Lesson 26: t-Distributions & One-Sample t

(a) Warm-Up

Back in Lesson 16 you found probabilities about a sample mean x̄ using z = (x̄ − μ) / (σ/√n). That formula has a hidden luxury baked into it: it assumes you know the population standard deviation σ.

Be honest — when would you ever actually know σ? If you don't even know the population mean μ (that's the whole reason you're sampling), how would you magically know its standard deviation? You almost never do. In real studies you estimate σ from the same sample, using the sample standard deviation s.

That seems harmless, but it isn't free. s is itself a noisy estimate — it bounces around from sample to sample, especially when n is small. Pretending s is as trustworthy as a known σ would make us overconfident: our intervals would be too narrow and our p-values too small. Statisticians fixed this in 1908 with a new curve — the t-distribution — that builds the extra uncertainty right in. That's today's lesson, and it powers nearly every "mean" problem on the AP exam.

(b) Core Concept

Why t instead of z

When we make inferences about a population mean and we do not know σ, we replace it with the sample SD s. The standardized statistic becomes

t = (x̄ − μ) / (s/√n)

This looks just like a z-score, but the denominator now contains a random quantity, s. Because s varies from sample to sample, t has more variability than a z-score would. Its distribution is no longer the standard Normal — it's a t-distribution.

A t-distribution is:

• Symmetric and bell-shaped, centered at 0 (just like the standard Normal).
• Heavier in the tails than the Normal curve — more area sits far from center, reflecting the extra uncertainty from estimating σ.
• Indexed by a single number, the degrees of freedom, df = n − 1. Each df gives a slightly different curve.
• Approaching the Normal curve as df → ∞. With small samples the tails are noticeably fat; by df = 30 or so the t-curve is nearly Normal, and in the limit they coincide.

The "heavier tails" is the whole point: to capture the same 95% of the area, a t-distribution has to reach farther from center than the Normal does. So the critical value t is always larger than the matching z — and it shrinks toward z* as n grows.

[GRAPH: Two density curves overlaid on the same axes, both symmetric and
centered at 0, horizontal axis from −4 to 4. A solid "Normal (z)" curve is
taller at the peak and pinches in close to the axis near ±3. A dashed
"t, df = 5" curve is slightly shorter at the peak and visibly thicker in
both tails — more area beyond ±2. A short caption arrow points to the right
tail: "t has more area out here → larger t* needed." A second note reads
"As df → ∞, the dashed t-curve merges into the solid Normal curve."]

For reference, the right-tail critical values for 95% confidence show the pattern clearly:

Every t* exceeds 1.960, and they march down toward it as df climbs.

The three conditions (every t procedure)

Before any t-interval or t-test, check:

1. Random. The data come from a random sample or randomized experiment.
2. 10% condition (independence). When sampling without replacement, n ≤ 0.10N so observations are effectively independent.
3. Normal / Large Sample. The sampling distribution of x̄ must be approximately Normal. You satisfy this any one of three ways:

- The problem states the population is Normal, or

- The sample is large, n ≥ 30 (Central Limit Theorem), or

- The sample is small but a graph of the sample data (dotplot, histogram, boxplot) shows no strong skew and no outliers.

That third condition is the one AP readers scrutinize. With a small n, you must actually appeal to a graph — you cannot just assert Normality.

The one-sample t-interval

To estimate an unknown mean μ:

x̄ ± t·(s/√n) where df = n − 1 and t matches the confidence level.

The piece t*·(s/√n) is the margin of error, and s/√n is the standard error of the mean.

Full PANIC interval. A laptop maker tests the battery life (hours) of a random sample of 20 units of a new model. The sample mean is x̄ = 11.4 hours with s = 1.8 hours; a dotplot of the 20 values is roughly symmetric with no outliers. Construct a 95% confidence interval for the true mean battery life.

P — Parameter. Let μ = the true mean battery life (hours) of all units of this laptop model. We want a 95% confidence interval for μ.

A — Assumptions.

