Lesson 27: Two-Sample Inference for Means

(a) Warm-Up

In Lesson 26 you ran inference on a single mean: one sample, one x̄, one μ. But just like with proportions, the questions that matter most are comparisons. Does the new fertilizer grow taller plants than the old one? Is reaction time faster with a focus app than without? Do morning commutes take longer than evening ones?

Here's the twist that makes means trickier than proportions, and it's the heart of this whole lesson. There are two completely different ways two groups of numbers can show up:

1. Two separate, independent groups — 20 plants get fertilizer A, a different 20 plants get fertilizer B. The two samples have nothing to do with each other.
2. Paired data — the same 20 plants are measured before and after a treatment, or subjects are matched into natural pairs (twins, left/right hand, identical plots split in two).

These two situations need two different procedures. Run the wrong one and your standard error, your degrees of freedom, and your p-value are all wrong — and the AP exam tests this exact decision constantly. By the end of this lesson, spotting "independent vs. paired" should be automatic.

(b) Core Concept

Two roads, one fork

Everything in this lesson branches at one decision:

Are the two sets of numbers independent (two separate groups), or paired (each value in one group is naturally tied to a specific value in the other)?

• Independent → use the two-sample t procedure (2-SampTInt / 2-SampTTest), with parameter μ₁ − μ₂.
• Paired → compute the differences d for each pair, then run a one-sample t procedure (TInterval / T-Test) on those differences, with parameter μ_d.

Get the fork right and the rest is mechanical. Get it wrong and everything downstream collapses. We'll build both roads fully.

ROAD 1 — Independent two-sample t

The parameter and statistic. Two populations, each with its own true mean: μ₁ and μ₂. From each we draw an independent sample and compute x̄₁ and x̄₂. Our estimate of the difference μ₁ − μ₂ is simply x̄₁ − x̄₂.

The standard error. Because the two samples are independent, their variances add:

SE = sqrt( s₁²/n₁ + s₂²/n₂ )

Notice we use each group's own s and n. AP Statistics uses the unpooled two-sample t — we do NOT combine the two s values into one pooled estimate. (Pooling assumes the two populations have equal variances, an assumption AP doesn't make. If your calculator asks "Pooled: Yes/No," always choose No.)

Degrees of freedom. The unpooled two-sample t doesn't follow a clean t-distribution, so technology uses the Welch–Satterthwaite approximation — usually a messy decimal. On the AP exam you may either (1) report the calculator's df (preferred, more accurate), or (2) use the conservative df = min(n₁−1, n₂−1) if you're working by hand. The conservative df gives a slightly wider interval / slightly larger p-value, so it's a safe fallback. In worked answers we'll show the calculator df.

The confidence interval:

(x̄₁ − x̄₂) ± t*·sqrt( s₁²/n₁ + s₂²/n₂ )

The test statistic (the null says μ₁ − μ₂ = 0, so the hypothesized difference drops out of the numerator):

t = (x̄₁ − x̄₂) / sqrt( s₁²/n₁ + s₂²/n₂ )

Conditions — check for BOTH groups:

1. Random: Both samples are random samples, OR subjects were randomly assigned to two treatment groups.
2. Independent (10%): The two groups are independent of each other (NOT paired), and when sampling without replacement, each sample is < 10% of its population.
3. Normal / Large Sample: For each group, the population is Normal, OR the sample is large (n ≥ 30), OR a graph of the sample data shows no strong skew/outliers. Check this separately for each group.

FULL PANIC — Independent two-sample TEST

Scenario. A school pilots a new teaching method. A random sample of 22 students taught with the new method scores a mean of 84.6 (s = 7.2) on the final; an independent random sample of 24 students taught the traditional way scores a mean of 79.8 (s = 8.1). Do these data give convincing evidence at α = 0.05 that the new method produces a higher mean score? (Sample dotplots show roughly symmetric data with no outliers.)

P — Parameter. Let μ₁ = true mean final score for students taught with the new method, and μ₂ = true mean score for the traditional method. We test:

• H₀: μ₁ = μ₂ (equivalently μ₁ − μ₂ = 0)
• Hₐ: μ₁ > μ₂

A — Assumptions/Conditions.

