm, $n = 100$ , $\bar X = 165$ cm. Construct the 95% CI for average height.
Select the correct option
95% CI: $[163.04;\; 166.96]$ cm ( $z_{0.025}=1.960$ , $E=1.960$ )95% CI: $[163.36;\; 166.64]$ cm (used $z=1.645$ from 90% level)95% CI: $[162.42;\; 167.58]$ cm (used $z=2.576$ from 99% level)95% CI: $[155.40;\; 174.60]$ cm (forgot $\sqrt{n}$ : used $E=1.96\times10$ )
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
$E = 1.960 \times 10/\sqrt{100} = 1.960$ . 95% CI: $[165 - 1.96;\; 165 + 1.96] = [163.04;\; 166.96]$ cm.
Ex. 102.2Application
Using the same data as exercise 102.1, construct the 90% and 99% CIs and compare the three levels.
Select the correct option
90% CI: $[163.36;\; 166.64]$ cm and 99% CI: $[162.42;\; 167.58]$ cm — greater confidence implies wider interval90% CI: $[163.04;\; 166.96]$ cm and 99% CI: $[163.36;\; 166.64]$ cm — quantiles were swapped90% CI: $[163.36;\; 166.64]$ cm and 99% CI: $[163.36;\; 166.64]$ cm — confidence level does not alter CI90% CI: $[162.42;\; 167.58]$ cm and 99% CI: $[163.04;\; 166.96]$ cm — greater confidence narrows CI
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
90% CI: $E = 1.645 \times 10/10 = 1.645$ . CI: $[163.36;\; 166.64]$ cm. 99% CI: $E = 2.576 \times 1 = 2.576$ . CI: $[162.42;\; 167.58]$ cm. Greater confidence implies wider interval.
Ex. 102.3Application
Weekly work hours: $n = 25$ , $\bar X = 45$ h, $s = 8$ h. Construct the 95% CI using the t distribution.
Select the correct option
95% CI: $[41.70;\; 48.30]$ h ( $t_{0.025,24}=2.064$ , $E=3.30$ )95% CI: $[41.86;\; 48.14]$ h (used $z=1.960$ instead of $t_{24}$ )95% CI: $[44.34;\; 45.66]$ h (used $t_{0.025,24}=2.064$ but divided by $n=25$ instead of $\sqrt{n}$ , $\mathrm{SE}=8/25=0.32$ )95% CI: $[42.26;\; 47.74]$ h (used $t_{0.05,24}=1.711$ , one-tailed quantile)
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
95% CI with $t_{0.025,\,24} = 2.064$ : $45 \pm 2.064 \times 1.6 = 45 \pm 3.30$ . CI: $[41.70;\; 48.30]$ weekly hours.
Show step-by-step (with the why)
1. Identify: $n = 25$ , $\bar X = 45$ , $s = 8$ , 95% level.
2. Degrees of freedom: $\nu = n - 1 = 24$ . Quantile: $t_{0.025,\,24} = 2.064$ .
3. Standard error: $\mathrm{SE} = s/\sqrt{n} = 8/5 = 1.6$ .
4. Margin: $E = 2.064 \times 1.6 = 3.30$ .
5. CI: $[45 - 3.30;\; 45 + 3.30] = [41.70;\; 48.30]$ hours.
6. Trick: With $n = 25$ and 95% level, the $t$ quantile is 2.064 — slightly larger than 1.960 from $z$ . The difference is small, but underestimating uncertainty with $z$ would be inadequate for small $n$ .
Ex. 102.4ApplicationAnswer key
A measuring device has $\sigma = 15$ units (known). What is the minimum $n$ to estimate the mean with maximum margin of error of 3 units at 95% confidence?
Select the correct option
$n = 97$ measurements ( $n\geq(1.960\times15/3)^2=96.04$ , round up) $n = 62$ measurements (used $z=1.645$ from 90% level: $(1.645\times15/3)^2=67.65\to68$ , did not round correctly) $n = 96$ measurements (computed 96.04 correctly but did not round up) $n = 25$ measurements (computed $n=(\sigma/E)^2=(15/3)^2=25$ , forgot $z^2$ factor)
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
$n \geq (z_{\alpha/2}\,\sigma/E)^2 = (1.960 \times 15/3)^2 = (9.80)^2 = 96.04$ . Rounds: $n = 97$ measurements.
Ex. 102.5ApplicationAnswer key
Tuition of $n = 12$ private colleges in a city: $\bar X = 1,500$ dollars, $s = 200$ dollars. 95% CI.
Select the correct option
95% CI: USD $[1372.90; 1627.10]$ ( $t_{0.025,11}=2.201$ , $E=127.1$ )95% CI: USD $[1392.00; 1608.00]$ (used $z=1.960$ ignoring that $\sigma$ is unknown)95% CI: USD $[1334.20; 1665.80]$ (used $t_{0.025,11}=2.201$ but divided $s$ by $n=12$ instead of $\sqrt{12}$ )95% CI: USD $[1353.60; 1646.40]$ (used $t_{0.025,12}=2.179$ , df $n$ instead of $n-1$ )
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
$t_{0.025,\,11} = 2.201$ . $\mathrm{SE} = 200/\sqrt{12} = 57.74$ . $E = 2.201 \times 57.74 = 127.1$ . 95% CI: USD $1,372.90 to USD$ 1,627.10.
Ex. 102.6ApplicationAnswer key
With $\sigma = 10$ and 95% CI currently with $n = 16$ (amplitude 9.8), what $n$ is needed to reduce amplitude to 5?
