A Statistics Mystery

Detective Data: The Case of the Uncertain Mean

Presented by Top Edu Prep · Written by Minho Kim · AP Statistics — Unit 7: Inference for Quantitative Data: Means

Every good detective story starts with a locked door. Ours starts with a locked number — the true population mean, $\mu$ . We can never see it directly. We only get a small pile of evidence: a sample.

Worse still, we usually don't even know how spread out the population really is — the true standard deviation, $\sigma$ , is locked away too. All we have is an estimate from our sample, $$s$$ .

Detective Minho has four case files on the desk today. Each one is solved with the same trusted tool: the t-distribution. Open a file below to begin.

Case File 01

The t-Distribution

Why we need a new curve when σ is unknown, and what "degrees of freedom" really means.

Case File 02

The Lone Sample

One sample, one claim to test. The one-sample t-test and t-interval.

Case File 03

The Paired Suspects

Before and after. Same subjects, measured twice — the matched pairs t-test.

Case File 04

Two Rivals

Two separate groups face off. The two-sample t-test and interval.

Case File 01 — The t-Distribution

Borrowed certainty: what happens when σ goes missing.

Open Case

Detective Minho opens the thinnest file on the desk. Inside: a handful of measurements from a sample, and a note that reads, "Population standard deviation unknown."

Normally, a detective would use the reliable z-distribution — but that tool only works when σ, the true population spread, is known. Without it, Minho must estimate σ using the sample's own standard deviation, $$s$$ . That estimate is a good guess, but a guess adds uncertainty.

To handle that extra uncertainty honestly, statisticians built a new curve: the t-distribution. It looks like the normal curve, but with fatter, taller tails — more room reserved for the unexpected.

◆ The Idea, in Plain Terms

When σ is unknown, we substitute the sample standard deviation $$s$$ for it. This is called standardizing with s instead of σ.
Because $$s$$ is only an estimate (it changes from sample to sample), our test statistic is less predictable than a z-score. The t-distribution's wider tails reflect that extra unpredictability.
How much wider the tails are depends on the sample size, through a number called degrees of freedom (df).
As the sample size $$n$$ grows, $$s$$ becomes a more trustworthy estimate of σ, so the t-distribution shrinks toward the normal curve. With a huge sample, t and z are almost identical.

\[ df = n - 1 \]

One degree of freedom is "spent" estimating the sample mean, leaving n − 1 independent pieces of information to estimate the spread.

◆ Evidence Board: t* vs. z* at 95% Confidence

Notice how the critical value $$t^*$$ shrinks toward $$z^*=1.960$$ as degrees of freedom increase.

df	5	10	15	20	30	z* (∞)
t*	2.571	2.228	2.131	2.086	2.042	1.960

Clue #1

Which interval is wider?

A researcher builds two 95% confidence intervals from samples of the same size — one using a t-distribution with df = 10 (because σ is unknown), and one using the normal distribution (imagine σ happened to be known). Which margin of error is larger?

1

Look up the critical values:

t^*_{df=10} = 2.228

vs.

\(z^* = 1.960\)

.

2

Since the margin of error is

critical\ value \times SE

, and 2.228 > 1.960, the t-interval has the larger margin of error.

Verdict: The t-interval is wider. Not knowing σ costs us precision — that's the price of borrowed certainty.

Clue #2

Find the critical value

A detective needs a 90% confidence interval from a sample of size n = 15. What is t*?

1

Degrees of freedom:

\(df = n - 1 = 15 - 1 = 14\)

.

2

For a 90% confidence level, the middle 90% is captured, leaving 5% in each tail: look up

\(t^*\)

with area 0.05 to the right, df = 14.

3

From the t-table (or technology):

t^* \approx 1.761

.

Verdict: t* = 1.761 for a 90% CI with df = 14.

Clue #3

Higher confidence, bigger number

Same detective now wants 99% confidence with a sample of size n = 25. Find t* and compare it to Clue #2.

1

\(df = 25 - 1 = 24\)

.

2

99% confidence leaves 0.5% in each tail. Looking up

\(t^*\)

at df = 24:

t^* \approx 2.797

.

