INF-3: Confidence Intervals for a Population Mean (Small Sample)
Module 3 · Statistical Inference
Section 1: Introduction
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You’ve just mastered INF-2, where you learned to build confidence intervals using the z-distribution — as long as n ≥ 30 or σ was known. But what happens when you’re a medical researcher running a pilot study with only 10 patients? Or a food scientist measuring a new ingredient in just 8 batches? The sample is too small for the CLT guarantee, and σ is unknown. Can you still build a reliable confidence interval?
In 1908, a statistician named William Sealy Gosset — writing under the pseudonym “Student” to protect his employer’s trade secrets — solved exactly this problem. He derived a new distribution that accounts for the extra uncertainty introduced by estimating σ from a small sample. We call it the t-distribution, and it is the tool that makes small-sample inference possible.
This lesson builds directly on INF-2. The CI formula keeps the same shape; we replace z* with t* and σ with s. The key new skill is knowing when to use t instead of z — and understanding why.
After this lesson, you will be able to:
Explain why the z-distribution is inadequate when σ is unknown and n is small
Construct a confidence interval for μ using the t-distribution:
Determine the correct degrees of freedom and look up t* from a t-table
Apply the t vs. z decision rule: which distribution to use and when
Interpret a t-interval correctly using the same frequentist logic from INF-2
Section 2: Prerequisites
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Small-sample inference is a refinement of the large-sample methods you mastered in INF-2.
From INF-2: Confidence Interval Structure. Point estimate () ± Margin of Error ().
From INF-2: Standard Error (SE). when is unknown.
From INF-1: Normality Condition. For small samples (), the population must be approximately normal for our probability models to work.
Z-Values vs. T-Values: Both measure distance from the mean in standard errors, but t-values are larger to account for the extra uncertainty of small samples.
Retrieval Checkpoint
In INF-2, we used for a 95% confidence interval. If we are forced to use the t-distribution because our sample size is small (e.g., ), do you expect the new critical value to be larger or smaller than 1.96?
Success Factor:
The most common error in this lesson is using instead of when looking up values in the t-table. If your sample size is 15, your degrees of freedom are 14. Always subtract one.
Retrieval Warm-up — from earlier lessons
A random sample of apples from an orchard has a mean weight of g. The population standard deviation is known: g. What is the standard error of the sample mean?
From INF-2, a 95% confidence interval for a population mean uses the formula . A 99% CI from the same sample would be:
Section 3: Core Concepts
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C1 — Why the -Distribution Falls Short
In INF-2, when was unknown and , we substituted for in the standard error formula. This worked because for large samples, reliably approximates — the approximation error is small. But for small samples (), can vary substantially from . Using in a -formula pretends we know exactly, which we don’t — and the -distribution has no way to account for that extra source of uncertainty.
The result: -intervals computed with small samples are too narrow. They appear more precise than they really are. The actual coverage rate — the fraction of such intervals that truly capture — falls below the stated confidence level. A “95% CI” might only capture less than 95% of the time — the actual shortfall depends on and can be substantial for very small samples.
This is not a minor correction. For , the difference between and the correct (, 95%) is large enough to make the -interval meaningfully misleading. The -distribution was invented precisely to fix this.
C2 — The -Distribution: Shape and Properties
The -distribution (also called Student’s ) is a family of bell-shaped, symmetric distributions — one for each value of its parameter, the degrees of freedom (). Like the standard normal, it is centered at 0. Unlike the standard normal, it has heavier tails: more probability mass in the extremes.
The t-Distribution
The t-distribution with degrees of freedom is a symmetric, bell-shaped distribution centered at 0. Compared to the standard normal :
It has heavier tails — more probability in the extremes
Its peak is slightly lower than the normal
As , the t-distribution converges to
In practice: use the t-distribution whenever is unknown and the population is approximately normal.
The heavier tails reflect our extra uncertainty. When is small and we’re estimating with , extreme outcomes are more likely than the normal distribution would predict. The -distribution builds that in automatically.
Figure 1: t-distribution (blue) vs. standard normal (orange). Adjust the df slider to watch the t-distribution converge toward the normal. Toggle critical values to see how t* is always larger than z* for the same confidence level — and by how much.
Before reading further, use the slider to explore — then check your observations against the questions below.
Set . Record the 95% critical value shown above the curves.
Increase to , then , then .