Select the correct option
$n = 62$ (amplitude 5 requires $n=(2\times1.960\times10/5)^2=61.5$ , round up) $n = 64$ (quadrupled $n=16$ directly, but resulting amplitude would be 4.9, not 5) $n = 32$ (doubled $n=16$ , but this reduces amplitude by $\sqrt{2}\approx1.41$ , not half) $n = 97$ (used sample size formula with $E=3$ instead of $E=2.5$ )
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
Amplitude = $2E = 2 \times z_{\alpha/2} \times \sigma/\sqrt{n}$ . If $n$ quadruples, $\sqrt{n}$ doubles and amplitude falls in half. For amplitude of 5: $n = (2 \times 1.960 \times 10/5)^2 = (7.84)^2 = 61.5$ . Rounds: $n = 62$ .
Ex. 102.7ApplicationAnswer key
If you double the sample size, what is the percentage effect on the margin of error? And if you want to halve the margin, how much should you multiply $n$ ?
Select the correct option
Doubling $n$ reduces $E$ by 29.3%; to halve $E$ , multiply $n$ by 4Doubling $n$ halves $E$ ; to halve $E$ , multiply $n$ by 2Doubling $n$ reduces $E$ by 50%; to halve $E$ , multiply $n$ by 2Doubling $n$ reduces $E$ by 29.3%; to halve $E$ , multiply $n$ by 2
Select an option first
Solve online OpenIntro Statistics · §4.1
Show solution
By doubling $n$ , $\sqrt{n}$ multiplies by $\sqrt{2} \approx 1.414$ , so $E$ divides by $\sqrt{2}$ : reduction of 29.3%. To halve $E$ , multiply $n$ by 4.
Ex. 102.8Application
Notebook battery time: $n = 36$ , $\bar X = 200$ min, $s = 30$ min. 95% CI.
Select the correct option
95% CI: $[189.85;\; 210.15]$ min ( $t_{0.025,35}=2.030$ , $E=10.15$ )95% CI: $[190.20;\; 209.80]$ min (used $z=1.960$ instead of $t_{35}$ )95% CI: $[184.40;\; 215.60]$ min (used $t_{0.025,35}$ but divided $s$ by $n=36$ instead of $\sqrt{36}$ )95% CI: $[191.43;\; 208.57]$ min (used $t_{0.05,35}=1.690$ , one-tailed quantile)
Select an option first
Solve online OpenStax Statistics · §8.2
Show solution
$\mathrm{SE} = s/\sqrt{n} = 30/\sqrt{36} = 5$ . 95% CI with $t_{0.025,\,35} = 2.030$ : $[200 - 10.15;\; 200 + 10.15] = [189.85;\; 210.15]$ min.
Ex. 102.9Application
Weight of dogs of a specific breed: $n = 9$ , $\bar X = 15$ kg, $s = 1.2$ kg. 95% CI.
Select the correct option
95% CI: $[14.08;\; 15.92]$ kg ( $t_{0.025,8}=2.306$ , $E=0.922$ )95% CI: $[14.22;\; 15.78]$ kg (used $z=1.960$ instead of $t_8$ , underestimating uncertainty)95% CI: $[13.60;\; 16.40]$ kg (used $t_{0.025,8}$ but divided $s$ by $n=9$ instead of $\sqrt{9}$ )95% CI: $[14.27;\; 15.73]$ kg (used $t_{0.025,9}=2.262$ with $\nu=n$ instead of $n-1$ )
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
With $t_{0.025,\,8} = 2.306$ and $\mathrm{SE} = 0.4$ : 95% CI = $[14.08;\; 15.92]$ kg.
Show step-by-step (with the why)
1. The 95% CI formula is $\bar X \pm t_{0.025,\,n-1} \times s/\sqrt{n}$ .
2. With $n = 9$ , $t_{0.025,\,8} = 2.306$ .
3. $\mathrm{SE} = 1.2/\sqrt{9} = 0.4$ kg.
4. $E = 2.306 \times 0.4 = 0.922$ kg.
5. CI: $[15 - 0.922;\; 15 + 0.922] = [14.08;\; 15.92]$ kg.
6. Note: With only 9 observations, the $t$ quantile is substantially larger than 1.96. This reflects greater uncertainty in estimating $\sigma$ with few data.
Ex. 102.10Understanding
The statement "the 95% CI for $\mu$ is [45; 55]" means:
Select the correct option
The parameter mu is in this interval with 95% probabilityIf we repeated the procedure with many samples, 95% of the calculated intervals would contain muThe collected sample is in the interval with 95% probabilityThere is a 5% chance we made a calculation error
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
In the frequentist view, $\mu$ is fixed — it has no probability of being anywhere. What has 95% probability is the construction procedure: 95% of samples produce an interval that covers $\mu$ .
Ex. 102.11UnderstandingAnswer key
Which of the following actions simultaneously produces a narrower CI AND greater confidence?
Select the correct option
Increase the confidence level from 95% to 99%, keeping n fixedIncrease n keeping the confidence level fixedDecrease the sample standard deviation sIncrease the confidence level from 95% to 99% and quadruple n
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
Narrower CI: increasing $n$ narrows the CI (dividing by $\sqrt{n}$ ). Higher confidence level widens. For the narrowest interval with 99% confidence, one must increase $n$ enough to compensate for the larger quantile. Quadrupling $n$ compensates exactly for a 2× factor in amplitude — enough to keep the 99% CI as narrow as the 95% CI with original $n$ .
Ex. 102.12Application
A tax auditor wants to estimate the mean value of invoices with $\sigma = 500$ dollars, margin of error USD 100, and 99% confidence. What is the minimum $n$ ?
Select the correct option
$n = 166$ invoices ( $n\geq(2.576\times500/100)^2=165.9$ , round up) $n = 97$ invoices (used $z=1.960$ from 95% level instead of $z=2.576$ from 99%) $n = 165$ invoices (computed 165.9 but did not round up) $n = 25$ invoices (computed $(\sigma/E)^2=(500/100)^2=25$ , forgot $z^2$ factor)
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
$n \geq (2.576 \times 500/100)^2 = (12.88)^2 = 165.9$ . Rounds: $n = 166$ invoices.
