INF-3 Solutions: Confidence Intervals for a Population Mean (Small Sample)

IP1 — Full t-Interval Construction (Variants A–E)

For each variant, always check conditions first: (1) population approximately normal or n large enough, (2) σ unknown, (3) independent random sample.

Variant A (ER wait times, \(\bar{x} = 47.3\) min, \(s = 12.4\) min, \(n = 18\), 95% CI):

• \(df = 17, \quad t^* = 2.110\).
• \(\text{SE} = 12.4/\sqrt{18} = 2.923, \quad E = 2.110 \times 2.923 = 6.167\)
• 95% CI: \((47.3 - 6.167,\ 47.3 + 6.167) = (41.133,\ 53.467)\) min.
• Interpretation: We are 95% confident the true mean ER wait time is between 41.13 and 53.47 minutes.

Variant B (calcium content, \(\bar{x} = 84.6\) mg/L, \(s = 9.3\) mg/L, \(n = 11\), 90% CI):

• \(df = 10, \quad t^* = 1.812\).
• \(\text{SE} = 9.3/\sqrt{11} = 2.804, \quad E = 1.812 \times 2.804 = 5.081\)
• 90% CI: \((79.519,\ 89.681)\) mg/L.

Variant C (sleep hours, \(\bar{x} = 6.8\) hrs, \(s = 1.1\) hrs, \(n = 20\), 99% CI):

• \(df = 19, \quad t^* = 2.861\).
• \(\text{SE} = 1.1/\sqrt{20} = 0.2460, \quad E = 2.861 \times 0.2460 = 0.704\)
• 99% CI: \((6.096,\ 7.504)\) hrs.

Variant D (weight loss, \(\bar{x} = 3.7\) kg, \(s = 1.6\) kg, \(n = 25\), 95% CI):

• \(df = 24, \quad t^* = 2.064\).
• \(\text{SE} = 1.6/\sqrt{25} = 0.320, \quad E = 2.064 \times 0.320 = 0.660\)
• 95% CI: \((3.040,\ 4.360)\) kg.

Variant E (test scores, \(\bar{x} = 68.4\), \(s = 8.9\), \(n = 14\), 90% CI):

• \(df = 13, \quad t^* = 1.771\).
• \(\text{SE} = 8.9/\sqrt{14} = 2.379, \quad E = 1.771 \times 2.379 = 4.213\)
• 90% CI: \((64.187,\ 72.613)\).

IP2 & IP3 — Generated Problems

Solutions for IP2 and IP3 are embedded in each generator's output. For IP3: if σ is known, use z regardless of n; if σ is unknown and n < 30, use t with df = n − 1.

IP4 — Find the Error (Variants A–E)

Variant A (z* used with n = 12 and unknown σ): Error — when σ is unknown and n < 30, the correct distribution is t, not z. Using z* = 1.96 ignores the extra variability from estimating σ with s, producing an interval that is too narrow.

Variant B (df = n = 18 instead of df = 17): Error — degrees of freedom should be \(df = n - 1 = 17\), not n = 18. Using df = 18 yields a t* that is slightly too small, producing an overconfident (too-narrow) interval.

Variant C (P(\mu \in [L, U]) = 0.95): Error — probability trap. Once the interval is computed, μ is a fixed constant; the probability is either 0 or 1. The 95% refers to the long-run frequency of the procedure — about 95% of all such intervals built from repeated samples would capture the true μ.

Variant D ("the t-interval is wrong because it's wider"): Error — the t-interval is supposed to be wider. The t-distribution has heavier tails than the normal precisely because we're estimating σ with s. The extra width is honest: it correctly accounts for the additional uncertainty. A narrower interval here would be overconfident.

Variant E (t-interval used with n = 50 and σ known): Error — when σ is known, always use z regardless of sample size. The t-distribution is only needed when σ must be estimated from the data. Using t when σ is known is conservative but technically incorrect.

IP5 — Synthesis: Lead Concentration

Data: \(n = 22\) water samples, \(\bar{x} = 8.4\) ppb, \(s = 3.1\) ppb, \(\sigma\) unknown.