At which does the gap between and first drop below 0.10?
You will revisit this question in Challenge 2 — record your answer before continuing. As grows, the heavier tails of the -distribution lighten toward the normal, which is why the -approximation becomes acceptable for large samples.
C3 — Degrees of Freedom: Why ?
The -distribution is indexed by degrees of freedom: . But why and not ?
Here’s the intuition. To compute (the sample standard deviation), we first compute the sample mean and then measure how far each observation deviates from it. But once we know and of the data values, the last value is completely determined — it’s not free to vary. So there are only truly independent pieces of information about the spread. We’ve “used up” one degree of freedom by estimating with .
Degrees of Freedom
For a one-sample t-interval, the degrees of freedom are:
where n is the sample size. This is the row you look up in the t-table.
The single most common arithmetic error in this lesson is using instead of . For : , not 12. Always subtract 1 before looking up . Getting this wrong gives you the wrong critical value and therefore the wrong CI.
Quick Check — Degrees of Freedom
A researcher uses observations to construct a t-interval. How many degrees of freedom does she have?
C4–C5 — Reading the -Table
The critical value depends on two things: the degrees of freedom () and the confidence level. The table below gives critical values for a wide range of and confidence levels. As increases, decreases toward the value — you can see the convergence in the bottom rows (the row matches the standard normal ).
Student's t-Distribution Table
Critical values (t*) for given degrees of freedom (df) and tail area.
df
Confidence
80%
90%
95%
98%
99%
99.9%
0.10 (1)0.20 (2)
0.05 (1)0.10 (2)
0.025 (1)0.05 (2)
0.01 (1)0.02 (2)
0.005 (1)0.01 (2)
0.0005 (1)0.001 (2)
1
3.078
6.314
12.706
31.821
63.657
636.619
2
1.886
2.920
4.303
6.965
9.925
31.599
3
1.638
2.353
3.182
4.541
5.841
12.924
4
1.533
2.132
2.776
3.747
4.604
8.610
5
1.476
2.015
2.571
3.365
4.032
6.869
6
1.440
1.943
2.447
3.143
3.707
5.959
7
1.415
1.895
2.365
2.998
3.499
5.408
8
1.397
1.860
2.306
2.896
3.355
5.041
9
1.383
1.833
2.262
2.821
3.250
4.781
10
1.372
1.812
2.228
2.764
3.169
4.587
11
1.363
1.796
2.201
2.718
3.106
4.437
12
1.356
1.782
2.179
2.681
3.055
4.318
13
1.350
1.771
2.160
2.650
3.012
4.221
14
1.345
1.761
2.145
2.624
2.977
4.140
15
1.341
1.753
2.131
2.602
2.947
4.073
16
1.337
1.746
2.120
2.583
2.921
4.015
17
1.333
1.740
2.110
2.567
2.898
3.965
18
1.330
1.734
2.101
2.552
2.878
3.922
19
1.328
1.729
2.093
2.539
2.861
3.883
20
1.325
1.725
2.086
2.528
2.845
3.850
21
1.323
1.721
2.080
2.518
2.831
3.819
22
1.321
1.717
2.074
2.508
2.819
3.792
23
1.319
1.714
2.069
2.500
2.807
3.768
24
1.318
1.711
2.064
2.492
2.797
3.745
25
1.316
1.708
2.060
2.485
2.787
3.725
26
1.315
1.706
2.056
2.479
2.779
3.707
27
1.314
1.703
2.052
2.473
2.771
3.690
28
1.313
1.701
2.048
2.467
2.763
3.674
29
1.311
1.699
2.045
2.462
2.756
3.659
30
1.310
1.697
2.042
2.457
2.750
3.646
40
1.303
1.684
2.021
2.423
2.704
3.551
50
1.299
1.676
2.009
2.403
2.678
3.496
60
1.296
1.671
2.000
2.390
2.660
3.460
80
1.292
1.664
1.990
2.374
2.639
3.416
100
1.290
1.660
1.984
2.364
2.626
3.390
∞
1.282
1.646
1.962
2.330
2.581
3.300
The bottom row () shows the normal critical values — notice that approaches from above as increases. Hover over any cell to highlight its row and column, making it easy to read the correct value.
Quick Check — Reading the t-Table
A sample of observations has unknown. You are constructing a 95% CI. What is the correct critical value ?