Ex. 102.13Application
Monthly food spending of $n = 20$ families: $\bar X = 500$ dollars, $s = 50$ dollars. 95% CI with Student's t.
Select the correct option
95% CI: USD $[476.60; 523.40]$ ( $t_{0.025,19}=2.093$ , $E=23.4$ )95% CI: USD $[480.10; 519.90]$ (used $z=1.960$ instead of $t_{19}$ )95% CI: USD $[470.15; 529.85]$ (used $t_{0.025,20}=2.086$ , df $n$ instead of $n-1$ , but computed incorrectly)95% CI: USD $[462.55; 537.45]$ (divided $s=50$ by $n=20$ instead of $\sqrt{20}$ )
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
With $t_{0.025,\,19} = 2.093$ : $E = 2.093 \times 50/\sqrt{20} = 2.093 \times 11.18 = 23.4$ . 95% CI: USD $476.60 to USD$ 523.40.
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.
Select the correct option
$t_{0.025,8}=2.306$ and $t_{0.025,99}\approx1.984$ ; with $n=9$ the CI is wider because $t$ is larger and $\sqrt{n}$ is smaller $t_{0.025,8}=1.860$ and $t_{0.025,99}=1.960$ ; the difference is small and CI is nearly equal $t_{0.025,8}=2.306$ and $t_{0.025,99}\approx1.984$ ; CIs have similar amplitude because large $n$ compensates $t_{0.025,8}=2.262$ and $t_{0.025,99}=1.960$ ; both use $z=1.960$ for $n\geq 30$
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
For $n = 9$ : $t_{0.025,\,8} = 2.306$ ; for $n = 100$ : $t_{0.025,\,99} \approx 1.984$ . With $n = 9$ , the CI is much wider: the combination of large $t$ and small $\sqrt{n}$ greatly amplifies uncertainty.
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.
Select the correct option
95% CI: $[4.84;\; 5.29]$ L ( $\bar X=5.063$ , $s\approx0.267$ , $t_{0.025,7}=2.365$ )95% CI: $[4.88;\; 5.24]$ L (used $z=1.960$ instead of $t_7$ )95% CI: $[4.78;\; 5.35]$ L (used $t_{0.025,7}$ but divided $s$ by $n=8$ instead of $\sqrt{8}$ )95% CI: $[4.91;\; 5.22]$ L (used $t_{0.025,8}=2.306$ with $\nu=n$ instead of $n-1$ )
Select an option first
Solve online OpenStax Statistics · §8.2
Show solution
Mean: 5.0625 L, $s \approx 0.267$ L. 95% CI with $t_{0.025,\,7} = 2.365$ : $[4.84;\; 5.29]$ L.
Show step-by-step (with the why)
1. Calculate the mean: $\bar X = (5.2 + 4.8 + 5.5 + 4.9 + 5.1 + 5.3 + 4.7 + 5.0)/8 = 40.5/8 = 5.0625$ L.
2. Calculate the sample standard deviation (squared deviations): $(0.1375)^2 + (-0.2625)^2 + (0.4375)^2 + (-0.1625)^2 + (0.0375)^2 + (0.2375)^2 + (-0.3625)^2 + (-0.0625)^2$ .
3. Sum of squares: $\approx 0.0189 + 0.0689 + 0.1914 + 0.0264 + 0.0014 + 0.0564 + 0.1314 + 0.0039 = 0.4988$ .
4. $s^2 = 0.4988/7 = 0.0713$ ; $s \approx 0.267$ L.
5. $\mathrm{SE} = 0.267/\sqrt{8} = 0.0944$ . With $t_{0.025,\,7} = 2.365$ : $E = 2.365 \times 0.0944 = 0.223$ .
6. CI: $[4.84;\; 5.29]$ L. Trick: With $n = 8$ , the $t$ quantile is 2.365 — well above 1.96. Uncertainty in $\sigma$ is heavy when $n$ is small.
Ex. 102.16Application
Blood glucose: $\sigma = 25$ mg/dL (from previous studies). What $n$ guarantees a margin of error of 5 mg/dL at 95%?
Select the correct option
$n = 97$ ( $n\geq(1.960\times25/5)^2=96.04$ , round up) $n = 96$ (computed 96.04 but did not round up) $n = 68$ (used $z=1.645$ from 90% level instead of 95%) $n = 25$ (computed $(\sigma/E)^2=(25/5)^2=25$ , forgot $z^2$ factor)
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
$n \geq (1.960 \times 25/5)^2 = (9.80)^2 = 96.04$ . Rounds: $n = 97$ . With 97 samples, the margin of error becomes exactly 5 mg/dL at 95%.
Ex. 102.17Modeling
A union of metalworkers collected salaries of $n = 25$ employees: $\bar X = 1,800$ dollars, $s = 200$ dollars. The union claims that the true average salary is below USD 1,883. Does the 95% CI support that claim?
Select the correct option
95% CI: USD $[1717; 1883]$ ; since USD 1883 is the upper limit, the data are marginally consistent with the union's claim95% CI: USD $[1722; 1878]$ (used $z=1.960$ ); USD 1883 falls outside CI, refuting the claim95% CI: USD $[1640; 1960]$ (divided $s$ by $n=25$ instead of $\sqrt{25}$ ); USD 1883 is within95% CI: USD $[1756; 1844]$ (used $t_{0.025,24}=2.064$ but $\mathrm{SE}=200/25=8$ , forgot root)
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
95% CI: $1800 \pm 2.064 \times 200/\sqrt{25} = 1800 \pm 82.6$ . CI: USD $1,717 to USD$ 1,883. The minimum wage (USD 1,412 in 2024) is not in the CI, indicating the sector pays above minimum. The strike is not supported based on this CI if the demand is above USD 1,883; if below, the data are consistent with the demand.