• Random: Both samples are random samples. ✓
• Independent: Two separate, independent groups of students; each student is in only one group. Each sample is plausibly < 10% of all such students. ✓
• Normal/Large Sample: Dotplots of each group are roughly symmetric with no outliers, so the t procedure is appropriate. ✓

N — Name the procedure. A two-sample t-test for μ₁ − μ₂ (unpooled), 2-SampTTest.

I — Test.

x̄₁ − x̄₂ = 84.6 − 79.8 = 4.8

SE = sqrt( 7.2²/22 + 8.1²/24 )
   = sqrt( 51.84/22 + 65.61/24 )
   = sqrt( 2.35636 + 2.73375 )
   = sqrt( 5.09011 )
   = 2.25613

t = 4.8 / 2.25613 = 2.128

df (technology, Welch) = 43.96
one-sided p-value = P(t > 2.128 | df = 43.96) = 0.0195

TI-84: STAT → TESTS → 2-SampTTest
Input: x̄1=84.6, Sx1=7.2, n1=22, x̄2=79.8, Sx2=8.1, n2=24
       μ1 > μ2,  Pooled: No
Output: t = 2.1275,  p = 0.0195,  df = 43.964

(Conservative-df check: with df = min(21, 23) = 21, p ≈ 0.0226 — still < 0.05, same decision.)

C — Conclusion in context. Since the p-value 0.0195 < α = 0.05, we reject H₀. We have convincing evidence that the new teaching method produces a higher mean final score than the traditional method.

ROAD 2 — Matched-pairs t

When the data are paired, the two columns are NOT independent — each value in column 1 is tied to a specific value in column 2 (same subject, or matched partners). Treating them as two independent groups throws away the pairing and inflates the standard error. Instead:

Step 1 — Collapse each pair to a single difference d = (value 1) − (value 2). You now have one list of differences.

Step 2 — Run a ONE-sample t procedure on the differences. The parameter is μ_d, the true mean difference. Everything from Lesson 26 applies — it's a one-sample problem now.

• Interval: d̄ ± t·(s_d / sqrt(n)) with df = n − 1 (n = number of pairs*).
• Test of H₀: μ_d = 0: t = d̄ / (s_d / sqrt(n)), df = n − 1.

Conditions (one-sample, on the differences):

1. Random: The pairs come from a random sample, or treatments were randomly assigned within each pair.
2. 10%: The number of pairs is < 10% of the population of pairs (if sampling without replacement).
3. Normal/Large Sample: The differences are from a Normal population, OR n ≥ 30 pairs, OR a graph of the differences shows no strong skew/outliers. (Check the differences, not the two original columns.)

FULL PANIC — Matched-pairs analysis

Scenario. A researcher tests whether a focus app lowers reaction time. The same 12 subjects each complete a reaction-time task without the app and with the app (order randomized). Reaction times (ms):

Do the data give convincing evidence at α = 0.05 that the app lowers mean reaction time? (A positive d means the app reduced the time.)

P — Parameter. Let μ_d = the true mean difference (without − with) in reaction time. We test:

• H₀: μ_d = 0
• Hₐ: μ_d > 0 (without is higher → the app lowered the time)

A — Assumptions/Conditions.

• Random: The order of the two conditions was randomized for each subject. ✓
• Independent / 10%: The 12 subjects are < 10% of all potential subjects; reasonable. ✓
• Normal/Large Sample: A dotplot of the 12 differences is roughly symmetric with no strong outliers, so the t procedure is appropriate. ✓ (This is paired data — same subjects measured twice — so we analyze the differences with a one-sample procedure.)

N — Name the procedure. A matched-pairs t-test, i.e., a one-sample t-test on the differences (T-Test on the list d).

I — Test.

