Math 115: Introduction to Statistics – 13. Inference for Paired Means

From Testing to Estimation

Still paired data: one within-pair difference $d$ per unit, and we ask about $\mu_d$, the mean of those differences.

The 13a test answered “Is the mean paired difference zero?”

Now: “How big is the mean paired difference?” → a confidence interval for $\mu_d$.

Same paired machinery as 13a — work with the within-pair differences, then use the same $SE = s_d / \sqrt{n}$ and $df = n - 1$. New piece: a t critical value $t^*$ replaces the test statistic. Just like 11b’s one-mean CI, with the “mean” being the mean of the differences.

Recap: Textbook Prices

How do textbook prices compare between the UCLA bookstore and Amazon? Sample of 68 UCLA-course textbooks priced at both stores; $d$ = UCLA − Amazon.

n	$\bar{x}_d$ ($)	$s_d$ ($)
68	3.58	13.42

From 13a: $T \approx 2.2$, p-value $\approx 0.031$ → rejected $H_0$ at $\alpha = 0.05$ (with a caution about skew + outliers). Today: how big is the typical price gap?

CI Formula for Paired Mean

When conditions are met:

\[\text{CI} = \bar{x}_d \pm t^*_{df} \times \frac{s_d}{\sqrt{n}}\]

$\bar{x}_d$ = sample mean of differences (the point estimate)
$t^*_{df}$ = t critical value with $df = n - 1$ (from Jamovi)
$s_d / \sqrt{n}$ = standard error of the mean difference

Identical to one-mean CI from 11b, with $\bar{x} \to \bar{x}_d$ and $s \to s_d$. The “reduce to differences” trick gives back the full one-mean toolkit.

Conditions for the CI

Same conditions as the hypothesis test in 13a — applied to the differences:

Independence: different books from different courses → pairs independent of each other ✓
Normality: $n = 68 \geq 30$ → CLT applies to the mean of differences ✓; right skew + a few large outliers → proceed with caution.

Same caution as 13a. With $n = 68$ the CLT provides cover for the sampling distribution of $\bar{x}_d$, but the borderline result invites careful interpretation.

Computing the 95% CI for Textbooks

From 13a: $\bar{x}_d = 3.58$ dollars, $s_d = 13.42$ dollars, $n = 68$, $df = 67$.

Standard error: $SE = \dfrac{s_d}{\sqrt{n}} = \dfrac{13.42}{\sqrt{68}} \approx 1.63$ dollars.

t critical value: $t^*_{67} \approx 1.996$ (95% confidence; very close to $z^* = 1.96$ since $df$ is large).

Margin of error: $ME = t^* \times SE \approx 1.996 \times 1.63 \approx 3.25$ dollars.

95% CI: $\bar{x}_d \pm ME = 3.58 \pm 3.25 \approx (0.33, 6.83)$ dollars.

Pull the pieces together for textbooks. From thirteen A we have the mean difference, about three dollars fifty-eight cents, and the standard deviation of the differences, about thirteen dollars forty-two cents, across sixty-eight books. The standard error is that standard deviation over the square root of the sample size — thirteen forty-two over the square root of sixty-eight — which comes to about one dollar sixty-three. It’s worth recomputing rather than just borrowing the number, since this same standard-error formula shows up in every one of these intervals. Degrees of freedom is n minus one, sixty-seven. For ninety-five percent confidence at this many degrees of freedom, the t critical value t-star is about one point nine nine six — close to the z-star value of one point nine six, though still a bit larger, as t-star always is at finite degrees of freedom. The margin of error is t-star times the standard error — one point nine nine six times one dollar sixty-three, about three dollars twenty-five cents. The ninety-five percent confidence interval is the mean difference plus or minus the margin of error — three dollars fifty-eight cents plus or minus three dollars twenty-five cents — giving an interval from about thirty-three cents up to about six dollars eighty-three cents per book.

Interpreting the CI

95% CI: $(\$0.33, \$6.83)$ per book.

Interpretation: We are 95% confident that, on average, a textbook costs between $0.33 and $6.83 more at the UCLA bookstore than on Amazon.

Does the CI include 0? No — zero sits just below the entire interval.

This agrees with the 13a hypothesis test: rejected $H_0$ ↔︎ zero is outside the CI. But the interval is wide — the data only pin the gap down to somewhere between about $0.30 and $7 per book.

Scope — generalization: not a documented random sample → the result covers only these books, not all UCLA Fall 2018 textbooks.

Scope — causation: not a randomized experiment → we can’t say the bookstore causes the higher price.

