B.6 Confidence Interval (Women's Heights)

Step 1 — write down the interval's two moving parts. A confidence interval for a mean with known \(\sigma\) has the form

$$ (\bar{x} - EBM,\ \bar{x} + EBM), \qquad EBM = z_{\alpha/2} \frac{\sigma}{\sqrt{n}} $$

Step 2 — ask which part depends on the sample. The error bound \(EBM\) is built from \(z_{\alpha/2}\) (fixed by the \(90\%\) confidence level), \(\sigma = 3.3\) inches (given, and a property of the population), and \(n = 10\) (fixed by the lab's instructions). Not one of those three comes from your ten numbers, so every student in the room computes the same \(EBM\). The sample mean \(\bar{x}\) is the only quantity that reads your particular ten heights.

Answer: the intervals differ because \(\bar{x}\) differs — one student's ten women happened to be taller than another's. The width of every interval in the class is identical, \(2 \times EBM\), because width depends only on the confidence level, the population standard deviation, and the sample size. So the thirty intervals on the board are thirty copies of the same ruler, slid to thirty different positions along the height axis. That picture is worth holding onto: what the class is about to measure is not whether the ruler is the right length, but how often a ruler of that length, dropped at a randomly chosen spot, happens to cover \(63.4\).

Step 1 — identify what is random. The population mean \(\mu\) is a fixed number: for these hundred women it is \(63.4\) inches, and it was \(63.4\) before anyone sampled and will be \(63.4\) afterwards. It has no distribution, so no probability can be assigned to where it "is".

Step 2 — identify what was random. The sample was. Ten women were drawn at random, and those ten produced \(\bar{x}\), which produced the endpoints \(62.91\) and \(66.35\). The interval is the random object; \(\mu\) is the fixed target.

Step 3 — locate the error. The sentence attaches the \(90\%\) to \(\mu\)'s position, as though \(\mu\) were rattling around and might land inside. Once the endpoints are written down, the interval either covers \(63.4\) or it does not. In this case it does, so the "probability" is 1, not \(0.90\) — and had the sample been unluckier it would have been 0. Neither is \(90\%\).

Answer: the sentence is wrong because it treats a fixed constant as random. A true version attaches the \(90\%\) to the method: "This interval was produced by a procedure that captures the true population mean for \(90\%\) of all possible samples of ten." Or, said the way a report usually says it, "we are \(90\%\) confident that the mean height of women in this population is between \(62.91\) and \(66.35\) inches" — where "confident" is deliberately not the word "probable", because the confidence belongs to the process that built the interval, not to the interval sitting in front of you.

Step 1 — find the sample mean. The ten values total \(646.3\).

$$ \bar{x} = \frac{646.3}{10} = 64.63 \text{ inches} $$

Step 2 — find the error bound. With \(CL = 90\%\), \(z_{\alpha/2} = 1.645\); the lab supplies \(\sigma = 3.3\) and \(n = 10\).

$$ EBM = 1.645 \cdot \frac{3.3}{\sqrt{10}} = 1.645 \cdot \frac{3.3}{3.1623} = 1.645 \cdot 1.0436 \approx 1.72 \text{ inches} $$

Step 3 — assemble the interval.

$$ (\bar{x} - EBM,\ \bar{x} + EBM) = (64.63 - 1.72,\ 64.63 + 1.72) = (62.91,\ 66.35) $$

Answer: Quinn's \(90\%\) confidence interval is \((62.91, 66.35)\) inches, and since \(62.91 < 63.4 < 66.35\), it does capture \(\mu\). Notice how close the call was on one side. This sample's mean, \(64.63\), sits \(1.23\) inches above the population mean — a fairly unlucky draw, since it happened to include the two tallest women in the table — and the interval still reached back far enough to cover \(63.4\) with about half an inch to spare. Had the sample been only a little more extreme, the interval would have missed entirely, and they would have had no way to tell from the ten numbers in front of them. That is the failure mode you are about to see several examples of on the board, and none of those students did anything wrong either.

Step 1 — find the new \(z\). At \(CL = 95\%\), \(\alpha = 0.05\) and \(\alpha/2 = 0.025\), so \(z_{0.025} = 1.96\).