• Random: a random sample of 20 units. ✓
• 10%: 20 units is no more than 10% of all units the company produces (N ≥ 200, clearly true). ✓
• Normal/Large Sample: n = 20 < 30, but the dotplot is roughly symmetric with no outliers, so it's reasonable that x̄ is approximately Normal. ✓

N — Name. One-sample t-interval for a mean, df = n − 1 = 19.

I — Interval.

t*  (95%, df = 19) = 2.093          ← from the t-table / invT
SE  = s/√n = 1.8/√20 = 1.8/4.47214 = 0.40249
ME  = t* · SE = 2.093 × 0.40249 = 0.84241
CI  = 11.4 ± 0.8424 = (10.558, 12.242)

C — Conclusion in context. We are 95% confident that the interval from 10.56 to 12.24 hours captures the true mean battery life of all units of this laptop model.

Confirm on the calculator:

TI-84: STAT → TESTS → TInterval
Inpt: Stats
x̄ = 11.4,  Sx = 1.8,  n = 20,  C-Level = 0.95
Calculate
Output:  (10.558, 12.242)
         x̄ = 11.4,  Sx = 1.8,  n = 20

The calculator's (10.558, 12.242) matches our by-hand interval exactly.

The one-sample t-test

To test a claim H0: μ = μ0:

t = (x̄ − μ0) / (s/√n) with df = n − 1, p-value from tcdf.

Unlike the proportion test, there's no special "use the null value in the SE" rule here — the standard error is always s/√n, because s is our only estimate of σ either way.

Full PANIC test. A city claims the mean morning commute is 30 minutes. A traffic analyst suspects it's actually longer. She times a random sample of 15 morning commutes and gets x̄ = 33.2 minutes, s = 6.5 minutes; a boxplot of the 15 times shows no outliers and only mild skew. Test at α = 0.05.

P — Parameter and hypotheses. Let μ = the true mean morning commute time (minutes).

H0: μ = 30 vs. Ha: μ > 30 (one-sided, right tail — "longer").

A — Assumptions.

• Random: a random sample of 15 commutes. ✓
• 10%: 15 commutes is ≤ 10% of all morning commutes in the city. ✓
• Normal/Large Sample: n = 15 < 30, but the boxplot shows no outliers and only mild skew, so it's reasonable to treat x̄ as approximately Normal. ✓

N — Name. One-sample t-test for a mean, df = 14, α = 0.05.

I — Test statistic and p-value.

SE = s/√n = 6.5/√15 = 6.5/3.87298 = 1.67829
t  = (x̄ − μ0)/SE = (33.2 − 30)/1.67829 = 3.2/1.67829 = 1.9067
df = 15 − 1 = 14
p-value = tcdf(1.9067, 1E99, 14) = 0.0387      ← right tail, Ha: μ > 30

C — Conclusion in context. Because the p-value ≈ 0.0387 < α = 0.05, we reject H0. There is convincing statistical evidence that the true mean morning commute time is greater than 30 minutes.

Confirm on the calculator:

TI-84: STAT → TESTS → T-Test
Inpt: Stats
μ0 = 30,  x̄ = 33.2,  Sx = 6.5,  n = 15
μ:  > μ0      (right tail, matches Ha)
Calculate
Output:  t = 1.9067,  p = 0.0387,  x̄ = 33.2,  Sx = 6.5,  n = 15

The calculator's t = 1.9067 and p = 0.0387 match our by-hand work exactly.

If you're given raw data instead of summary stats, choose Inpt: Data, point the routine at your list (e.g., List: L1), and the calculator computes x̄ and Sx for you before running the test.

(c) Worked Examples

Example 1 — Full PANIC t-interval (easy–medium)

Problem. A barista records the brew temperature (°F) of a random sample of 12 espresso shots: x̄ = 152.5, s = 4.2. A dotplot of the 12 temperatures is roughly symmetric with no outliers. Construct a 90% confidence interval for the true mean brew temperature.