Now what does that interval actually tell us? We’re ninety-five percent confident that, on average, a textbook costs between about thirty-three cents and six dollars eighty-three cents more at the UCLA bookstore than on Amazon. The interval excludes zero — it sits just above it — which agrees with the hypothesis test in thirteen A. But notice how wide it is: the data only narrow the true gap to somewhere between a fraction of a dollar and about seven dollars a book. That width comes from the standard error — the spread of the differences relative to the sample size — and here the spread is fairly large, inflated in part by the skew and outliers we flagged. Two cautions on scope. On generalization: without random sampling, the result covers only these particular books, not all UCLA Fall twenty-eighteen textbooks. On causation: this isn’t a randomized experiment, so we can’t say the bookstore itself causes the higher price — it’s a comparison of the two stores’ prices, not a cause-and-effect claim.

Sleep — CI Setup

How much more sleep, on average, does Drug 2 produce than Drug 1? Same 10-patient study as 13a (Cushny & Peebles, 1905; used as Student’s 1908 t-test example).

Recall the variable: each drug’s value is that patient’s extra sleep vs. no drug; the paired difference $d$ = (Drug 2 extra) − (Drug 1 extra) — a difference in extra hours of sleep.

Recall: $n = 10$, $\bar{x}_d = 1.58$ hours, $s_d = 1.23$ hours, $df = 9$.

From 13a: $T \approx 4.06$, p-value $\approx 0.003$ → rejected $H_0$.

Are the conditions for the CI met? Independence: the patients are separate individuals tested on their own, so we assume one patient’s response doesn’t affect another’s — plausible, though the data don’t tell us how they were recruited ✓. Normality: $n = 10 < 30$, so we’d want the differences roughly normal — most cluster between 0 and 2.4, but the 4.6 outlier nudges us toward caution. Same caveat as 13a — t inference is somewhat robust, but proceed with care.

Back to the sleep example from thirteen A — the same ten patients who each tried both drugs. Remember what each number means: for a given drug, a patient’s value is the extra sleep they got compared to taking no drug. Our paired difference d subtracts one from the other — drug two’s extra sleep minus drug one’s — so d is a difference in extra hours of sleep, and a positive d means drug two helped that patient sleep more. Across those patients, the mean difference was one point five eight hours, with a standard deviation of the differences of one point two three hours. The hypothesis test gave T around four point oh six and a p-value near zero point zero zero three, so we rejected the null. The question now shifts from whether the drugs differ to by how much — and that’s a confidence interval. We check the conditions first, just as before. Advance when you are ready to see the answer. Conditions for the confidence interval are the same as for the hypothesis test in thirteen A. For independence, the patients are separate individuals tested on their own, so we assume one patient’s response doesn’t affect another’s — plausible, but an assumption, since the data don’t tell us how they were recruited. Normality needs more thought, since n equals ten is below thirty, so we want the differences themselves to look roughly normal. Most of them cluster between zero and two and a half hours, but the four point six outlier nudges us toward caution. As in thirteen A, the t-procedure is somewhat robust here, but we should proceed with care.

Sleep — CI & Conclusion

Recall: $\bar{x}_d = 1.58$, $s_d = 1.23$, $n = 10$, $df = 9$. Use $t^*_{9} \approx 2.262$ for 95% confidence (df is small, so $t^*$ is noticeably larger than 1.96).

Compute $SE$, then the margin of error and 95% CI for $\mu_d$. $SE = \dfrac{s_d}{\sqrt{n}} = \dfrac{1.23}{\sqrt{10}} \approx 0.389$; $ME = 2.262 \times 0.389 \approx 0.88$; 95% CI = $1.58 \pm 0.88 \approx (0.7, 2.46)$ hours.
Interpret and check consistency with the 13a hypothesis test. We’re 95% confident Drug 2 gives, on average, between about 0.7 and 2.46 hours more extra sleep than Drug 1 — roughly 40 minutes to 2.5 hours. The CI excludes 0, consistent with rejecting $H_0$ in 13a.
State the scope of inference. Scope: each patient is their own control (removes between-patient differences). But order wasn’t randomized (alternate nights) → order/carryover effects aren’t ruled out; and no random sample → strongest for patients like this group.