Ex. 102.18ModelingAnswer key
Body temperature of $n = 30$ healthy adults: $\bar X = 36.8$ °C, $s = 0.5$ °C. 95% CI. Is the canonical value of 37°C consistent with this data?
Select the correct option
95% CI: $[36.61;\; 36.99]$ °C; since 37°C is outside CI, the data challenge the canonical value95% CI: $[36.62;\; 36.98]$ °C (used $z=1.960$ ); same conclusion, 37°C outside95% CI: $[36.61;\; 37.01]$ °C (used $t_{0.025,30}=2.042$ with $\nu=n$ ); 37°C is within, so compatible95% CI: $[35.82;\; 37.78]$ °C (divided $s$ by $n=30$ instead of $\sqrt{30}$ ); 37°C is within
Select an option first
Solve online OpenIntro Statistics · §4.3
Show solution
95% CI: $[36.61;\; 36.99]$ °C. Since 37°C is outside CI (marginally), the data challenge the canonical value. For more robust conclusion, a formal test $H_0: \mu = 37$ would be appropriate.
Show step-by-step (with the why)
1. 95% CI for body temperature: $\bar X = 36.8$ °C, $s = 0.5$ °C, $n = 30$ .
2. $t_{0.025,\,29} = 2.045$ . $\mathrm{SE} = 0.5/\sqrt{30} = 0.0913$ .
3. $E = 2.045 \times 0.0913 = 0.187$ . CI: $[36.61;\; 36.99]$ °C.
4. The canonical value of 37°C is outside the CI (marginally). This suggests the true value may be slightly lower than canonical.
5. Fun fact: Recent large-sample studies confirm that human body temperature has declined historically — likely due to improved health conditions reducing chronic inflammation.
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.
Select the correct option
62 quarters (more than 15 years); practically infeasible for short-term economic policy16 quarters (4 years); feasible since $n\geq(1.645\times2/0.5)^2\approx43$ — used z from 90%97 quarters; computed $(1.960\times2/0.5)^2$ correctly but multiplied by 1.5 by mistake10 quarters; computed $(\sigma/E)^2=(2/0.5)^2=16$ , forgot $z^2$ factor, and rounded down
Select an option first
Solve online Statistical Thinking for the 21st Century · Ch. 9
Show solution
For $\delta = 0.5$ percentage point and $\sigma \approx 2$ pp (typical GDP variation), $n \geq (1.960 \times 2/0.5)^2 = (7.84)^2 \approx 61.5$ . Would need 62 quarters — more than 15 years of data to detect a 0.5 pp difference in GDP growth. With real economic variability, statistical power of short-term studies is limited.
Ex. 102.20Application
A 90% CI has amplitude $A_{90}$ . How many times wider is the 99% CI for the same sample and the same $\sigma$ ?
Select the correct option
About 1.566 times wider ( $z_{99}/z_{90}=2.576/1.645\approx1.566$ )About 1.314 times wider ( $z_{99}/z_{95}=2.576/1.960\approx1.314$ , used 95% quantile in denominator)About 2 times wider (popular incorrect estimate; actual factor is 1.566)Exactly 1.131 times wider ( $\sqrt{2.576/1.645}$ , incorrectly applied square root)
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
Increasing the confidence level from 90% to 99%, the quantile grows from 1.645 to 2.576. The margin of error increases by ratio $2.576/1.645 \approx 1.566$ , or about 56.6% wider. The CI becomes significantly wider for the same sample.
Ex. 102.21Application
Weekly screen time of teenagers: $\sigma = 3$ h, $n = 49$ , $\bar X = 12$ h. 95% CI. Is the value 10 h/week plausible?
Select the correct option
95% CI: $[11.16;\; 12.84]$ h; 10 h/week is outside CI — not plausible95% CI: $[11.16;\; 12.84]$ h; 10 h/week is outside CI, therefore plausible as a rejected value95% CI: $[6.12;\; 17.88]$ h (forgot $\sqrt{n}$ : $E=1.960\times3=5.88$ ); 10 h/week is within95% CI: $[11.29;\; 12.71]$ h (used $z=1.645$ from 90% level); 10 h/week also outside
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
95% CI: $\bar X \pm 1.960 \times \sigma/\sqrt{n} = 12 \pm 1.960 \times 3/\sqrt{49} = 12 \pm 0.840$ . CI: $[11.16;\; 12.84]$ hours. The value 10 hours is outside the CI — the data are inconsistent with the hypothesis of average 10 hours.
Ex. 102.22Application
Cholesterol: $\sigma = 8$ mg/dL. What $n$ for 99% CI with margin of 2 mg/dL?
Select the correct option
$n = 107$ ( $n\geq(2.576\times8/2)^2=106.2$ , round up) $n = 62$ (used $z=1.960$ from 95% level: $(1.960\times8/2)^2=61.5$ ) $n = 106$ (computed 106.2 but did not round up) $n = 16$ (computed $(\sigma/E)^2=(8/2)^2=16$ , forgot $z^2$ factor)
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
$E = 2.576 \times \sigma/\sqrt{n}$ . We want $E \leq 2$ mg/dL. $n \geq (2.576 \times 8/2)^2 = (10.304)^2 = 106.2$ . Rounds: $n = 107$ .
Ex. 102.23Application
Battery duration: $n = 40$ , $\bar X = 520$ h, $s = 60$ h. 95% CI. Does the manufacturer's claim of 500 hours average duration support these data?