C6 — Confidence Interval Formula
The -interval has the same structure as the -interval from INF-2. The only changes are replacing with and with :
Confidence Interval for μ (Small Sample, σ Unknown)
which gives the interval:
where is the critical value from the -table with and the desired confidence level.
C7 — When Can You Use the -Interval?
Three conditions must hold:
Random sample — observations must be independent (stated or assumed)
unknown — if were known, use the -interval regardless of
Population approximately normal — required because the CLT guarantee doesn’t apply for small . Check with a histogram or Q-Q plot; for the assumption matters most; for mild departures are acceptable.
What does “approximately normal” look like in practice? When you inspect a histogram or dot plot, look for:
Single peak — the data should mound in one place, not two or three
Rough symmetry — neither tail should be dramatically longer than the other
No extreme outliers — one or two points far beyond the rest can distort the t-interval, especially when is very small
You do not need a perfect bell curve. For , mild skew is generally tolerable. For , even moderate skew is enough to call the normality assumption into question and should be reported explicitly.
Quick Check — Checking the Conditions
A researcher has measurements of reaction time (ms). is unknown. A histogram of the sample looks roughly bell-shaped. Which statement best describes whether a t-interval is valid?
C8 — When to Use vs. : The Practical Decision Rule
Is known? → Use
unknown? → Use — always. The t-distribution is the correct procedure whenever must be estimated from the data.
unknown + ? → Use (the -approximation is also acceptable here, since for large , but is the principled choice)
unknown + + population clearly non-normal? → Neither method is valid without more information
The rule is simple: if is unknown, use . For large samples (), the -approximation is acceptable — and converge at high degrees of freedom, so the practical difference is small. But is the approximation; is the correct procedure. The distinction matters most when , where the gap between and is large enough to meaningfully affect the interval.
C9 — Interpreting the -Interval
The frequentist interpretation from INF-2 applies unchanged:
We are confident that the true population mean lies between and . This means: if we repeated this procedure many times, about of intervals constructed this way would contain the true .
What it does NOT mean: “There is a probability that lies in this interval.” The mean is a fixed constant — it either is or isn’t in your interval.
C10 — Why the -Interval is Wider (and That’s Correct)
For the same data, confidence level, and sample size, the -interval is always wider than the -interval, because for any finite . For example, at 95% confidence with : vs. .
Students sometimes worry that the -interval “must be wrong” because it’s wider than they expected. This is backwards. The -interval is wider because it honestly reflects our uncertainty about . The -interval applied to small samples would be artificially narrow — falsely precise. Wider = more honest when is estimated, not worse.
Section 4: Worked Examples
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Example 1 — Computing a t-Interval (Fully Worked)
A physician tests a new pain medication on randomly selected patients. The reduction in pain score (0–10 scale) after treatment yields points and points. The physician does not know the population SD . Construct a 95% confidence interval for the true mean reduction .
Step 1 — Check Conditions:
Random sample: ✓ (stated)
σ unknown: ✓ → use t-distribution
Population approximately normal: assumed ✓ (pain score reductions are typically symmetric)
→ CLT does not guarantee normality; the normality assumption of the population is required
Step 2 — Find df and t:*
From the t-table, for df = 11 and 95% confidence:
Step 3 — Compute SE and Margin of Error:
Step 4 — Build the Interval:
Interpretation: We are 95% confident that the true mean reduction in pain score is between 2.91 and 4.69 points. Notice that this interval does not include 0 — suggesting the medication has a real effect, though this is not a formal hypothesis test.
Note on z vs. t: If we had incorrectly used z* = 1.96, we would get E = 1.96 × 0.404 = 0.792, CI = (3.008, 4.592) — a narrower, but falsely precise interval.
Example 2 — Constructing a 90% t-Interval (Partially Scaffolded)
An environmental scientist collects 8 soil samples near an industrial site and measures lead concentration. The results yield x̄ = 22.5 ppm and s = 4.2 ppm. Construct a 90% CI for the true mean lead concentration. (σ unknown.)
Before seeing the solution: what is df here? Which row of the t-table will you look up? Take a moment to answer before continuing.
Step 1 — Conditions: n = 8 < 30, σ unknown → use t. Assume approximately normal population.
Step 2 — df and t:*
. From t-table: (90%, df = 7).