P. Let μ = the true mean brew temperature (°F) of all this machine's espresso shots.

A. Random ✓ (random sample). 10% ✓ (12 ≤ 10% of all shots the machine will make). Normal/Large Sample: n = 12 < 30, but the dotplot is roughly symmetric with no outliers, so x̄ is approximately Normal. ✓

N. One-sample t-interval, df = 11.

I.

t*  (90%, df = 11) = 1.7959
SE  = 4.2/√12 = 4.2/3.46410 = 1.21244
ME  = 1.7959 × 1.21244 = 2.1774
CI  = 152.5 ± 2.1774 = (150.32, 154.68)
TInterval (x̄=152.5, Sx=4.2, n=12, C=.90):  (150.32, 154.68)

C. We are 90% confident that the true mean brew temperature is between 150.32°F and 154.68°F.

Example 2 — Full PANIC t-test (medium)

Problem. A cereal company advertises a net weight of 16 oz. A consumer group suspects boxes are underfilled. A random sample of 10 boxes gives x̄ = 15.6 oz, s = 0.55 oz; a histogram of the 10 weights shows no strong skew or outliers. Test at α = 0.05.

P. Let μ = the true mean net weight (oz) of all boxes.

H0: μ = 16 vs. Ha: μ < 16 (left-tailed — "underfilled").

A. Random ✓. 10% ✓ (10 ≤ 10% of all boxes produced). Normal/Large Sample: n = 10 < 30, but the histogram shows no strong skew or outliers, so x̄ is approximately Normal. ✓

N. One-sample t-test, df = 9, α = 0.05.

I.

SE = 0.55/√10 = 0.55/3.16228 = 0.17393
t  = (15.6 − 16)/0.17393 = −0.4/0.17393 = −2.2998
p-value = tcdf(−1E99, −2.2998, 9) = 0.0235     ← left tail, Ha: μ < 16
T-Test (μ0=16, x̄=15.6, Sx=0.55, n=10, μ<μ0):  t = −2.2998,  p = 0.0235

C. Because the p-value ≈ 0.0235 < 0.05, we reject H0. There is convincing evidence that the true mean net weight is less than 16 oz — the boxes are underfilled on average.

Example 3 — Checking the Normal/Large Sample condition (short, conceptual)

Problem. For each scenario, state whether the Normal/Large Sample condition for a one-sample t procedure is satisfied, and why.

(a) n = 8; the population is stated to be Normally distributed.

(b) n = 45; the sample data are strongly right-skewed.

(c) n = 12; a dotplot of the sample shows a clear high outlier and right skew.

Solution.

• (a) Satisfied. With the population stated Normal, x̄ is exactly Normal for any n, even n = 8.
• (b) Satisfied. n = 45 ≥ 30, so by the Central Limit Theorem the sampling distribution of x̄ is approximately Normal despite the skew in the raw data. (Large n rescues skew.)
• (c) Not satisfied. n = 12 < 30, and the only way to justify Normality at small n is a graph free of strong skew and outliers — but this graph has both a clear outlier and right skew. A t procedure is not appropriate here.

Example 4 — Interpreting an interval (AP-style)

Problem. Researchers measure resting heart rate (bpm) for a random sample of 25 adults and report a 95% one-sample t-interval of (63.41, 73.39) bpm (from x̄ = 68.4, s = 12.1, df = 24, t = 2.064). (a) Interpret the interval. (b) Interpret the confidence level*. (c) A trainer claims the true mean resting heart rate is 60 bpm. Does the interval support that claim?

Solution.

• (a) Interval: We are 95% confident that the interval from 63.41 to 73.39 bpm captures the true mean resting heart rate of all adults in this population.
• (b) Confidence level: If we repeated this sampling process many times and built a 95% t-interval each time, about 95% of those intervals would capture the true mean resting heart rate. (The 95% describes the method's long-run capture rate — not the probability for this one interval.)
• (c) Claim: The value 60 bpm is not inside (63.41, 73.39), so 60 is not a plausible value for the true mean. The data give convincing evidence against the trainer's claim of 60 bpm; the true mean appears to be higher.

