11. Inference for One Mean
a. Hypothesis Test
Math 115
Where We’ve Been — and Where We’re Going
| Topic | Variables | Statistic | Example |
|---|---|---|---|
| 6 | One binary categorical | \hat{p} | Payday loans — fraction supporting regulation |
| 10 | Two binary categorical | \hat{p}_1 - \hat{p}_2 | CPR — survival rates by treatment |
| 11 (today) | One numerical | \bar{x} | Cherry Blossom — mean finish time |
| 12 | Numerical + binary categorical | \bar{x}_1 - \bar{x}_2 | Birth weights — by smoking habit |
Same framework throughout — only the statistic changes row to row.
From Proportions to Means
Topic 2 introduced the sample mean as a way to describe a numerical variable’s center. Today it becomes our statistic of interest for inference — with notation that mirrors the proportions pattern.
| Quantity | Sample (statistic) | Population (parameter) |
|---|---|---|
| Center | \bar{x} (“x-bar”) | \mu (“mu”) |
| Spread | s | \sigma (“sigma”) |
\bar{x} and s come from the sample; \mu and \sigma are population values we want to learn about. Both pairs will show up in the formulas coming next.
Cherry Blossom Run
Are runners getting faster or slower? Annual 10-mile race in Washington, D.C.
- 2006 mean finish time (all runners): 93.29 minutes
- Data: random sample of 100 runners from the 2017 race
Hypotheses (two-sided). Let \mu = true mean finish time for all 2017 10-mile runners.
- H_0: \mu = 93.29 (no change from 2006)
- H_A: \mu \neq 93.29 (mean has changed in either direction)
EDA: Cherry Blossom
| n | mean (\bar{x}) | sd (s) | min | max |
|---|---|---|---|---|
| 100 | 99.02 | 17.93 | 53.3 | 139.1 |
Observed: sample mean is \bar{x} = 99.02 minutes — about 5.7 minutes higher than the 2006 mean of 93.29. Sampling noise, or a real shift?
A New CLT Result: One Mean
When conditions are met, the sampling distribution of the sample mean is approximately normal with:
- Mean: \mu (the true population mean)
- Standard Error: SE = \dfrac{\sigma}{\sqrt{n}}
The catch: we don’t know the population SD \sigma.
Fix: substitute the sample SD s for \sigma, so SE \approx \dfrac{s}{\sqrt{n}}.
Try it: if we quadrupled the sample size (keeping s unchanged), what happens to SE?
SE shrinks by half: \sqrt{4n} = 2\sqrt{n}, so the denominator doubles. Halving SE costs four times the data.
Technical Conditions: One Mean
Two conditions for inference about \mu from a single sample:
| Condition | Rule |
|---|---|
| Independence | Sample observations independent — typically from a random sample (with n < 10\% of the population), otherwise justified from study context |
| Normality | n \geq 30: data may show mild skew, but no extreme outliers. n < 30: data should look roughly normal. |
Try it: do both conditions hold for the Cherry Blossom sample? (The 2017 10-mile field had 17,442 runners.)
Independence: random sample of 2017 runners, n = 100 is well under 10% of the 17,442 10-mile runners ✓. Normality: n = 100 \geq 30 ✓; histogram is roughly symmetric with no extreme outliers ✓. Both conditions met.
Test Statistic for One Mean
The T-statistic measures how far \bar{x} is from the null value \mu_0, in SE units:
T = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
Where:
- \bar{x} is the observed sample mean
- \mu_0 is the null value (from H_0: \mu = \mu_0)
- s / \sqrt{n} is the estimated SE
Same formula structure as Z — observed minus null over SE. But T uses s in place of \sigma, so under H_0 it doesn’t quite follow the standard normal. It needs its own reference distribution — coming up next.
The t-Distribution
Under H_0, T = (\bar{x} - \mu_0)/(s/\sqrt{n}) follows a t-distribution with degrees of freedom df = n - 1.