With conditions noted, build and interpret the ninety-five percent confidence interval. Work through the three steps, and advance when you are ready to see each answer. Start with the standard error: the standard deviation of the differences over the square root of the sample size — one point two three over the square root of ten — which is about zero point three nine hours. The t critical value for ninety-five percent confidence with nine degrees of freedom is about two point two six two — noticeably larger than the one point nine six we’d use with very large samples, because the t-distribution has thicker tails when degrees of freedom are small. Margin of error is t-star times the standard error — two point two six two times zero point three nine, about zero point eight eight hours. The ninety-five percent confidence interval is the mean difference of one point five eight plus or minus zero point eight eight, giving an interval from about zero point seven hours up to about two point four six hours. We’re ninety-five percent confident that the true mean extra sleep produced by drug two over drug one is between about zero point seven and two point four six hours per patient. The interval excludes zero, consistent with rejecting the null in thirteen A. The whole interval sits above zero — drug two reliably produces more sleep than drug one in this study — with the typical gap somewhere between forty minutes and two and a half hours. That’s a substantial range, reflecting the small sample size and the variability across patients. Finally, the scope. Because each patient took both drugs, each person serves as their own control, which removes the between-patient differences — that’s the strength of the paired design. But the drugs were given on alternate nights, and we don’t know that the order was randomized, so we can’t fully separate the drug effect from order or carryover effects — be cautious about saying the drugs alone caused the gap. And the patients weren’t a random sample of any broader population, so the conclusion is strongest for patients similar to this group.

Factors Affecting Tests and CIs

H-tests — larger $|T|$ → smaller p-value

$T = \dfrac{\bar{x}_d}{SE}$

Larger effect size $|\bar{x}_d|$ → larger $|T|$
Larger sample size $n$ (pairs) → smaller SE → larger $|T|$
Smaller variability $s_d$ → smaller SE → larger $|T|$

CIs — narrower = more precise

$\bar{x}_d \pm t^* \times SE$

Larger sample size $n$ → smaller SE and smaller $t^*$ → narrower CI
Smaller variability $s_d$ → smaller SE → narrower CI
Lower confidence level → smaller $t^*$ → narrower CI

Stepping back, let’s pull together the factors that move our two inference quantities for a paired mean. For the hypothesis test, the strength of evidence — how large T is, and how small the p-value — depends on three things. The first is effect size: a bigger mean difference pushes T further from zero. The second is sample size: more pairs shrink the standard error, and a smaller standard error inflates T. The third is variability: a smaller spread in the differences also shrinks the standard error and inflates T. For the confidence interval, those same factors act through the standard error, plus one more. More pairs shrink the standard error and also nudge t-star down, since more degrees of freedom move t-star toward z-star, and both effects narrow the interval. Less variability in the differences shrinks the standard error and narrows the interval. The confidence level is the factor unique to the interval: a lower level uses a smaller t-star and gives a narrower interval. That third factor — the spread of the differences — is the whole reason we pair. The next slide makes that concrete.

Why Pairing Pays Off

Imagine analyzing the textbooks the wrong way — as two independent samples of 68 prices (ignore which book is which).

Books vary enormously: prices run from about $7 to $229 (SD ≈ $57). But the two stores track each other — a pricey book is pricey at both, a cheap book cheap at both.

Which analysis gives the smaller SE — and why?

Paired wins, by a lot. Subtracting within each book removes the $7–$229 book-to-book variation, leaving only the $13 spread in the differences: paired $SE \approx \$1.63$ vs. $\approx \$9.52$ if treated as independent — about 6× larger. Ignoring the pairing would bury the real $3.58 gap in noise.

Here’s why pairing is worth the trouble — made concrete on the textbooks. Suppose we analyzed them the wrong way: as two independent samples of sixty-eight prices, throwing away the information about which book is which. The books themselves vary enormously — prices run from about seven dollars to two hundred twenty-nine, a standard deviation around fifty-seven dollars. But the two stores track each other closely: a pricey book is pricey at both, and a cheap book is cheap at both. Pause and think about which analysis gives the smaller standard error, and why. Pairing wins, and by a lot. When we take the within-book difference, that huge seven-to-two-hundred-twenty-nine-dollar variation subtracts away — both prices for a book rise and fall together — and we’re left with just the thirteen-dollar spread in the differences. The paired standard error is about one dollar sixty-three. If we had ignored the pairing and treated the two columns as independent samples, the standard error would be about nine dollars fifty-two — roughly six times larger. Same data either way, but ignoring the pairing would bury the real three-dollar-fifty-eight-cent gap in noise. That’s the payoff: pairing works when the two measurements move together, which is exactly the case here.

Strength of Evidence — Predict

Textbooks recap: $n = 68$ pairs, $\bar{x}_d \approx 3.58$, $s_d \approx 13.42$ → $T \approx 2.2$, p-value $\approx 0.031$.