(d) Common Mistakes

1. Using z instead of t. When σ is unknown and you estimate it with s, the critical value must come from the t-distribution with df = n − 1, not the Normal. Using z = 1.96 for a 95% interval makes your margin of error too small (under-covers). Reach for t whenever you see s.

1. Wrong degrees of freedom. It's df = n − 1, not n. With n = 20, df = 19. Using df = 20 (or, worse, df = n for some other procedure) gives the wrong t* and the wrong p-value. Single sample ⇒ df = n − 1, every time.

1. Using s when σ is actually given. Read carefully. If a problem genuinely states the population standard deviation σ (rare, but it happens in textbook setups), you'd use a z procedure, not t. The trigger for t is that σ is unknown and you're plugging in the sample SD s. Don't reflexively run a t-test if the problem handed you σ.

1. Mis-stating the Normal/Large Sample condition. With small n, you cannot just write "Normal ✓." You must either cite that the population is stated Normal, that n ≥ 30 (CLT), or that a graph of the sample shows no strong skew/outliers. "The sample is large enough" when n = 12 earns nothing.

1. Confusing the interval and test standard errors. For means, both the t-interval and t-test use the same standard error s/√n. (There is no "null value" substitution like there was for proportions.) Don't overthink it.

(e) Practice Problems

MC 1. Compared with the standard Normal distribution, a t-distribution with df = 8:

• (A) Has the same shape exactly.
• (B) Is skewed to the right.
• (C) Is symmetric but has heavier tails.
• (D) Has lighter tails and a taller peak.

MC 2. A one-sample t-interval is built from a sample of size n = 16. The degrees of freedom are:

• (A) 16
• (B) 15
• (C) 14
• (D) 17

MC 3. As the degrees of freedom increase, the t-distribution:

• (A) Becomes more skewed.
• (B) Approaches the standard Normal distribution.
• (C) Develops even heavier tails.
• (D) Shifts its center to the right.

MC 4. For a 95% confidence interval with df = 14, the critical value t* is closest to:

• (A) 1.645
• (B) 1.960
• (C) 2.145
• (D) 2.624

MC 5. Which situation calls for a t procedure rather than a z procedure for a mean?

• (A) The population σ is known and n = 50.
• (B) The population σ is unknown and estimated by the sample SD s.
• (C) The data are proportions, not means.
• (D) The sample size exceeds 30 and σ is known.

MC 6. A one-sample t-test for H0: μ = 50 vs. Ha: μ ≠ 50 gives t = 2.40 with df = 12. The p-value is closest to:

• (A) 0.017
• (B) 0.034
• (C) 0.966
• (D) 0.983

MC 7. A right-tailed one-sample t-test (Ha: μ > μ0) gives t = 1.85 with df = 9. The p-value is closest to:

• (A) 0.049
• (B) 0.098
• (C) 0.951
• (D) 0.025

MC 8. A 95% t-interval for a mean is (12.3, 18.7). Which statement is the correct interpretation?

• (A) 95% of the data lie between 12.3 and 18.7.
• (B) There is a 95% probability the true mean is between 12.3 and 18.7.
• (C) We are 95% confident the interval (12.3, 18.7) captures the true population mean.
• (D) 95% of all sample means fall between 12.3 and 18.7.

MC 9. Increasing the sample size n (holding s and x̄ fixed) does which of the following to a t-interval's margin of error?

• (A) Increases it.
• (B) Decreases it.
• (C) Leaves it unchanged.
• (D) Decreases t* only, with no effect on the margin.

Short Answer 10 (in context). A botanist measures the heights (cm) of a random sample of 8 seedlings: x̄ = 24.5, s = 3.1. A dotplot is roughly symmetric with no outliers. Construct and interpret a 95% confidence interval for the true mean seedling height (full PANIC).

Short Answer 11 (in context). A bottling line is supposed to dispense μ = 5.0 L. A technician suspects the true mean differs from 5.0 in either direction. A random sample of 22 fills gives x̄ = 4.85 L, s = 0.62 L; a histogram shows no strong skew or outliers. Carry out a full PANIC t-test at α = 0.05.