Select the correct option
95% CI: $[500.8;\; 539.2]$ h; 500 h is outside CI — the data marginally challenge the claim95% CI: $[501.4;\; 538.6]$ h (used $z=1.960$ ); 500 h also falls outside, same conclusion95% CI: $[500.8;\; 539.2]$ h; 500 h is outside CI, therefore the manufacturer's claim is proven95% CI: $[490.5;\; 549.5]$ h (divided $s$ by $n=40$ instead of $\sqrt{40}$ ); 500 h is within, data support manufacturer
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
95% CI: $[500.8;\; 539.2]$ hours. Since 500 is outside CI, the data marginally challenge the manufacturer's claim.
Show step-by-step (with the why)
1. Identify: $n = 40$ , $\bar X = 520$ hours, $s = 60$ hours.
2. With $n = 40$ , $t_{0.025,\,39} \approx 2.023$ .
3. $\mathrm{SE} = 60/\sqrt{40} = 9.49$ . $E = 2.023 \times 9.49 = 19.2$ hours.
4. CI: $[500.8;\; 539.2]$ hours.
5. The manufacturer claims 500 hours. Since 500 is outside the CI (marginally), the data are marginally inconsistent with the claim. A formal test would be appropriate.
6. Note: When $n \geq 30$ , the difference between $t$ and $z$ is small (2.023 vs. 1.960). For practical purposes, many use $z$ when $n \geq 30$ .
Ex. 102.24UnderstandingAnswer key
What happens to the 95% CI when you increase $n$ from 25 to 100, keeping everything else constant?
Select the correct option
Increases confidence level but narrows CIMaintains confidence level but narrows CIIncreases confidence level and widens CIHas no effect on CI if the population is normal
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
Increasing $n$ from 25 to 100 quadruples $n$ , doubling $\sqrt{n}$ . The margin of error $E \propto 1/\sqrt{n}$ falls in half. The confidence level (95%) does not change — it is chosen by the researcher, not by sample size.
Ex. 102.25Modeling
The INSS collected $n = 100$ retirement approval cases: $\bar X = 37$ days, $s = 15$ days. The legal goal is 45 days. Construct the 95% CI and interpret relative to the goal.
Select the correct option
95% CI: $[34.02;\; 39.98]$ days; completely below 45 days — goal compatible with data95% CI: $[34.06;\; 39.94]$ days (used $z=1.960$ ); same conclusion, goal compatible95% CI: $[34.02;\; 39.98]$ days; since 45 is outside CI, the goal is not being met95% CI: $[22.06;\; 51.94]$ days (divided $s=15$ by $n=100$ instead of $\sqrt{100}$ ); 45 is within
Select an option first
Solve online OpenIntro Statistics · §4.3
Show solution
INSS Report: 95% CI with $t_{0.025,\,99} \approx 1.984$ . $\mathrm{SE} = 15/\sqrt{100} = 1.5$ days. CI: $[37 - 2.976;\; 37 + 2.976] = [34.02;\; 39.98]$ days. Goal: 45 days. The CI is completely below 45 days — data are strongly consistent with goal achievement. Can report: "With 95% confidence, the true average time for benefits approval is between 34 and 40 days, below the goal of 45 days."
Ex. 102.26Challenge
For data on monthly income 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 appropriate to describe "typical" income? Why?
Select the correct option
For skewed income, the CI for the median (using order statistics: $X_{(6)}$ to $X_{(15)}$ for $n=20$ ) is more robust to outliers and more representative of typical incomeFor skewed income, the CI for the mean is more representative because the mean is always more statistically efficient than the medianBoth CIs are equivalent for $n=20$ ; the choice between mean and median is only aestheticThe CI for the median uses the 5% and 95% percentiles of ordered data, independent of $n$
Select an option first
Solve online OpenIntro Statistics · §4.3
Show solution
For the mean: 95% CI with $t_{0.025,\,n-1}$ . For the median: CI based on order statistics — the limits are $X_{(j)}$ and $X_{(k)}$ where $j$ and $k$ come from binomial tables. For $n = 20$ , the 95% CI for the median uses the 6th and 15th ordered values. For income (skewed distribution with long right tail), the median is more representative than the mean, and the CI for the median is more robust to income outliers.
Ex. 102.27Proof
Formally derive the $(1-\alpha)$ CI for $\mu$ with $\sigma$ known from the symmetry property of the standard normal distribution. Clearly identify what is random and what is fixed in $P(L \leq \mu \leq U) = 1 - \alpha$ .
Select the correct option
$L=\bar X - z_{\alpha/2}\sigma/\sqrt{n}$ and $U=\bar X + z_{\alpha/2}\sigma/\sqrt{n}$ are random (functions of $\bar X$ ); $\mu$ is fixed. Randomness is in the interval, not the parameter $\mu$ is random and $L, U$ are fixed; probability describes where $\mu$ may fall $L$ and $U$ are deterministic after collecting data; $P(L\leq\mu\leq U)=0$ or $1$ (not 0.95)Both $\mu$ and $\bar X$ are random; probability reflects uncertainty of both
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
Derivation: by symmetry of $\mathcal{N}(0,1)$ , $P(-z_{\alpha/2} \leq Z \leq z_{\alpha/2}) = 1 - \alpha$ . Substituting $Z = (\bar X - \mu)/(\sigma/\sqrt{n})$ and rearranging: $P(\bar X - z_{\alpha/2}\sigma/\sqrt{n} \leq \mu \leq \bar X + z_{\alpha/2}\sigma/\sqrt{n}) = 1-\alpha$ . The endpoints $L = \bar X - z_{\alpha/2}\sigma/\sqrt{n}$ and $U = \bar X + z_{\alpha/2}\sigma/\sqrt{n}$ are statistics (functions of data), while $\mu$ is the fixed parameter.
Ex. 102.28Proof
Prove 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?