Verdict: t* = 2.797 — much larger than 1.761 from Clue #2. Higher confidence always demands a bigger critical value, which means a wider interval.

📋 Case Notes — Unit Recap

Use the t-distribution whenever σ is unknown and estimated by s (almost always, in real life).
$$df = n-1$$ for a single sample.
Bigger df → t-distribution looks more like the normal curve.
Higher confidence level → larger t* → wider interval, holding everything else fixed.

Solved

Solve It Yourself

Detective's Challenge

A sample has n = 21 observations. How many degrees of freedom does its t-distribution have?

Case File 02 — The Lone Sample

One sample. One claim to test. Can the evidence back it up?

Open Case

Café Central hangs a sign: "95 mg of caffeine per cup — guaranteed." Detective Minho isn't so sure. He buys 12 cups, tests each one in the lab, and now holds a single sample of evidence against a single claim.

This is the simplest kind of case: one sample, one number to check. The tool for the job is the one-sample t-test — and if Minho instead wants to estimate the true mean rather than test a claim, he builds a one-sample t-interval.

◆ Conditions to Check First

Random

The sample was collected using random sampling (or a randomized experiment).

Normal

The population is Normal, or the sample is large (n ≥ 30), or a graph of the data shows no strong skew or outliers.

Independent

Individual observations are independent. If sampling without replacement, the sample is less than 10% of the population.

\[ t = \dfrac{\bar{x} - \mu_0}{s / \sqrt{n}} \qquad df = n-1 \]

Confidence interval: $ \bar{x} \pm t^{*}\dfrac{s}{\sqrt{n}} $

Case Evidence

Testing Café Central's Claim

Café Central claims μ = 95 mg. Minho's sample: n = 12 cups, mean $\bar{x} = 91.2$ mg, s = 5.8 mg. Test at α = 0.05. Assume conditions are met.

1

Hypotheses: $H_0: \mu = 95$, $H_a: \mu \neq 95$.

2

Standard error: $SE = \dfrac{5.8}{\sqrt{12}} = 1.674$.

3

Test statistic: $t = \dfrac{91.2 - 95}{1.674} = -2.270$, with $df = 11$.

4

Two-tailed p-value: $p \approx 0.0443$. Compare to α = 0.05.

Verdict: Since p ≈ 0.0443 < 0.05, reject H₀. There is convincing evidence the true mean caffeine content differs from 95 mg — Café Central's sign is misleading.

Case Evidence

Estimating Battery Life

A tech reviewer tests n = 20 phones of the same model. Sample mean battery life $\bar{x} = 4.6$ hours, s = 1.2 hours. Build a 95% confidence interval for the true mean battery life.

1

$df = 20 - 1 = 19$, so $t^* = 2.093$ for 95% confidence.

2

Standard error: $SE = \dfrac{1.2}{\sqrt{20}} = 0.268$.

3

Margin of error: $ME = 2.093 \times 0.268 = 0.562$.

4

Interval: $4.6 \pm 0.562 = (4.038,\ 5.162)$.

Verdict: We are 95% confident the true mean battery life is between 4.04 and 5.16 hours.

Quick Check

A Smaller Clue

n = 16, $\bar{x} = 50$ , s = 8, claimed $\mu_0 = 45$. Test at α = 0.05, two-tailed.

1

$SE = \dfrac{8}{\sqrt{16}} = 2$.

2

$t = \dfrac{50-45}{2} = 2.5$, $df = 15$, so $t^* = 2.131$.

3

Since $|t| = 2.5 > 2.131$, the p-value (≈ 0.0245) is less than 0.05.

Verdict: Reject H₀. The evidence suggests the true mean is not 45.

📋 Case Notes — Unit Recap

One sample, one hypothesized value μ₀ → one-sample t-test.
Always check Random, Normal/Large Sample, Independent before calculating.
Small p-value (less than α) → reject H₀; otherwise, fail to reject.
A confidence interval estimates μ directly, with no hypothesized value needed.