The 12 differences: 14, −3, 15, 15, 3, 14, 12, 6, 17, 8, 2, 16

d̄ = 119/12 = 9.9167
s_d = 6.5569      (sample SD of the differences)
n = 12  →  df = 11

SE = s_d / sqrt(n) = 6.5569 / sqrt(12) = 6.5569 / 3.4641 = 1.89280

t = d̄ / SE = 9.9167 / 1.89280 = 5.239
one-sided p-value = P(t > 5.239 | df = 11) ≈ 0.00014

TI-84: enter the 12 differences in L3 (or compute L1 − L2 → L3)
STAT → TESTS → T-Test
Input: Data, List: L3, μ0 = 0, μ > μ0
Output: t = 5.2391,  p = 1.4×10⁻⁴,  d̄ = 9.9167,  Sx = 6.5569,  n = 12

C — Conclusion in context. Since the p-value ≈ 0.0001 < α = 0.05, we reject H₀. We have convincing evidence that the focus app lowers the mean reaction time. (We could also report a 95% CI for μ_d: 9.9167 ± 2.201(1.89280) = (5.75, 14.08) ms — the whole interval is positive, agreeing with the test.)

The decision box — INDEPENDENT vs. PAIRED, how to tell

Ask: "Is each number in one group naturally tied to one specific number in the other group?"

| If… | then it's… | use… | parameter |

|---|---|---|---|

| Two separate groups of subjects (different people/things in each) | Independent | 2-SampTTest / 2-SampTInt on x̄₁, x̄₂ | μ₁ − μ₂ |

| The same subjects measured twice (before/after, two conditions) | Paired | one-sample t on the differences | μ_d |

| Subjects matched into pairs (twins, split plots, left/right) | Paired | one-sample t on the differences | μ_d |

Tells for paired data: same subjects measured before and after; "each person did both"; natural pairs (twins, two halves of one field, two eyes); the two columns must have equal sample sizes and rows that line up one-to-one.

Tells for independent data: two different random samples; two treatment groups from random assignment; the two sample sizes can differ (n₁ ≠ n₂); there's no way to match a specific subject in group 1 to a specific subject in group 2.

If n₁ ≠ n₂, it cannot be paired (you can't pair up unequal lists). But equal sample sizes do not automatically mean paired — you still need a real one-to-one tie between the rows.

(c) Worked Examples

Worked Example 1 — Independent two-sample CI (easy)

Problem. Two brands of rechargeable battery are tested. A random sample of 15 Brand-A batteries lasts a mean of 8.4 hours (s = 1.2); an independent random sample of 18 Brand-B batteries lasts a mean of 7.1 hours (s = 1.5). Build a 95% confidence interval for μ_A − μ_B. (Both samples look roughly symmetric.)

Strategy. Two separate groups → independent two-sample t interval, unpooled SE.

Solution.

x̄_A − x̄_B = 8.4 − 7.1 = 1.3
SE = sqrt( 1.2²/15 + 1.5²/18 ) = sqrt( 0.096 + 0.125 ) = sqrt(0.221) = 0.47011
df (technology) = 30.96  →  t* = 2.0396
95% CI = 1.3 ± 2.0396(0.47011) = 1.3 ± 0.959 = (0.341, 2.259)

TI-84: 2-SampTInt → x̄1=8.4, Sx1=1.2, n1=15, x̄2=7.1, Sx2=1.5, n2=18, C=.95, Pooled:No
Output: (0.3412, 2.2588),  df = 30.963

(Conservative df = min(14,17) = 14, t = 2.145 → CI (0.292, 2.308), slightly wider.)*

Interpretation. We are 95% confident the true difference in mean battery life (A − B) is between 0.34 and 2.26 hours. Because the interval is entirely above 0, we have convincing evidence Brand A lasts longer on average.

Worked Example 2 — Independent two-sample test (medium)

Problem. A commuter compares two routes. A random sample of 30 trips on Route A averages 24.5 min (s = 4.2); an independent sample of 28 trips on Route B averages 27.1 min (s = 5.0). Is there convincing evidence at α = 0.05 that Route A is faster on average?

Strategy. Two independent samples → 2-SampTTest, Hₐ: μ_A < μ_B.