Select the correct option
The three properties are: (1) $\bar X \sim \mathcal{N}(\mu,\sigma^2/n)$ ; (2) $(n-1)S^2/\sigma^2 \sim \chi^2_{n-1}$ ; (3) $\bar X$ and $S^2$ are independent. The ratio of standard normal by square root of chi-square/df is the definition of $t_{n-1}$ .The three properties are: normality of $X_i$ , $n>30$ , and $\sigma$ unknown. The $t$ distribution is only an approximation valid for large $n$ Only need to establish that $\bar X$ is normal (CLT) and that $S^2$ estimates $\sigma^2$ . Independence between $\bar X$ and $S^2$ is automaticProof is not possible — the $t$ distribution was discovered empirically by Student (1908) without formal proof
Select an option first
OpenIntro Statistics · §4.2
Show solution
See the referenced source for detailed resolution.
Ex. 102.29Challenge
Show the duality between CI and hypothesis test: "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 bilateral test for all values of $\mu_0$ .
Select the correct option
Rejecting $H_0: \mu=\mu_0$ at level $\alpha$ is equivalent to $\mu_0$ being outside the $(1-\alpha)$ CI; the CI contains exactly all non-rejectable $\mu_0$ values, serving as simultaneous bilateral testRejecting $H_0: \mu=\mu_0$ at level $\alpha$ is equivalent to $\mu_0$ being inside the $(1-\alpha)$ CI; values inside are suspiciousDuality holds only for one-tailed tests; bilateral tests use CI of $1-2\alpha$ Duality is only approximate — CI and hypothesis test use different distributions and are not mathematically equivalent
Select an option first
Solve online OpenIntro Statistics · §4.4
Show solution
By the duality theorem: rejecting $H_0: \mu = \mu_0$ at level $\alpha$ is equivalent to $\mu_0$ being outside the $(1-\alpha)$ CI. Therefore, all values of $\mu_0$ that would NOT be rejected by a bilateral test at level 5% form exactly the 95% CI. The CI can be used as a graphical tool for multiple testing: simultaneously test all hypotheses of the form $H_0: \mu = c$ for every $c$ .
Ex. 102.30Application
Among $n = 200$ ENEM students from a public school, 65 scored above 700 on the essay. Construct the 95% CI for the true proportion.
Select the correct option
95% CI: $[26.0\%;\; 39.0\%]$ ( $\hat p=0.325$ , $\mathrm{SE}=0.0331$ , $z=1.960$ )95% CI: $[27.9\%;\; 37.1\%]$ (computed $\hat p=0.325$ but used $\mathrm{SE}=\sqrt{0.325/200}$ without $(1-\hat p)$ factor)95% CI: $[25.0\%;\; 40.0\%]$ (used $z=2.576$ from 99% level instead of 95%)95% CI: $[28.1\%;\; 36.9\%]$ (used $z=1.645$ from 90% level)
Select an option first
Solve online OpenStax Statistics · §8.3
Show solution
For proportion, use Wilson's CI: $\hat p \pm z_{\alpha/2}\sqrt{\hat p(1-\hat p)/n}$ . With $\hat p = 65/200 = 0.325$ , $\mathrm{SE} = \sqrt{0.325 \times 0.675/200} = 0.0331$ . 95% CI: $0.325 \pm 1.960 \times 0.0331 = 0.325 \pm 0.065$ . CI: $[26.0\%;\; 39.0\%]$ .
Ex. 102.31Application
A survey wants to estimate the average number of weekly work hours with $\sigma = 7$ h, maximum margin of error of 2 h, and 90% confidence. What is the minimum $n$ ?
Select the correct option
$n = 34$ ( $n\geq(1.645\times7/2)^2=33.2$ , round up) $n = 47$ (used $z=1.960$ from 95% level instead of $z=1.645$ from 90%) $n = 33$ (computed 33.2 but did not round up) $n = 13$ (computed $(\sigma/E)^2=(7/2)^2=12.25$ , forgot $z^2$ factor)
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
With $\sigma = 7$ and 90%: $n \geq (z_{0.05} \times \sigma/E)^2 = (1.645 \times 7/2)^2 = (5.7575)^2 \approx 33.2$ . Rounds: $n = 34$ .
Ex. 102.32Application
A camp director collects data from 16 participants on letters received per week: $\bar X = 7.9$ , $s = 2.8$ . Construct the 95% CI.
Select the correct option
95% CI: $[6.41;\; 9.39]$ letters/week ( $t_{0.025,15}=2.131$ , $E=1.492$ )95% CI: $[6.52;\; 9.28]$ letters/week (used $z=1.960$ instead of $t_{15}$ )95% CI: $[5.15;\; 10.65]$ letters/week (divided $s=2.8$ by $n=16$ instead of $\sqrt{16}$ )95% CI: $[6.56;\; 9.24]$ letters/week (used $t_{0.025,16}=2.120$ , df $n$ instead of $n-1$ )
Select an option first
Solve online OpenStax Statistics · §8.2
Show solution
95% CI with $t_{0.025,\,15} = 2.131$ : $E = 2.131 \times 2.8/\sqrt{16} = 2.131 \times 0.7 = 1.492$ . CI: $[7.9 - 1.492;\; 7.9 + 1.492] = [6.41;\; 9.39]$ letters/week.
Ex. 102.33Application
An accounting firm surveys 100 people on weekly leisure hours. Result $\bar X = 23.6$ h with $\sigma = 7.0$ h (known). Construct the 95% CI.
Select the correct option
95% CI: $[22.23;\; 24.97]$ h ( $z=1.960$ , $E=1.372$ , $\sigma$ known)95% CI: $[22.23;\; 24.97]$ h using $t_{0.025,99}\approx1.984$ — result nearly equal, but $t$ is unnecessary when $\sigma$ is known95% CI: $[16.60;\; 30.60]$ h (forgot $\sqrt{n}$ : $E=1.960\times7.0=13.72$ )95% CI: $[22.47;\; 24.73]$ h (used $z=1.645$ from 90% level)
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
95% CI: $23.6 \pm 1.372$ . CI: $[22.23;\; 24.97]$ weekly hours.