For each variant (others held constant), does the evidence become stronger or weaker?

The researchers had sampled 272 textbooks instead of 68 (same $\bar{x}_d$, same $s_d$). Stronger. Quadrupling $n$ halves SE → $|T|$ doubles. The original p-value of about 0.03 would drop well below 0.001.
The observed mean difference had been $7 per book instead of about $3.58 (same $n$, same $s_d$). Stronger. Larger $|\bar{x}_d|$ → larger $|T|$ directly.
The differences had been more variable: $s_d \approx \$25$ instead of $\$13.42$ (same $n$, same $\bar{x}_d$). Weaker. Larger $s_d$ → larger SE → smaller $|T|$. Pairing controls for between-subject variability, but the random scatter from book to book in the price difference itself still hurts.

Some prediction practice, this time about the strength of evidence from the hypothesis test — how large T is and how small the p-value. The original textbook study had sixty-eight pairs, a mean difference about three dollars fifty-eight cents, and a standard deviation of the differences about thirteen dollars forty-two cents — giving T about two point two zero with a p-value about zero point zero three one in the thirteen A test. For each variant, predict whether that hypothesis-test evidence becomes stronger or weaker. Pause and think before advancing. Quadrupling the sample to two hundred seventy-two textbooks shrinks the standard error by half, which doubles T — so the evidence becomes stronger. A p-value of about zero point zero three one would drop well below zero point zero zero one. A mean difference of seven dollars per book widens the numerator of T directly, so the evidence becomes stronger. A larger spread in the differences — a standard deviation of twenty-five dollars — pushes the standard error up, which shrinks T, so the evidence becomes weaker. Worth flagging — pairing controls for between-subject variability, but the random scatter from book to book in the price difference itself still hurts the test.

CI Width — Predict

Textbooks recap: 95% CI for $\mu_d$ = $(0.33,\ 6.83)$ dollars — width about $6.5.

For each variant (others held constant), does the CI become narrower or wider?

The researchers had sampled 272 textbooks instead of 68 (same $\bar{x}_d$, same $s_d$). Narrower. Larger $n$ → smaller SE and slightly smaller $t^*$ → narrower CI.
They build a 90% CI instead of 95%. Narrower. Lower confidence → smaller $t^*$ → narrower CI. (Less confidence buys precision.)
The differences had been more variable: $s_d \approx \$25$ instead of $\$13.42$. Wider. Larger $s_d$ → larger SE → wider CI.

The Pattern, One Last Time

Across the model-based topics (6, 10, 11, 12, 13), every inference question has the same shape:

Question	Tool
Is there a difference / departure from null?	Test statistic = (observed − null) / SE; reference distribution (normal or t)
How big is the effect?	CI = statistic ± multiplier × SE

Only two things change topic-to-topic:

The statistic (point estimate): $\hat{p}$, $\hat{p}_1 - \hat{p}_2$, $\bar{x}$, $\bar{x}_1 - \bar{x}_2$, $\bar{x}_d$.
The SE formula, which captures sampling variability for that statistic.

Conditions, reference distribution, and decision rules follow the same pattern every time.

Standing back across the model-based topics — six, ten, eleven, twelve, and thirteen — every inference question we’ve asked has the same shape. The first question, is there a difference or a departure from the null, is answered by a test statistic of observed minus null over standard error, compared to a reference distribution — normal or t. The second question, how big is the effect, is answered by a confidence interval — statistic plus or minus a multiplier times the standard error. Only two things change topic to topic. The statistic itself — p-hat, the difference of two p-hats, x-bar, the difference of two x-bars, or x-bar sub d for paired data. And the standard error formula, which captures the sampling variability of whichever statistic we’re working with. Conditions, reference distribution, and decision rules all follow the same pattern every time.

Looking Ahead

That completes our work on single and paired groups: one proportion, two proportions, one mean, two independent means, and paired means. (Means and proportions aren’t gone for good — they resurface in later topics.)

Next: contingency tables — testing for an association between two categorical variables with more than two levels each. New reference distribution: chi-squared.

References

Introduction to Modern Statistics (2e) textbook by Mine Çetinkaya-Rundel and Johanna Hardin
Section 21.3
Cushny, A. R. & Peebles, A. R. (1905). The action of optical isomers: II hyoscines. The Journal of Physiology, 32, 501–510 — original source of the sleep data.
Student (1908). The probable error of a mean. Biometrika, 6(1) — the paper that introduced the t-test, using the Cushny & Peebles data as its worked example.