Solution.

P:  H₀: μ_A = μ_B    Hₐ: μ_A < μ_B
A:  random trips; routes independent; both n large (≥30 and ≈28) ✓
N:  two-sample t-test (unpooled)
I:  SE = sqrt(4.2²/30 + 5.0²/28) = sqrt(0.588 + 0.89286) = sqrt(1.48086) = 1.21690
    t = (24.5 − 27.1)/1.21690 = −2.6/1.21690 = −2.137
    df (tech) = 52.91 → one-sided p = P(t < −2.137) = 0.0186

TI-84: 2-SampTTest → ... μ1 < μ2, Pooled:No → t = −2.1366, p = 0.0186, df = 52.908

Conclusion. Since p = 0.0186 < 0.05, reject H₀. There is convincing evidence Route A has a lower mean travel time than Route B.

Worked Example 3 — Matched-pairs t (AP-style)

Problem. Eight patients have their systolic blood pressure measured before and after taking a new drug. Build a 95% CI for the mean reduction (μ_d, before − after).

Strategy. Same patients measured twice → paired. Run a one-sample t interval on the 8 differences.

Solution.

d: 7, 3, 3, 9, 5, 5, 2, 9
d̄ = 43/8 = 5.375     s_d = 2.7223     n = 8 → df = 7
SE = 2.7223/sqrt(8) = 2.7223/2.8284 = 0.96247
t* (df=7, 95%) = 2.3646
95% CI = 5.375 ± 2.3646(0.96247) = 5.375 ± 2.276 = (3.099, 7.651)

TI-84: put the 8 differences in L1 → TInterval → Data, L1, C=.95
Output: (3.0991, 7.6509),  d̄ = 5.375,  Sx = 2.7223,  n = 8

Interpretation. We are 95% confident the true mean reduction in systolic BP (before − after) is between 3.10 and 7.65 points. The interval is entirely above 0, so the drug is associated with a genuine mean decrease.

Worked Example 4 — Classify: independent or paired?

Problem. For each scenario, state whether to use an independent two-sample t or a matched-pairs t.

(i) A trainer records the 40-yard dash time of 15 athletes, has them do a 6-week program, then records each athlete's time again.

(ii) Researchers randomly assign 50 volunteers — 25 to a low-carb diet, 25 to a low-fat diet — and compare mean weight loss.

(iii) To compare two tire brands, each of 20 cars gets Brand A on the left side and Brand B on the right side; tread wear is measured on each tire.

(iv) Two independent random samples — 40 men and 35 women — report mean weekly screen time.

Solution.

• (i) Paired — the same athletes measured before and after. Analyze the 15 differences with a one-sample t.
• (ii) Independent — two separate groups of different people from random assignment. Two-sample t on μ_lowcarb − μ_lowfat.
• (iii) Paired — each car gives a natural pair (left vs. right on the same car), controlling for the car. Analyze the 20 differences (A − B).
• (iv) Independent — two distinct samples with unequal sizes (40 vs. 35); there's no way to tie a specific man to a specific woman. Two-sample t.

Key tell: if the two lists must line up one-to-one (same subject or matched partner), it's paired; if they're two separate groups (and especially if n₁ ≠ n₂), it's independent.

(d) Common Mistakes

1. Running a two-sample test on paired data (or vice versa). The #1 trap. If the same subjects are measured twice, you must collapse to differences and run a one-sample procedure on μ_d. Treating paired data as two independent groups uses the wrong (too large) standard error and the wrong df, usually killing a real effect. Conversely, you can't "pair up" two genuinely separate groups — there's nothing to subtract row-by-row.

2. Using the wrong standard error. For independent groups the SE is sqrt(s₁²/n₁ + s₂²/n₂) — variances add under the radical. Don't write sqrt(s₁²/n₁) + sqrt(s₂²/n₂) (adding SDs), and don't pool the variances — AP uses the unpooled t (Pooled: No).