Step 3 — SE and Margin of Error:
Step 4 — Interval:
Interpretation: We are 90% confident the true mean lead concentration at this site is between 19.69 and 25.31 ppm.
Example 3 — 99% Confidence Interval (Minimally Scaffolded)
A nutritionist measures daily caloric intake for randomly selected college students. She finds calories and calories. is unknown. Construct a 99% CI.
Hint: df = 19. Look up the 99% critical value in the t-table.
Show Solution
. From t-table: (99%, df = 19).
Interpretation: We are 99% confident the true mean daily caloric intake is between 1952 and 2348 calories.
Example 4 — Choosing the Right Distribution (Application Twist)
A quality control engineer measures shaft diameters from n = 25 machined parts: x̄ = 50.3 mm and s = 0.8 mm. σ is unknown. A colleague suggests using z* = 1.96 because “n = 25 is close enough to 30.” Evaluate this claim and construct the correct 95% CI.
The colleague’s reasoning is incorrect. The rule for using z is that σ is known — not that n is “large enough.” Here σ is unknown, so we must use t.
. From t-table: (95%, df = 24).
Correct t-interval:
Incorrect z-interval (for comparison):
The difference is small here (n = 25, close to 30), but the principle is absolute: σ unknown → use t. The error grows as n decreases. For n = 6, df = 5 and t* = 2.571 — a 31% wider margin of error than z* = 1.96.
Section 5: Guided Practice
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Work through each problem step by step. The dropdowns give immediate feedback — wrong answers explain what went wrong. The generator problems (GP4) create fresh problems each time you click.
Problem 1 — Degrees of Freedom and Critical Values (C1, C2)
Each variant gives you , , , and a confidence level. Work through: (a) degrees of freedom, (b) the correct t*, (c) the resulting interval.
Sleep duration study. A sleep researcher samples n = 12 college students. Mean nightly sleep: hrs, hrs. unknown. Construct a 95% CI.
Step (a): What are the degrees of freedom?
Step (b): What is for these degrees of freedom and confidence level?
Step (c): What is the 95% CI? (SE = 0.9/√12 ≈ 0.260, E = 2.201 × 0.260 ≈ 0.572)
Plant growth experiment. A botanist grows n = 8 seedlings under a new lighting protocol. Mean height after 30 days: cm, cm. unknown. Construct a 90% CI.
Problem 2 — Distribution Selection and CI Interpretation (C3, C4)
For each scenario: (a) select whether to use t or z, then (b) select the ONE correct interpretation of the resulting confidence interval.
Scenario: A nutritionist records sodium content for n = 20 randomly selected prepared meals. She does not know σ. She finds mg and mg. Using the correct method, she computes a 95% CI of (1,742, 1,938) mg.
Part (a): Which distribution should be used?
Part (b): Which is the correct interpretation?
Scenario: A factory uses a machine whose fill weight variability has been characterized over thousands of runs: g (known with high precision). Quality control draws n = 15 containers and finds g. A 90% CI is computed as (249.42, 251.38) g.
Part (a): Which distribution should be used?
Part (b): Correct interpretation?
Scenario: A professor surveys n = 45 students about weekly study hours. σ is unknown. She finds hrs and hrs. She computes a 95% CI using the z-distribution: (16.43, 20.17) hrs.
Part (a): Which distribution should be used?
Part (b): Correct interpretation?
Scenario: A pharmacist weighs n = 8 tablets from a new batch. mg, mg. σ unknown. A 90% CI is (494.89, 511.51) mg.
Part (a): Which distribution should be used?
Part (b): Correct interpretation?
Scenario: A historian measures the width of n = 12 Roman coins from a single mint. mm, mm. σ unknown. A 99% CI is (13.74, 17.86) mm.
Part (a): Which distribution should be used?
Part (b): Correct interpretation?
Problem 3 — Comparative Analysis of z vs. t Intervals (C3)
Setup: A researcher measures anxiety scores for n = 16 participants before a public speaking event. Results: , , σ unknown. She considers two approaches for a 95% CI.
z-based interval (treating s as if it were σ, which is incorrect here): SE = 5.8/ = 1.450. Using z* = 1.96: E = 1.96 × 1.450 = 2.842. CIz = (39.758, 45.442).
Problem 4 — Parameterized Practice: Constructing the t-Interval (C3)
A new problem is generated each time. The t* value is provided — focus on computing the correct margin of error and interval bounds.