Short Answer 12. A study reports a one-sample t-test with n = 30, x̄ = 0.42 s, s = 0.08 s, testing H0: μ = 0.45 vs. Ha: μ < 0.45. Compute df, the test statistic, and the p-value, and state the decision at α = 0.05.

Short Answer 13. Explain in one or two sentences why the critical value t for a 95% interval is always larger than z = 1.96.

Short Answer 14. A sample of n = 18 is strongly right-skewed with a clear high outlier. A student wants to run a one-sample t-interval. Is the Normal/Large Sample condition met? Explain.

Short Answer 15. A 90% t-interval for a mean is (40.2, 47.8). What is the point estimate x̄, and what is the margin of error?

(f) FRQ Practice (10 points)

Statistical Practices 3 (Analyze) + 4 (Interpret)

A nutrition watchdog is concerned that a popular brand of canned soup contains more sodium than the 140 mg per serving printed on the label. An inspector selects a random sample of 16 cans and measures the sodium content (mg) per serving. The sample has a mean of x̄ = 146.2 mg and a standard deviation of s = 11.8 mg. A dotplot of the 16 measurements is roughly symmetric with no outliers.

(a) State the hypotheses for an appropriate significance test, defining any parameter. (2 pts)

(b) Identify the appropriate procedure and verify that the conditions for inference are met. (3 pts)

(c) Calculate the test statistic, the degrees of freedom, and the p-value. (3 pts)

(d) At the α = 0.05 significance level, state your conclusion in the context of the problem. (2 pts)

Model Response

(a) Parameter and hypotheses.

Let μ = the true mean sodium content (mg per serving) of all cans of this brand of soup.

H0: μ = 140 vs. Ha: μ > 140

The alternative is one-sided (right-tailed) because the watchdog suspects the true mean is more than the labeled 140 mg.

(b) Procedure and conditions.

Procedure: a one-sample t-test for a mean (σ is unknown, estimated by s).

• Random: The 16 cans are a random sample. ✓
• 10% condition: 16 cans is no more than 10% of all cans of this brand (N ≥ 160, clearly true). ✓
• Normal/Large Sample: n = 16 < 30, but the dotplot of the 16 measurements is roughly symmetric with no outliers, so it is reasonable that the sampling distribution of x̄ is approximately Normal. ✓

(c) Test statistic, df, and p-value.

SE = s/√n = 11.8/√16 = 11.8/4 = 2.95
t  = (x̄ − μ0)/SE = (146.2 − 140)/2.95 = 6.2/2.95 = 2.1017
df = n − 1 = 16 − 1 = 15
p-value = tcdf(2.1017, 1E99, 15) = 0.0264      ← right tail, Ha: μ > 140
T-Test (μ0=140, x̄=146.2, Sx=11.8, n=16, μ>μ0):  t = 2.1017,  p = 0.0264

(d) Conclusion in context.

Because the p-value ≈ 0.0264 < α = 0.05, we reject H0. There is convincing statistical evidence that the true mean sodium content of this brand of soup is greater than 140 mg per serving (more than the label states).

Point-by-Point Rubric (sums to 10)

Where Students Lose Points

• Part (a): Defining the parameter as a sample quantity ("the mean of the 16 cans") instead of the true population mean — μ is about all cans, not the sample.
• Part (b): Calling it a z-test, or using z*/normalcdf anywhere — σ is unknown, so it must be a t procedure. Also: writing "Normal ✓" at n = 16 with no appeal to the dotplot loses the Normal/Large Sample point.
• Part (c): Using df = 16 instead of 15 — wrong df ⇒ wrong p-value, and forfeits the df point. Reporting the left-tail area (≈ 0.974) instead of the right tail loses the p-value point.
• Part (d): "Reject H0" with no context, saying "we accept Ha" or "we proved the soup has more sodium," or failing to compare the p-value to α explicitly.
• Bare-calculator answers: Writing only t = 2.10, p = 0.026 with no SE setup, df, or labeled conditions can cap part (c) — show SE = s/√n, state df = n − 1, and name the T-Test routine.