3. Forgetting to actually analyze the differences for paired data. Some students recognize the pairing, then still plug x̄₁, x̄₂, s₁, s₂ into a two-sample routine. You must compute the list d, then use d̄ and s_d. The parameter is μ_d, and df = (number of pairs) − 1.

4. Mis-stating the conditions. For paired data, the Normal/Large-Sample condition is about the differences, not the two original columns. For independent data, you must check Normal/Large Sample separately for both groups, and the two groups must be independent of each other (not paired).

5. Reporting the wrong df or hypotheses. For matched pairs, df = n − 1 (one-sample), not some two-sample formula. State H₀: μ_d = 0 for pairs (not μ₁ = μ₂). For independent groups, use the calculator's Welch df or the conservative min(n₁−1, n₂−1) — never n₁ + n₂ − 2 (that's the pooled df AP doesn't use).

(e) Practice Problems

1 (MC). Which standard error is correct for an independent two-sample t for μ₁ − μ₂?

• (A) sqrt(s₁²/n₁) + sqrt(s₂²/n₂)
• (B) sqrt( s₁²/n₁ + s₂²/n₂ )
• (C) sqrt( s_p²(1/n₁ + 1/n₂) ) with a pooled s_p
• (D) s_d / sqrt(n)

2 (MC). For independent samples with s₁ = 12, n₁ = 20, s₂ = 15, n₂ = 25, the standard error of x̄₁ − x̄₂ is closest to:

• (A) 2.85 (B) 4.02 (C) 8.10 (D) 16.20

3 (MC). A study measures the same 18 runners' heart rates before and after a race. The correct procedure is:

• (A) A two-sample t-test on μ₁ − μ₂.
• (B) A one-sample t-test on the 18 differences (μ_d).
• (C) A two-proportion z-test.
• (D) A chi-square test.

4 (MC). When AP students compute an independent two-sample t by hand, the safe conservative degrees of freedom is:

• (A) n₁ + n₂ − 2
• (B) n₁ + n₂ − 1
• (C) min(n₁ − 1, n₂ − 1)
• (D) max(n₁ − 1, n₂ − 1)

5 (MC). Independent samples: x̄₁ = 50, s₁ = 8, n₁ = 16 and x̄₂ = 46, s₂ = 10, n₂ = 20. The test statistic t for H₀: μ₁ = μ₂ is closest to:

• (A) 0.50 (B) 1.33 (C) 2.00 (D) 4.00

6 (MC). For a matched-pairs t-test with 25 pairs and d̄ = 3.2, s_d = 4.0, the test statistic and df are:

• (A) t = 4.0, df = 24
• (B) t = 4.0, df = 25
• (C) t = 0.8, df = 24
• (D) t = 4.0, df = 49

7 (MC). Two studies compare the same two diets. Study X uses two separate groups of 30 people each; Study Y measures the same 30 people on both diets (with a washout period). Which is true?

• (A) Both should use a two-sample t.
• (B) Both should use a matched-pairs t.
• (C) Study X → two-sample t; Study Y → matched-pairs t.
• (D) Study X → matched-pairs t; Study Y → two-sample t.

8 (MC). A two-sample t-test on a TI-84 asks "Pooled: Yes or No." For AP Statistics you should choose:

• (A) Yes, always.
• (B) No, always (AP uses the unpooled t).
• (C) Yes if n₁ = n₂.
• (D) It doesn't affect the answer.

9 (Short answer, classify). For each, state independent or paired and name the procedure: (i) 12 married couples each report hours of sleep — compare husbands vs. wives; (ii) 60 plots of land — 30 randomly get fertilizer A, 30 get fertilizer B — compare mean yield.

10 (Short answer, in context). Two coffee shops' wait times: Shop A (random sample of 12) has x̄ = 4.2 min, s = 1.1; Shop B (independent random sample of 14) has x̄ = 5.0 min, s = 1.4. Construct a 90% CI for μ_A − μ_B (calculator df = 23.85) and state whether there's convincing evidence of a difference.