Show step-by-step (with the why)
1. Identify: $n = 100$ , $\bar X = 23.6$ , $\sigma = 7.0$ , 95% level.
2. Since $\sigma$ is known, use $z_{0.025} = 1.960$ .
3. $E = 1.960 \times 7.0/\sqrt{100} = 1.960 \times 0.7 = 1.372$ .
4. CI: $[23.6 - 1.372;\; 23.6 + 1.372] = [22.23;\; 24.97]$ hours.
5. Trick: With $n = 100$ and $\sigma$ known, the $z$ pivot is appropriate. The CI has amplitude of 2.74 hours.
Ex. 102.34Application
A sample of 8 candy sachets has average weight $\bar X = 2$ ounces and $s = 0.12$ ounces. Construct the 90% CI for true mean weight. Is the stated value of 2 ounces plausible?
Select the correct option
90% CI: $[1.92;\; 2.08]$ ounces ( $t_{0.05,7}=1.895$ ); 2 ounces is within — plausible90% CI: $[1.93;\; 2.07]$ ounces (used $z=1.645$ from 90% level instead of $t_7$ ); 2 ounces also within90% CI: $[1.77;\; 2.23]$ ounces (divided $s=0.12$ by $n=8$ instead of $\sqrt{8}$ ); 2 ounces within but CI very wide90% CI: $[1.90;\; 2.10]$ ounces (used $t_{0.025,7}=2.365$ , bilateral 95% quantile instead of 90%)
Select an option first
Solve online OpenStax Statistics · §8.2
Show solution
90% CI with $t_{0.05,\,7} = 1.895$ : $\mathrm{SE} = 0.12/\sqrt{8} = 0.0424$ . $E = 1.895 \times 0.0424 = 0.0804$ . CI: $[2 - 0.0804;\; 2 + 0.0804] = [1.92;\; 2.08]$ ounces. The value 2.0 is in the CI — consistent with stated weight.
Ex. 102.35Understanding
What happens to the CI when you increase the sample size $n$ , keeping confidence level and $\sigma$ fixed?
Select the correct option
Increasing n increases confidence but widens CIIncreasing n maintains confidence and narrows CIIncreasing n has no effect on CI if sigma is largeIncreasing n decreases confidence and narrows CI
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
The confidence level is determined by the researcher — increasing $n$ does not change it. What changes is the margin of error: $E \propto 1/\sqrt{n}$ . With larger $n$ , the CI becomes narrower (more precise) while maintaining the same confidence level.
Ex. 102.36Application
Sleep of 20 campers: $\bar X = 7.9$ h, $s = 3$ h. 95% CI for true average sleep.
Select the correct option
95% CI: $[6.50;\; 9.30]$ h ( $t_{0.025,19}=2.093$ , $E=1.404$ )95% CI: $[6.59;\; 9.21]$ h (used $z=1.960$ instead of $t_{19}$ )95% CI: $[5.62;\; 10.18]$ h (divided $s=3$ by $n=20$ instead of $\sqrt{20}$ )95% CI: $[6.56;\; 9.24]$ h (used $t_{0.025,20}=2.086$ , df $n$ instead of $n-1$ )
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
95% CI with $t_{0.025,\,19} = 2.093$ : $\mathrm{SE} = 3/\sqrt{20} = 0.671$ . $E = 2.093 \times 0.671 = 1.404$ . CI: $[7.9 - 1.404;\; 7.9 + 1.404] = [6.50;\; 9.30]$ hours.
Ex. 102.37ApplicationAnswer key
The standard deviation of heights of young men is approximately $\sigma = 2.5$ inches (known). How many men are needed to estimate average height with a margin of error of 1 inch at 95%?
Select the correct option
$n = 25$ ( $n\geq(1.960\times2.5/1)^2=24.01$ , round up) $n = 24$ (computed 24.01 but did not round up) $n = 17$ (used $z=1.645$ from 90% level: $(1.645\times2.5)^2=16.9$ ) $n = 6$ (computed $(\sigma/E)^2=(2.5/1)^2=6.25$ , forgot $z^2$ factor)
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
$n \geq (z_{\alpha/2} \times \sigma/E)^2 = (1.960 \times 2.5/1)^2 = (4.9)^2 = 24.01$ . Rounds: $n = 25$ young men.
Ex. 102.38Application
A study of 21 students recorded average sleep of $\bar X = 7.24$ h with $s = 1.93$ h. Construct the 90% CI and check if 7 hours of sleep is a plausible value.
Select the correct option
90% CI: $[6.51;\; 7.97]$ h ( $t_{0.05,20}=1.725$ , $E=0.727$ ); 7 h is within — plausible90% CI: $[6.36;\; 8.12]$ h (used $t_{0.025,20}=2.086$ , bilateral 95% quantile instead of 90%)90% CI: $[6.55;\; 7.93]$ h (used $z=1.645$ instead of $t_{20}$ )90% CI: $[7.08;\; 7.40]$ h (divided $s=1.93$ by $n=21$ instead of $\sqrt{21}$ )
Select an option first
Solve online OpenStax Statistics · §8.2
Show solution
90% CI with $t_{0.05,\,20} = 1.725$ : $\mathrm{SE} = 1.93/\sqrt{21} = 0.421$ . $E = 1.725 \times 0.421 = 0.727$ . CI: $[7.24 - 0.727;\; 7.24 + 0.727] = [6.51;\; 7.97]$ hours. The value 7 hours is within the CI, so the data are consistent with the hypothesis of average sleep of 7 hours.
Ex. 102.39Modeling
Records from 35 flights show average of 11.6 empty seats per flight, with $s = 4.1$ . Construct the 95% CI and interpret from an overbooking policy perspective.