Step 2 — recompute the error bound. Only \(z\) changed; \(\sigma/\sqrt{n} = 1.0436\) as before.

$$ EBM = 1.96(1.0436) \approx 2.05 \text{ inches} $$

Step 3 — assemble the interval.

$$ (64.63 - 2.05,\ 64.63 + 2.05) = (62.58,\ 66.68) $$

Answer: Mateo's \(95\%\) interval is \((62.58, 66.68)\) inches, which is \(4.09\) inches wide against the \(90\%\) interval's \(3.43\) — about \(19\%\) wider. What the extra confidence bought him is a procedure that misses one time in twenty instead of one in ten. What it cost is precision: the claim now spans over four inches, and a claim that wide is harder to act on. Both intervals sit at the same center, \(64.63\), because the confidence level moves only the endpoints. It is worth noticing that the wider interval is not a better estimate of \(\mu\) — the point estimate did not change at all. Widening only makes the net easier to catch a fish with; it does not tell you where the fish is.

Step 1 — read the gap. The sample overestimates the population's spread by about \(50\%\) (\(4.91\) against \(3.27\)). Nothing went wrong: these ten women included both of the table's tallest values, \(71.6\) and \(69.3\), and its shortest, \(55.2\). Ten values simply do not pin down a spread, and \(s\) from a sample of ten is a noisy estimate of \(\sigma\) in a way that \(\bar{x}\) is not nearly as bad an estimate of \(\mu\).

Step 2 — see what a \(t\) interval would have done. With \(\sigma\) unknown you would use \(t_{0.05}\) at \(df = n - 1 = 9\), which is \(1.833\) rather than \(1.645\), and you would put \(s = 4.91\) in place of \(\sigma\):

$$ EBM = 1.833 \cdot \frac{4.91}{\sqrt{10}} = 1.833(1.5527) \approx 2.85 \text{ inches} $$ $$ (64.63 - 2.85,\ 64.63 + 2.85) = (61.78,\ 67.48) $$

Answer: Erin's \(t\) interval would have been \(5.69\) inches wide against the \(z\) interval's \(3.43\) — two thirds wider from the same ten numbers. Both changes push the same way: \(t\) is fatter than \(z\) at nine degrees of freedom, and this particular \(s\) came in high. That widening is not a defect of the \(t\) procedure; it is the honest price of not knowing \(\sigma\), and it is exactly why §7.2 exists. This lab dodges that price by handing you \(\sigma\), which is also why the whole class can compare intervals of identical width. Real work rarely gets that gift.

Step 1 — check the widths. Every interval built in this lab must be \(2 \times EBM = 2(1.7166) \approx 3.43\) inches wide. Subtracting: \(64.63 - 61.19 = 3.44\); \(66.99 - 63.55 = 3.44\); \(65.27 - 61.83 = 3.44\); \(63.19 - 59.75 = 3.44\). All four match to rounding, so all four followed the instructions.

Step 2 — recover each sample mean. The sample mean sits at the midpoint, \(\bar{x} = \frac{\text{lower} + \text{upper}}{2}\):

$$ \frac{61.19 + 64.63}{2} = 62.91, \quad \frac{63.55 + 66.99}{2} = 65.27, \quad \frac{61.83 + 65.27}{2} = 63.55, \quad \frac{59.75 + 63.19}{2} = 61.47 $$

Step 3 — check each against \(\mu = 63.4\). Anh's contains it (\(61.19 < 63.4 < 64.63\)). Cody's does not — his lower endpoint \(63.55\) is already above \(63.4\). Lucía's contains it. Jamal's does not — his upper endpoint \(63.19\) falls short.

Answer: all four are valid \(90\%\) intervals, with sample means \(62.91\), \(65.27\), \(63.55\) and \(61.47\) inches, and two of the four capture \(\mu\). Two misses out of four is nowhere near \(90\%\) coverage, and it is also not evidence that anything is broken — four intervals is far too few to measure a rate. That is precisely why the lab collects the whole class's intervals rather than asking you to judge from your own. Notice also that the two misses failed in opposite directions, one entirely above \(\mu\) and one entirely below, which is what you should expect from a symmetric procedure.

Step 1 — answer the counting question. \(37\) of the \(100\) values lie in \((62.91, 66.35)\), which is \(37\%\). That is not close to \(90\%\); it is not close to half.

Step 2 — see why the number is so small. The interval is \(2(1.7166) \approx 3.43\) inches wide, and it is that width because it was sized to a standard error of \(\sigma/\sqrt{n} = 3.3/\sqrt{10} \approx 1.04\) inches. Individual heights are spread out by \(\sigma = 3.3\) inches, so an interval a little over three inches wide covers only about one standard deviation's worth of the population.