Section 6: Independent Practice
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Work through each problem independently. Problems are interleaved across the lesson’s concepts. Show your work in the scratch space, then check the solution.
Problem 1 — Full t-Interval (Concept Check)
Construct the t-interval for each scenario. Check conditions first.
Emergency room wait times. A random sample of n = 18 visits gives min, min. σ unknown. Construct a 95% CI.
Interpretation: We are 95% confident the true mean wait time is between 41.1 and 53.5 minutes.
Calcium content in groundwater samples. n = 11, mg/L, mg/L. σ unknown. Construct a 90% CI.
Show Solution
df = 10,
SE = 9.3/ ≈ 2.804, E = 1.812 × 2.804 ≈ 5.081
CI:(79.52, 89.68) mg/L
Nightly sleep hours. n = 20, hrs, hrs. σ unknown. Construct a 99% CI.
Show Solution
df = 19,
SE = 1.1/ ≈ 0.246, E = 2.861 × 0.246 ≈ 0.704
CI:(6.10, 7.50) hours
Weight loss after 8-week program. n = 25, kg, kg. σ unknown. Construct a 95% CI.
Show Solution
df = 24,
SE = 1.6/ = 0.320, E = 2.064 × 0.320 = 0.660
CI:(3.04, 4.36) kg
Standardized test scores. n = 14, , . σ unknown. Construct a 90% CI.
Show Solution
df = 13,
SE = 8.9/ ≈ 2.379, E = 1.771 × 2.379 ≈ 4.213
CI:(64.19, 72.61)
Problem 2 — Generated: End-to-End t-Interval
A new problem is generated each time. All the values you need are provided — construct the full confidence interval.
Problem 3 — Generated: Choose the Distribution, Then Compute
Each generated problem requires you to first decide whether to use t or z, then construct the CI.
Problem 4 — Find the Error in Confidence Interval Computations
Each scenario shows a completed analysis with one error. Identify it.
Scenario: A psychologist constructs a CI for mean reaction time with n = 12 participants ( ms, ms). She reports: “Using z* = 1.96 and SE = 48/√12 = 13.86, I get E = 27.15, so CI = (282.85, 337.15) ms.”
What is the researcher’s error?
Show Corrected Solution
Correct CI uses df = 11, , E = 2.201 × 13.86 ≈ 30.50.
Corrected CI = (279.50, 340.50) ms — about 6.5 ms wider per side than the incorrect z-based result.
Scenario: A biologist constructs a 95% CI for mean leaf length from n = 18 plants ( cm, cm). She reports: “df = 18, t* = 2.101 from the t-table, SE = 1.4/√18 ≈ 0.330, E = 2.101 × 0.330 = 0.693, CI = (7.907, 9.293) cm.”
What is the researcher’s error?
Show Corrected Solution
df = n − 1 = 17 (not 18). From t-table: = 2.110 (95%, df = 17).
E = 2.110 × 0.330 = 0.696. Corrected CI = (7.904, 9.296) cm.
Scenario: A researcher builds a CI from 22 observations and reports: “I am 95% confident that μ is in (14.2, 18.8). This means P(μ ∈ (14.2, 18.8)) = 0.95 — there is a 95% probability the population mean is in this specific interval.”
What is the researcher’s error?
Show Correction
The correct statement: “We are 95% confident the true population mean is between 14.2 and 18.8.” The 95% describes the long-run capture rate of the method, not a probability about this specific interval. μ is a fixed (though unknown) constant.
Scenario: A data analyst computes both a t-interval (39.5, 45.7) and a z-interval (39.8, 45.4) for the same dataset. He concludes: “The z-interval is better because it’s narrower and more precise. The t-interval must contain an error since it’s wider than the z-interval.”
What is the analyst’s error?
Show Correction
When σ is unknown, the t-interval’s wider margin is correct — it honestly accounts for the extra uncertainty from estimating σ with s. A narrower z-interval is not more precise; it is falsely narrow and will undercover the true μ more often than advertised.
Scenario: A manager collects n = 60 employee satisfaction scores (, , σ unknown). She uses a t-interval: df = 59, t* = 2.001, CI = (6.74, 7.66). Her supervisor says: “With n = 60, you should have used z* = 1.96 — your t* = 2.001 is wrong.”
Who made the error?