Solved

Solve It Yourself

Detective's Challenge

df = 15 and the test statistic is t = 2.5. Using t* = 2.131 for a two-tailed test at α = 0.05, should we reject H₀?

Case File 03 — The Paired Suspects

Same subjects, two measurements. The difference tells the story.

Open Case

A new file lands on the desk — this one has two columns: Before and After. Eight people joined a diet program, and Minho has each person's weight before and after eight weeks.

This isn't two separate groups — it's the same subjects, measured twice. When data comes in natural pairs like this, we don't run a two-sample test. Instead, we collapse each pair into a single difference, and run a one-sample t-test on those differences.

◆ The Idea, in Plain Terms

Matched pairs data comes from the same individual measured twice (before/after), or from naturally linked pairs (twins, left shoe/right shoe).
Step 1: compute the difference for every pair, $d_i = x_{1i} - x_{2i}$ .
Step 2: treat those differences as a single sample and run an ordinary one-sample t-test, testing whether the mean difference is 0 (or some other claimed value).
Pairing controls for individual differences between subjects — it isolates the effect we actually care about.

\[ t = \dfrac{\bar{d} - 0}{s_d / \sqrt{n}} \qquad df = n - 1 \]

n = number of pairs (not individual measurements), $\bar{d}$ = mean of the differences, $s_d$ = standard deviation of the differences.

Case Evidence

The Diet Program

Weights (lbs) for 8 participants, before and after an 8-week program:

Before	180	165	200	175	190	160	205	170
After	174	160	193	171	183	155	198	166
Diff (B−A)	6	5	7	4	7	5	7	4

Test whether the program produces true mean weight loss (α = 0.05, one-tailed: $H_a: \mu_d > 0$).

1

Mean difference: $\bar{d} = 5.625$ lbs. Standard deviation of differences: $s_d = 1.303$ lbs.

2

$SE = \dfrac{1.303}{\sqrt{8}} = 0.461$.

3

$t = \dfrac{5.625}{0.461} = 12.22$, $df = 7$.

4

One-tailed p-value is essentially 0 (far beyond $t^* = 1.895$).

Verdict: Reject H₀. This is an open-and-shut case — overwhelming evidence of true mean weight loss.

Case Evidence

The Reaction Time Puzzle

Six volunteers have their reaction time (milliseconds) measured before and after drinking a cup of coffee. Differences (Before − After): 5, −3, 2, 5, 3, 2. Test $H_a: \mu_d > 0$ at α = 0.05.

1

$\bar{d} = 2.333$ ms, $s_d = 2.944$ ms, $n = 6$.

2

$SE = \dfrac{2.944}{\sqrt{6}} = 1.202$, $t = \dfrac{2.333}{1.202} = 1.942$, $df = 5$.

3

One-tailed critical value $t^* = 2.015$; p-value ≈ 0.0549.

Verdict: Since p ≈ 0.0549 > 0.05, fail to reject H₀. The trail runs a little cold — this sample doesn't give quite enough evidence that coffee speeds reaction time. "Fail to reject" is not the same as "proven no effect": with a slightly larger sample, this case might resolve differently.

📋 Case Notes — Unit Recap

Recognize matched pairs: same subjects measured twice, or naturally linked pairs.
Reduce every pair to one difference, then run a one-sample t-test on the differences.
$$df = n-1$$ , where n is the number of pairs.
A large p-value doesn't prove "no difference" — it only means the sample didn't provide convincing evidence.

Solved

Solve It Yourself

Detective's Challenge

Which situation calls for a matched pairs t-test?

Case File 04 — Two Rivals

Two independent groups. Is the gap between them real?

Open Case

The final file compares two rival study methods. Group A used flashcards; Group B used practice tests. The two groups are made of completely different students — nobody appears in both groups.

Because the two samples are independent (no pairing, no overlap), Minho reaches for the two-sample t-test to compare the two population means directly.

◆ Conditions to Check First

Random

Both samples are randomly selected, or come from a randomized experiment.

Normal

Both populations are Normal, or both samples are large, or graphs show no strong skew/outliers.