Select the correct option
95% CI: $[10.19;\; 13.01]$ seats ( $t_{0.025,34}=2.032$ , $E=1.41$ ); overbooking should cover at least 13 seats95% CI: $[10.24;\; 12.96]$ seats (used $z=1.960$ ); similar conclusion95% CI: $[10.19;\; 13.01]$ seats; to minimize complaints just cover lower limit of 10 seats95% CI: $[8.47;\; 14.73]$ seats (divided $s=4.1$ by $n=35$ instead of $\sqrt{35}$ )
Select an option first
Solve online OpenStax Statistics · §8.2
Show solution
95% CI: $[10.19;\; 13.01]$ empty seats per flight. With average of 11.6 empty seats, airlines should reduce overbooking to cover at least 13 seats.
Show step-by-step (with the why)
1. Identify: $n = 35$ , $\bar X = 11.6$ seats, $s = 4.1$ seats.
2. With $n = 35$ , use $t_{0.025,\,34} \approx 2.032$ .
3. $\mathrm{SE} = 4.1/\sqrt{35} = 0.693$ .
4. $E = 2.032 \times 0.693 = 1.408$ .
5. CI: $[11.6 - 1.41;\; 11.6 + 1.41] = [10.19;\; 13.01]$ empty seats.
6. Interpretation: In 95% of samples of this size, the CI constructed this way would capture the true average number of empty seats.
Ex. 102.40ModelingAnswer key
A survey of 400 drivers revealed that 320 always wear seatbelts. Construct the 95% CI for the true proportion of drivers who always wear seatbelts.
Select the correct option
95% CI: $[76.1\%;\; 83.9\%]$ ( $\hat p=0.80$ , $\mathrm{SE}=0.020$ , $z=1.960$ )95% CI: $[74.9\%;\; 85.1\%]$ (used $z=2.576$ from 99% level)95% CI: $[77.0\%;\; 83.0\%]$ (computed $\mathrm{SE}=\sqrt{0.80/400}$ without $(1-\hat p)$ factor)95% CI: $[75.7\%;\; 84.3\%]$ (used $z=1.645$ from 90% level)
Select an option first
Solve online OpenStax Statistics · §8.3
Show solution
95% CI for proportion with $\hat p = 320/400 = 0.80$ : $\mathrm{SE} = \sqrt{0.80 \times 0.20/400} = 0.020$ . $E = 1.960 \times 0.020 = 0.0392$ . CI: $[0.761;\; 0.839]$ . Interpretation: we are 95% confident that between 76.1% and 83.9% of drivers always wear seatbelts.
Ex. 102.41Modeling
Heights of men: group A ( $n_1 = 100$ , $\bar X_1 = 71$ inches, $\sigma_1 = 3$ ) and group B ( $n_2 = 100$ , $\bar X_2 = 68$ inches, $\sigma_2 = 3$ ). Construct the 95% CI for the difference of means $\mu_1 - \mu_2$ .
Select the correct option
95% CI: $[2.17;\; 3.83]$ inches ( $\mathrm{SE}_{\text{diff}}=0.424$ , CI does not include 0 — difference significant)95% CI: $[1.17;\; 4.83]$ inches (used $\mathrm{SE}=3/\sqrt{100}=0.3$ from only one group, not the difference)95% CI: $[2.41;\; 3.59]$ inches (used $\mathrm{SE}_{\text{diff}}=\sqrt{3^2/100}=0.3$ without summing two variances)95% CI: $[1.71;\; 4.29]$ inches (used $\mathrm{SE}_{\text{diff}}=\sqrt{3^2+3^2}/\sqrt{100}$ , summed deviations instead of variances)
Select an option first
Solve online OpenStax Statistics · §8.1
Show solution
Standard error of difference of means: $\mathrm{SE}_{\text{diff}} = \sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2} = \sqrt{3^2/100 + 3^2/100} = \sqrt{0.09 + 0.09} = \sqrt{0.18} \approx 0.424$ . 95% CI: $(\bar X_1 - \bar X_2) \pm 1.960 \times 0.424 = (71 - 68) \pm 0.831 = [2.17;\; 3.83]$ cm. The CI does not include 0, indicating the difference is statistically significant.
Ex. 102.42Challenge
Describe the bootstrap percentile method for constructing a 95% CI for the mean. What are the advantages over the Student's t-based CI? In which situations would bootstrap be especially recommended?
Select the correct option
Bootstrap percentile: re-sample $B=2000$ times, calculate $\bar X^*$ in each replicate, use 2.5% and 97.5% percentiles; recommended when population is not normal or $\sigma$ unknown and $n$ smallBootstrap percentile: calculate mean and standard deviation of $B$ samples, then use normal formula $\bar X \pm 1.96\cdot s^*/\sqrt{B}$ ; equivalent to z CI with many dataBootstrap percentile is equivalent to t CI for any distribution; main advantage is avoiding quantile tablesBootstrap percentile uses 5% and 95% percentiles of bootstrap means for 95% CI, with no tail adjustment
Select an option first
Solve online OpenIntro Statistics · §4.2
Show solution
The bootstrap percentile: (1) re-samples 2000 times with replacement from $\{X_1,\ldots,X_n\}$ ; (2) calculates $\bar X^*$ in each replicate; (3) uses 2.5% and 97.5% percentiles of bootstrap means as limits of 95% CI. Advantages: requires no normality assumption nor knowledge of $\sigma$ ; valid for any statistic (median, ratio, correlation). Limitation: requires computation; for very small samples ( $n < 15$ ), may have inadequate coverage.

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 with 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 statistic and the CI for mean

Case 1: σ\sigmaσ known (z pivot)

Case 2: σ\sigmaσ unknown (t pivot)

Reference quantiles

Margin of error and minimum sample size

Worked examples

Sources

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

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