Show Explanation
The manager is correct. The t-distribution is always valid when σ is unknown, regardless of n. For large n (≥ 30), z is an acceptable approximation because t* ≈ z* at high df. But using t is not wrong — it is technically more principled. The supervisor’s claim is backwards.
Problem 5 — Multi-Step Synthesis: Lead Contamination Study
An environmental scientist collects water samples from a monitoring station. She measures lead concentration: ppb, ppb. is unknown.
The t-interval is wider: 9.775 − 7.025 = 2.750 vs. 9.695 − 7.105 = 2.590.
For n = 22, the difference is modest. But the t-interval is the correct choice when σ is unknown.
Part (c): Her colleague argues: “Both intervals are essentially the same — why bother with t?” Write a short response (3–5 sentences) explaining why the distinction matters, especially for smaller samples.
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Your response is saved locally and not submitted anywhere.
Part (d): Suppose the team had collected water samples instead of (four times as many). Which two quantities in the CI formula would change, and in which direction?
Mixed Review — Retrieval from Earlier Lessons
These problems draw on concepts from earlier in the course. Attempting them without re-reading prior lessons is the point — retrieval practice strengthens long-term memory more than re-reading.
Review Problem 1 — CLT and Standard Error (INF-1)
A bottling machine fills bottles with mL and mL. A quality-control engineer randomly selects bottles per hour.
(a) What is the standard error of the hourly mean ? (b) The CLT requires for non-normal populations. Does it apply here if the fill distribution is skewed? (c) If the population is approximately normal, what is ? (Use .)
Show Solution
(a) mL.
(b) No — with , the CLT does not guarantee approximate normality of unless the population itself is approximately normal. The engineer must check whether fill amounts are roughly bell-shaped.
(c) Assuming the population is approximately normal, .
About 6.7% of hourly samples would average below 497 mL — noticeable and worth investigating.
Review Problem 2 — Large-Sample z-CI Construction and Interpretation (INF-2)
A dietitian surveys adults and finds mean daily fibre intake g with g. The population SD is unknown.
(a) Construct a 95% CI for the true mean fibre intake using the large-sample approximation. (b) A nutritionist says: “There is a 95% probability that the true mean is in this interval.” Identify the error and restate correctly.
Show Solution
(a) Since , substitute for :
(b) The error: treating as random. The true mean fibre intake is a fixed (unknown) constant — it does not have a 95% probability of being anywhere. The correct statement: “We are 95% confident that the true mean daily fibre intake is between 16.7 and 20.1 g — meaning this procedure captures in about 95% of all samples constructed this way.”
Section 7: Mastery Check
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No hints or scaffolds here. These questions measure whether you’ve genuinely understood the concepts — not just recognized the right formula.
Question 1a — Feynman: Why t Instead of z?
Explain this to a classmate who understands INF-2 but hasn’t seen this lesson:
Why do we use the t-distribution instead of z when σ is unknown and the sample is small? What goes wrong if we use z anyway?
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Show Model Answer
When σ is unknown, we substitute s — but s varies from sample to sample, introducing extra uncertainty that the z-distribution ignores. For small samples this matters: a z-interval pretends we know σ exactly, which we don’t. The result is an interval that is too narrow — it appears more precise than it really is. The actual coverage rate falls below the stated confidence level. The t-distribution fixes this by having heavier tails, producing wider intervals that honestly reflect the uncertainty from estimating σ with s.
Question 1b — Feynman: Degrees of Freedom
Same classmate. Now explain:
What does “degrees of freedom” mean in the context of the t-distribution? Why is df = n − 1 and not n?
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Show Model Answer
Degrees of freedom measures how many independent pieces of information contributed to estimating the spread. To compute s, we first need — and once we know and n − 1 of the data values, the last value is completely determined by the constraint that all values must average to . It has no freedom to vary. So only n − 1 deviations are truly independent. Using df = n in the t-table would underestimate uncertainty; df = n − 1 gives the correct, wider critical value.
Question 2 — Iron Content CI
A food scientist tests n = 9 batches of a new energy bar for iron content. Results: mg, mg. She does not know . Construct a 99% CI.
Interpretation: We are 99% confident the true mean iron content is between 2.53 and 3.87 mg.