Part (a) — 95% t-interval:

• \(df = 21, \quad t^* = 2.080\).
• \(\text{SE} = 3.1/\sqrt{22} = 0.661, \quad E = 2.080 \times 0.661 = 1.375\)
• 95% CI: \((8.4 - 1.375,\ 8.4 + 1.375) = (7.025,\ 9.775)\) ppb.

Part (b) — z-based comparison (incorrect):

• \(E_z = 1.96 \times 0.661 = 1.295, \quad \text{CI}_z = (7.105,\ 9.695)\) ppb.
• The t-interval (7.025, 9.775) is wider — it extends 0.080 ppb further in each direction.

Part (c) — Why the distinction matters: With n = 22, t* = 2.080 while z* = 1.96 — a 6% difference in the critical value. For smaller samples this gap grows dramatically (e.g., at n = 6, t* ≈ 2.571 vs. z* = 1.96 for 95% CI — a 31% difference). Reporting the z-interval here understates uncertainty by assuming σ is known precisely. If this data were being used to compare lead levels against a safety threshold, using the incorrectly narrow z-interval could lead to a false conclusion that concentrations are safe when the honest t-interval suggests otherwise.

Feynman Test — Explain t vs. z + Why df = n − 1

Model answer: The z-distribution assumes we know the population standard deviation σ exactly. In most real studies, σ is unknown and we estimate it from the sample using s. This introduces a second source of variability: not only does x̄ vary from sample to sample, but so does s. The t-distribution accounts for this by having heavier tails — giving wider critical values — which translates into wider (more honest) confidence intervals.

We use df = n − 1 rather than n because once we've computed the sample mean x̄, the n deviations \(x_i - \bar{x}\) are constrained: they must sum to zero. This means only n − 1 of them can vary freely. One degree of freedom is "spent" computing x̄, leaving n − 1 independent pieces of information about the spread. As df → ∞, this constraint becomes negligible and the t-distribution converges to the standard normal.

Apply Question — Energy Bars

Data: \(n = 9\), \(\bar{x} = 3.2\) g protein, \(s = 0.6\) g, 99% CI, \(\sigma\) unknown.

• Distribution: t (σ unknown, n < 30).
• \(df = 8, \quad t^* = 3.355\) (99% CI).
• \(\text{SE} = 0.6/\sqrt{9} = 0.200, \quad E = 3.355 \times 0.200 = 0.671\)
• 99% CI: \(3.2 \pm 0.671 = (2.529,\ 3.871)\) g.

Interpretation: We are 99% confident the true mean protein content is between 2.529 g and 3.871 g per bar.

Error Analysis — Wrong df

The researcher used df = 15 but the sample has n = 15, so the correct df is 14, not 15.

With df = 14 and 90% CI: \(t^* = 1.761\) (vs. 1.753 they used with df = 15).

• \(\text{SE} = s/\sqrt{15}\) (same as before).
• Corrected E is slightly larger (t* = 1.761 > 1.753), giving a slightly wider interval.
• The error causes the interval to be marginally too narrow — df = n overestimates df and therefore underestimates t*.

Path A — The Analyst: Lake Champlain pH Study

Data: \(\bar{x} = 4.79\), \(s = 0.33\), \(n = 15\), 95% CI, \(\sigma\) unknown.

Step 1 — Check conditions:

• Random sample: stated ✓
• σ unknown: stated ✓ → use t
• Population approximately normal: pH readings from a lake tend to be approximately normally distributed ✓ (or n = 15 is borderline; state the assumption)

Step 2 — Compute the CI:

• \(df = 14, \quad t^* = 2.145\) (95% CI).
• \(\text{SE} = 0.33/\sqrt{15} = 0.0852, \quad E = 2.145 \times 0.0852 = 0.183\)
• 95% CI: \(4.79 \pm 0.183 = (4.607,\ 4.973)\).

Step 3 — Does the interval suggest pH < 4.5? No. The entire interval (4.607, 4.973) lies above 4.5. If the true mean were below 4.5, we'd expect our sample mean to fall lower than 4.79 and the CI to include values below 4.5. This interval gives no evidence that the lake pH is critically low.