Step 3 — size the interval that would do the job. To capture the middle \(90\%\) of individual heights you would use the spread of \(X\) itself, not of \(\bar{X}\):

$$ \mu \pm z_{0.05}\sigma = 63.4 \pm 1.645(3.3) = 63.4 \pm 5.43 = (57.97,\ 68.83) $$

Answer: the confidence interval captures \(37\%\) of the data, not \(90\%\). The interval that would capture about \(90\%\) of individual heights runs from roughly \(58.0\) to \(68.8\) inches — nearly \(11\) inches wide, more than three times the confidence interval, because it has to accommodate the full scatter of individual women rather than the much smaller scatter of ten-woman averages. The \(\sqrt{n}\) in the denominator of the standard error is the entire difference between the two, and it is why a bigger sample makes a confidence interval narrower without making the population any less varied. Two intervals, two questions: one asks where the average is, the other asks where the women are.

Step 1 — treat the thirty intervals as thirty trials. Each interval either hits or misses, hits are supposed to happen with probability \(0.90\), and the students sampled independently. That is a binomial setting with \(n = 30\) and \(p = 0.90\).

Step 2 — find the expected count and the wobble around it.

$$ \mu_{\text{hits}} = np = 30(0.90) = 27, \qquad \sigma_{\text{hits}} = \sqrt{npq} = \sqrt{30(0.90)(0.10)} = \sqrt{2.7} \approx 1.64 $$

Step 3 — measure the shortfall in units of that wobble.

$$ \frac{25 - 27}{1.64} \approx -1.22 $$

Answer: yes, \(83\%\) is entirely consistent with a correctly working \(90\%\) procedure, and Cruz should say so. Twenty-five hits sits about \(1.2\) standard deviations below the expected twenty-seven, and a result that close to the middle is unremarkable — a class doing this lab would land at twenty-five or below a fair fraction of the time by luck alone. The useful move here is not to compare \(83\%\) against \(90\%\) directly, which invites the answer "no, it is seven points off", but to compare the shortfall against the size of the random wobble a class of thirty carries. With only thirty intervals that wobble is roughly \(5\) percentage points, so \(83\%\) and \(93\%\) are equally ordinary results and neither one is evidence about the method. This is also the answer to question 4: with a hundred intervals the wobble shrinks to about \(3\) points, and the observed percent would cluster more tightly around \(90\%\).

Step 1 — change the confidence level, hold \(n = 10\). At \(CL = 99\%\), \(\alpha/2 = 0.005\) and \(z_{0.005} = 2.576\).

$$ EBM = 2.576 \cdot \frac{3.3}{\sqrt{10}} = 2.576(1.0436) \approx 2.69, \qquad \text{width} = 5.38 \text{ inches} $$

Step 2 — change the sample size, hold \(CL = 90\%\).

$$ EBM = 1.645 \cdot \frac{3.3}{\sqrt{40}} = 1.645 \cdot \frac{3.3}{6.3246} = 1.645(0.5217) \approx 0.86, \qquad \text{width} = 1.72 \text{ inches} $$

Answer: raising the confidence to \(99\%\) widens the interval from \(3.43\) to \(5.38\) inches, while quadrupling the sample to forty narrows it from \(3.43\) to \(1.72\). Only one of these is a genuine improvement in precision, and it is the sample size — quadrupling \(n\) exactly halves the width, since \(n\) sits under a square root. Raising the confidence level does not make you better informed; it makes you safer by making the claim vaguer, and the vaguer claim is the price. The only lever that improves precision and reliability together is collecting more data, which is also the only one that costs money — this is the trade-off behind the sample-size formula \(n = z^2\sigma^2/EBM^2\) from §7.1, read from the other direction.

Step 1 — answer the question as asked. No. Checking whether an interval contains \(\mu\) requires knowing \(\mu\), and a researcher who knew \(\mu\) would have had no reason to build an interval. When he and his husband, who teaches the other section, compare notes, neither of them can tell either: there is no internal signal — no width, no roundness of the numbers, no property of the ten heights — that distinguishes a hit from a miss from the inside. Every one of the four intervals in Try It Now B.6.6 looked equally respectable, and two of them missed.

Step 2 — say what can be claimed. What survives is a claim about the procedure: the interval was produced by a method that succeeds for \(90\%\) of all possible samples of ten. That is not a hedge, it is the actual content of the result, and it is enough to act on — which is why confidence intervals are used at all despite never being individually verifiable.

Step 3 — say what would settle it. Only measuring the whole population, which for these hundred women is possible and is exactly what this lab did. In the field it usually is not: if you could measure every woman in the population you would report \(\mu\) and skip the interval entirely.

Answer: Dr. Yazzie has no way of knowing for certain, and never will for that one interval. This lab is engineered to be the exception — it prints the whole population so that \(\mu\) can be checked — and the reason it is worth an afternoon is that it lets you watch the \(90\%\) come true across thirty intervals at once, in the only setting where you are allowed to look. Every use of a confidence interval after this one is question 8 again: one interval, no answer key, and a claim that rests entirely on the method having been applied honestly.