Question 3 — Error Analysis
A student constructs a 90% CI from n = 15 measurements (, ). Here is her work:
df = 15 (← is this correct?)
t*(df = 15, 90%) = 1.753
SE = 7.2/ = 1.859
E = 1.753 × 1.859 = 3.258
CI = (51.742, 58.258)
What is the error?
Show Corrected Solution
df = n − 1 = 14 (not 15). From t-table: (90%, df = 14)
E = 1.761 × 1.859 = 3.272
Corrected CI = (51.728, 58.272)
The error changes t* from 1.753 (df = 15) to 1.761 (df = 14) — a small but systematic error. With smaller samples, off-by-one df errors matter more.
Question 4 — Distribution Selection
A marine biologist collects seawater samples and measures salinity. She records ppt and ppt. is unknown and the population is approximately normal. Which distribution should she use for a 95% CI?
Question 5 — Computing the CI
Using the biologist’s data from Question 4 (, ppt, ppt, 95% CI), the critical value is (). Compute the margin of error and state the 95% CI.
Show Solution
ppt
ppt
95% CI: (33.726, 35.874) ppt
Interpretation: We are 95% confident the true mean salinity is between 33.7 and 35.9 ppt.
Question 6 — Interpretation
A toxicologist constructs a 90% CI for mean arsenic concentration from water samples: μg/L. Which interpretation is correct?
Self-Assessment
How confident are you with the concepts from this lesson?
Still confusedReady for the Boss Fight
Section 8: Boss Fight
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Choose your path. Both require the same core skills — the difference is the angle of attack.
🔬 The Analyst
You have real data from an environmental study. Construct and interpret a confidence interval, then answer a critical question about the results.
🏗️ The Architect
A colleague’s study report contains a statistical error. Identify the mistake, correct the analysis, and explain why it matters.
🔬 Path A: The Analyst — Lake Acidification Study
An environmental team is monitoring Lake Champlain for acid rain effects. They collected n = 15 water samples at random locations and measured pH. A preliminary analysis (computed from their raw data) gives pH units and pH units. σ is unknown. The ecological threshold for concern is pH < 4.5 (below this, fish populations are at risk).
Does the 95% CI provide evidence that mean pH is below 4.5?
Task 2: Communicate the Results
The team wants to present these results to city officials. Write a 2–3 sentence interpretation suitable for a non-statistical audience.
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Show Model Interpretation
We are 95% confident that the true mean pH of Lake Champlain water is between 4.61 and 4.97. Since this interval is entirely above the 4.5 ecological threshold, the data do not provide strong evidence of a crisis at this time. However, with only 15 samples, continued monitoring is warranted.
Reflection
What was the most challenging part of this analysis? What would you do differently if you could collect more data?
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🏗️ Path B: The Architect — Critiquing a Flawed Study
A junior researcher submits a report on patient recovery times. Her abstract reads: “We sampled n = 20 patients and measured recovery time in days. Results: days, days, σ unknown. We constructed a 90% confidence interval using z* = 1.645 and found CI = (7.90, 10.70) days. We are 90% confident recovery takes between 7.9 and 10.7 days.”
Task 1: Identify the Error
Task 2: Compute the Correct CI
Show Correct Calculation
df = 19, (90%, df = 19)
SE = 3.8/ = 3.8/4.472 ≈ 0.850
Et = 1.729 × 0.850 ≈ 1.470
Correct CI = (9.3 − 1.470, 9.3 + 1.470) = (7.830, 10.770) days
How does the correct CI compare to the reported one?
Task 3: Explain the Error to the Researcher
Write a 2–3 sentence explanation to the researcher explaining why her method was incorrect and what she should report instead.
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Show Model Explanation
Because σ is unknown, the z-distribution is only an approximation — the correct procedure with n = 20 and unknown σ is to use the t-distribution with df = 19. Using t* = 1.729 instead of z* = 1.645 gives a slightly wider interval: (7.83, 10.77) days, which more honestly reflects the uncertainty from the small sample. Please update Table 1 with the corrected interval.
Reflection
What would happen to the gap between t* and z* if n were 8 instead of 20? How would this change your assessment of the error’s severity?
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Section 9: Challenge Problems
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Ready for more? These go beyond the lesson objectives. They’re not required — but they build deeper understanding.
Challenge 1 — The ‘Why’ Behind df = n − 1
Most students accept df = n − 1 as a rule. This problem asks you to explain why it must be n − 1, not n.