Step 4 — Interpretation: "We are 95% confident the true mean pH of Lake Champlain at this location is between 4.607 and 4.973. The interval lies entirely above the acidic threshold of 4.5, suggesting the lake is not in the critical range — though the lower bound of 4.61 is still quite acidic relative to neutral pH 7."

Path B — The Architect: Critique and Redesign

The junior researcher used z* = 1.96 with n = 20 and unknown σ.

Error identification: When σ is unknown and n < 30, the t-distribution must be used. Using z* = 1.96 ignores the extra uncertainty from estimating σ with s, producing an interval that is too narrow (overconfident).

Corrected CI: Data: \(\bar{x} = 9.3\) days, \(s = 3.8\) days, \(n = 20\), 90% CI.

• \(df = 19, \quad t^* = 1.729\) (90% CI).
• \(\text{SE} = 3.8/\sqrt{20} = 0.850, \quad E = 1.729 \times 0.850 = 1.470\)
• 90% CI using t: \((9.3 - 1.470,\ 9.3 + 1.470) = (7.830,\ 10.770)\) days.

Comparison: The researcher's z-based CI would be: \(E_z = 1.645 \times 0.850 = 1.399\), giving (7.901, 10.699). The corrected t-interval is wider by 0.071 days on each side (making the entire interval 0.142 days wider in total width) — a meaningful difference when reporting clinical recovery times. The z-interval gives a false sense of precision not warranted by these data.

Why it matters here: For medical recovery data, understating uncertainty could affect treatment protocol decisions. The additional width of the t-interval honestly reflects that with only 20 patients and no known σ, there is more uncertainty than the z-interval admits.

C1 — Why df = n − 1 (Intuitive Derivation)

Suppose you have n = 4 data values and you've computed \(\bar{x} = 10\). The deviations from the mean must sum to zero: \((x_1 - \bar{x}) + (x_2 - \bar{x}) + (x_3 - \bar{x}) + (x_4 - \bar{x}) = 0\).

If you know x₁ = 8, x₂ = 11, x₃ = 13 and the mean is 10, then x₄ is forced: x₄ = 4×10 − 8 − 11 − 13 = 8. You have no freedom to choose x₄.

In general: once you fix x̄ and the first n − 1 values, the last value is completely determined. Only n − 1 of the n deviations are free to vary independently. Therefore the sample variance s² — which is built from these deviations — has only n − 1 degrees of freedom.

When we plug s into the CI formula to substitute for σ, we're using a quantity with n − 1 independent pieces of information. The t-distribution with df = n − 1 correctly accounts for this reduced independence.

C2 — t-Table Convergence

t* values at 95% confidence:

Key observation: Even at df = 29 (n = 30), t* = 2.045 remains meaningfully larger than z* = 1.96 — a 4.3% difference. At df = 5, the gap is 31%. The convergence is slow, especially in the lower tail region that matters for critical values. This is why the t vs. z distinction remains practically important even at moderate sample sizes, not just at n < 10.

At what df does t* first drop below 2.05? From our table: df = 25 gives t* = 2.060, df = 29 gives t* = 2.045 — so somewhere around df = 28–29. Full t-tables show the crossover below 2.05 occurs at approximately df = 26 (t* = 2.056). The precise value where t* = z* = 1.960 is only reached at df = ∞.

C3 — Overlapping CIs Preview (Variants A–E)

This challenge previews INF-6 (two-sample inference). The key intuition: non-overlapping CIs are strong evidence that the two population means differ; overlapping CIs suggest the means may be similar, but overlap alone does not confirm equivalence. Formal tests (two-sample t) are needed for definitive conclusions.

For each variant:

• If the intervals don't overlap: The data provide strong evidence that the two means are different. (The true means are unlikely to be equal if each interval completely excludes the other.)
• If the intervals overlap substantially: We cannot conclude the means differ. There is a plausible range of values consistent with both groups having the same mean.
• If the intervals barely overlap: Ambiguous — a formal two-sample t-test is needed. (Intervals can overlap and the means still be significantly different, depending on the test.)

This is a conceptual preview — no specific arithmetic is required for these variants. Focus on the logic: confidence intervals describe plausible values for each mean; comparison requires reasoning about whether those plausible ranges share common ground.