Independent

The two samples are independent of each other, and observations within each sample are independent (10% condition if needed).

\[ t = \dfrac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\dfrac{s_1^2}{n_1} + \dfrac{s_2^2}{n_2}}} \]

Confidence interval: $ (\bar{x}_1-\bar{x}_2) \pm t^{*}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}} $ — technology finds df (Welch's method); by hand, the conservative choice is $df = \min(n_1-1,\ n_2-1)$.

Case Evidence

Flashcards vs. Practice Tests

Group A (flashcards): n₁ = 10, mean score $\bar{x}_1 = 82.4$ , s₁ = 6.1. Group B (practice tests): n₂ = 12, $\bar{x}_2 = 76.8$ , s₂ = 7.3. Test at α = 0.05 whether the true mean scores differ.

1

Hypotheses: $H_0: \mu_1 = \mu_2$, $H_a: \mu_1 \neq \mu_2$.

2

$SE = \sqrt{\dfrac{6.1^2}{10} + \dfrac{7.3^2}{12}} = 2.857$.

3

$t = \dfrac{82.4-76.8}{2.857} = 1.960$, technology gives $df \approx 20.0$.

4

Two-tailed p-value ≈ 0.0641.

Verdict: Since p ≈ 0.064 > 0.05, fail to reject H₀. The case remains open — this sample doesn't provide quite enough evidence that the two study methods truly differ, even though flashcards scored higher on average.

Case Evidence

Weekly Study Hours

Group A: n₁ = 15, $\bar{x}_1 = 6.2$ hrs, s₁ = 1.5. Group B: n₂ = 18, $\bar{x}_2 = 5.4$ hrs, s₂ = 1.8. Build a 95% confidence interval for $\mu_1 - \mu_2$ .

1

Point estimate: $6.2 - 5.4 = 0.8$ hrs.

2

$SE = \sqrt{\dfrac{1.5^2}{15}+\dfrac{1.8^2}{18}} = 0.575$; technology df ≈ 31, so $t^* \approx 2.040$.

3

Margin of error: $2.040 \times 0.575 = 1.172$.

4

Interval: $0.8 \pm 1.172 = (-0.372,\ 1.972)$.

Verdict: The interval contains 0, so we cannot rule out "no real difference" between the groups' true mean study hours — consistent with the test result above.

📋 Case Notes — Unit Recap

Two-sample t applies to two separate, independent groups — never the same subjects twice.
No pooling required in AP Statistics: always use the two-sample (Welch) standard error formula.
A confidence interval for $\mu_1-\mu_2$ that contains 0 matches "fail to reject H₀" in a two-tailed test.

Solved

Solve It Yourself

Detective's Challenge

A two-sample t-test is the right tool when:

Master File — Case Closed

All four cases, solved. Here is the complete evidence file.

Closed

01 · The t-Distribution

\[ df = n-1 \]

Use when σ is unknown and estimated by s.
Fatter tails than normal; approaches normal as df increases.
Higher confidence level → larger t* → wider interval.

02 · One-Sample t-Test / Interval

\[ t = \dfrac{\bar{x}-\mu_0}{s/\sqrt{n}} \qquad \bar{x} \pm t^{*}\dfrac{s}{\sqrt{n}} \]

One sample, one hypothesized value μ₀.
df = n − 1.
Check Random, Normal/Large Sample, Independent.

03 · Matched Pairs t-Test

\[ t = \dfrac{\bar{d}}{s_d/\sqrt{n}} \]

Same subjects measured twice, or naturally linked pairs.
Reduce pairs to differences first, then run a one-sample t-test.
n = number of pairs, not number of measurements.

04 · Two-Sample t-Test / Interval

\[ t = \dfrac{(\bar{x}_1-\bar{x}_2)}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}} \]

Two separate, independent groups — never the same subjects twice.
No pooling in AP Statistics; use technology or the conservative df.
A CI for μ₁ − μ₂ containing 0 matches "fail to reject" at the matching α.

Case Closed

Every mean has a story. Every t-statistic is a clue. Well done, Detective — Unit 7 is solved.