11 (Short answer, in context). Seven students take a practice essay, get feedback, and rewrite it; the score change (rewrite − original) for each is: 2.1, −0.4, 1.8, 0.9, 1.2, 2.5, 0.6. Test at α = 0.05 whether the feedback changes mean scores (Hₐ: μ_d ≠ 0).

12 (Short answer). A student has data on the same 20 subjects' cholesterol before and after a diet, but runs a 2-SampTTest using the before-mean and after-mean as two groups. Explain what's wrong and what they should have done.

13 (MC). A 95% CI for μ₁ − μ₂ (independent) comes out to (−1.2, 3.6). The correct conclusion is:

• (A) μ₁ is convincingly larger than μ₂.
• (B) μ₂ is convincingly larger than μ₁.
• (C) No convincing evidence of a difference, since 0 is in the interval.
• (D) The two means are exactly equal.

14 (Short answer). Explain why the matched-pairs design (same subjects twice) often detects a real effect that a two-sample design with the same data would miss.

(f) FRQ Practice (10 points)

A psychology researcher believes that studying in silence improves recall compared with studying with background music. She recruits 10 students. Each student studies one word list in silence and a comparable word list with music playing (which list goes with which condition, and the order, are randomized for each student). The number of words recalled is recorded for each student under each condition.

(a) Carry out an appropriate significance test at α = 0.05 to determine whether there is convincing evidence that students recall more words, on average, in silence than with music. (8 points)

(b) Based on your conclusion in part (a), which type of error — Type I or Type II — could the researcher have made, and what would it mean in context? (2 points)

Model Response

Part (a).

First, compute the difference d = silence − music for each student:

P — Parameter. Let μ_d = the true mean difference in words recalled (silence − music) for students like these. We test:

• H₀: μ_d = 0
• Hₐ: μ_d > 0 (more recalled in silence)

A — Assumptions/Conditions.

• Random: For each student the order/assignment of conditions was randomized. ✓
• Independent / 10%: The 10 students are independent and are < 10% of the population of such students. ✓
• Normal/Large Sample: A dotplot of the 10 differences is roughly symmetric with no strong outliers, so the t procedure is reasonable. ✓ (This is paired data — same student under both conditions — so we analyze the differences.)

N — Name. A matched-pairs t-test (one-sample t-test on the differences d).

I — Test.

d: 3, 2, −1, 3, 3, 3, 1, 4, 1, 4
d̄ = 23/10 = 2.30      s_d = 1.5670      n = 10 → df = 9

SE = s_d/sqrt(n) = 1.5670/sqrt(10) = 1.5670/3.16228 = 0.49554

t = d̄/SE = 2.30/0.49554 = 4.641
one-sided p-value = P(t > 4.641 | df = 9) ≈ 0.00061

TI-84: differences in L1 → T-Test → Data, L1, μ0=0, μ > μ0
Output: t = 4.6414,  p = 6.1×10⁻⁴,  d̄ = 2.3,  Sx = 1.5670,  n = 10

C — Conclusion. Since the p-value ≈ 0.0006 < α = 0.05, we reject H₀. We have convincing evidence that students recall more words, on average, when studying in silence than with music.

Part (b). Because we rejected H₀, the only possible error is a Type I error — rejecting a true null. In context: concluding that silence improves mean recall when in reality there is no difference in mean recall between studying in silence and with music.

Rubric (10 points total)

Part (a) — 8 points:

Part (b) — 2 points:

Where students lose points:

• Running a two-sample t (2-SampTTest) on the silence and music columns — this is the biggest error. It's paired data; you must use the differences. Loses the design point, the SE point, and usually the test-statistic/df points (the wrong SE and df result).
• Wrong hypotheses — writing μ₁ = μ₂ instead of μ_d = 0, or ≠ / < instead of > for the alternative.
• Checking Normality on the two original columns instead of on the differences — costs a condition point.
• Wrong df — using a two-sample df instead of n − 1 = 9.
• Bare conclusion ("reject H₀") with no context or direction — loses the conclusion point.
• Part (b): saying Type II. Since we rejected, only a Type I error is possible. Naming the error without contextual meaning earns only 1 of 2.