Setup: Suppose you have n = 4 observations: . You’ve computed .
(a) If , , , and , what must be?
(b) How many of the four deviations are “free to vary independently”?
(c) In your own words, explain why the constraint “deviations sum to zero” means we must divide by n − 1 rather than n when computing sample variance.
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Show Model Answer
The sample variance uses n − 1 in the denominator because only n − 1 of the n deviations are independently determined — the last is forced by the others summing to zero. Dividing by n would underestimate the true population variance , because the constrained deviations “shrink” toward zero relative to what we’d see if we knew and could compute deviations from the true mean. Using n − 1 (the degrees of freedom) corrects this bias and makes an unbiased estimator of .
Challenge 2 — How Quickly Does t* Converge to z*?
As df increases, t* approaches z*. But how quickly? This challenge quantifies the convergence.
Task: Using the t-table from Section 3, record t* for 95% confidence at each df below, then answer the questions.
df
t* (95% confidence)
1
12.706
5
2.571
10
2.228
20
2.086
29
2.045
∞ (z)
1.960
(a) At df = 29 (n = 30), how much larger is t* = 2.045 compared to z* = 1.96?
(b) At df = 5 (n = 6), how much larger is t* = 2.571 compared to z* = 1.96?
Reflection: At what df does the difference between t* and z* become small enough that you’d consider the z-approximation “good enough”? Is there a single right answer? Why or why not?
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Challenge 3 — Two CIs, One Question: Do the Groups Differ? (Preview of INF-6)
This previews INF-6. The formal test for two groups (two-sample t-test) comes later, but you can reason about the evidence from confidence intervals alone.
Two training programs are tested. Program A (n = 12): 95% CI for mean performance score = (72.3, 81.7). Program B (n = 15): 95% CI = (68.1, 76.5). Do the CIs overlap? What does this suggest?
Do the CIs overlap?
What does this suggest?
Drug vs. placebo for sleep latency. Drug group (n = 10): 95% CI = (8.2, 14.4) minutes. Placebo group (n = 11): 95% CI = (18.7, 26.3) minutes. Do the CIs overlap?
Do the CIs overlap?
What does this suggest?
Comparing exercise programs. Group 1 (n = 14): 90% CI for weight loss = (2.1, 4.9) kg. Group 2 (n = 16): 90% CI = (3.8, 7.2) kg. Do they overlap?
Do the CIs overlap?
What does this suggest?
Two manufacturing methods. Method A (n = 8): 99% CI for mean cycle time = (12.2, 17.6) seconds. Method B (n = 10): 99% CI = (18.9, 25.1) seconds. Do they overlap?
Do the CIs overlap?
What does this suggest?
Instructional format study. Students who watched a lecture video (n = 18): 95% CI for quiz score = (71.4, 78.6). Students who read a text summary (n = 22): 95% CI = (69.8, 77.2). Do they overlap?
Do the CIs overlap?
What does this suggest?
Section 10: Solutions Reference
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Complete, step-by-step solutions for all problems in Sections 5–9 are available on the solutions page. Solutions include worked arithmetic, common mistakes to watch for, and interpretation guidance.
If you’re stuck: Re-read the relevant Core Concept in Section 3, then find the Worked Example that maps to that concept (e.g., Example 1 maps to Concept 1). If you’re unsure whether to use t or z, review the decision rule in C8. The solutions page shows the reasoning behind every step, not just the final answer.
Quick-Reference Formulas
Confidence Interval for (Small Sample / Unknown):
Degrees of Freedom ():
Population standard deviation ()
Sample size ()
Population shape
Distribution to use
Known
Any
Normal
-distribution
Known
Any
-distribution ‡
Unknown (use )
Any
Normal
-distribution
Unknown (use )
Any
-distribution †
† For with unknown , the -approximation is also acceptable because at large degrees of freedom. However, is always the principled choice when is estimated from the data.
‡ The case not shown: known, , non-normal population. The -interval relies on the sampling distribution of being approximately normal — for small from a clearly non-normal population, this holds only if the population itself is normal. If the population is skewed and is small, even known does not rescue the -interval. This edge case is uncommon in practice but worth knowing: always check the normality condition when , regardless of whether is known.
t-Distribution Reference Table
Student's t-Distribution Table
Critical values (t*) for given degrees of freedom (df) and tail area.