- Bell-shaped and symmetric like the normal
- Thicker tails (more probability of extreme values — accounting for the noise in s)
- Smaller n → thicker tails (noisier s)
- As df grows, t approaches the normal
A p-value comes from an area under the appropriate t-distribution — we’ll walk through one when we work through Cherry Blossom.
Statistic & Test Statistic
For each scenario, identify the statistic of interest and the test statistic (Z or T).
A pre-election poll asks 500 voters about a ballot measure. Is the percentage who support it different from 50%? Statistic: \hat{p} (single proportion). Test statistic: Z — normal model (topic 6).
A clinical trial randomly assigns patients to a new drug or placebo. Do the recovery rates differ between groups? Statistic: \hat{p}_1 - \hat{p}_2 (difference of proportions). Test statistic: Z — normal model (topic 10).
A nutrition study measures sodium content in 40 fast-food burgers. Is the mean sodium content different from 1200 mg? Statistic: \bar{x} (single mean). Test statistic: T — t-distribution with df = n - 1 (today).
Putting It Together for Cherry Blossom
Hypotheses (two-sided): H_0: \mu = 93.29 vs H_A: \mu \neq 93.29
Sample
- n = 100
- \bar{x} = 99.02 minutes
- s = 17.93 minutes
Conditions met (random sample; n = 100 \geq 30; no extreme outliers). Coming up: compute SE → compute T → get p-value → conclude.
Computing T for Cherry Blossom
SE: \dfrac{s}{\sqrt{n}} = \dfrac{17.93}{\sqrt{100}} \approx 1.793 minutes
Try it: plug into T = \dfrac{\bar{x} - \mu_0}{SE} with \bar{x} = 99.02, \mu_0 = 93.29, SE \approx 1.793.
T = \dfrac{99.02 - 93.29}{1.793} \approx 3.19 — the sample mean sits about 3.2 SEs above the 2006 value.
Calculating the P-value
Two-sided p-value = 2 × (smaller-tail area) \approx 0.0019
Conclusion for Cherry Blossom
Results: T \approx 3.19 (df = 99), p-value \approx 0.0019 — well below 0.05 → reject H_0.
Conclusion: Convincing evidence that the mean finish time for 2017 Cherry Blossom 10-mile runners differs from 93.29 minutes, the mean from 2006.
In plain terms: the 2017 sample mean was about 5.7 minutes higher than 93.29 — runners were noticeably slower, and a difference this large is very unusual by chance.
Scope: random sample of 2017 runners → conclusion extends to all 2017 10-mile runners.
Body Temperature — Setup
Is the average body temperature really 98.6°F? The 98.6 value comes from 19th-century work by Wunderlich; modern work suggests it’s lower.
Data: a random sample of n = 30 healthy adults with \bar{x} = 98.25^\circF, s = 0.73^\circF (sample stats from Shoemaker, 1996; based on Mackowiak et al., 1992).
State H_0 and H_A for \mu = true mean body temperature (two-sided). H_0: \mu = 98.6 H_A: \mu \neq 98.6
What is \bar{x} - \mu_0? In which direction? 98.25 - 98.6 = -0.35^\circF. Sample mean is lower than the 98.6 reference — consistent with the modern challenge to Wunderlich’s value.
Body Temperature — Conditions
Recall: n = 30, \bar{x} = 98.25, s = 0.73, \mu_0 = 98.6.
Check independence. Random sample of healthy adults ✓ — observations should be independent.
Check normality. n = 30 meets the rule of thumb (n \geq 30) ✓. With raw data we’d also look for strong skew or extreme outliers; body temperatures for healthy adults are typically near-symmetric, so we’ll assume that’s fine.
Body Temperature — T and Conclusion
Recall: \bar{x} = 98.25, s = 0.73, n = 30, \mu_0 = 98.6; conditions met.
Compute the degrees of freedom. df = n - 1 = 29
Compute the standard error. SE = s / \sqrt{n} = 0.73 / \sqrt{30} \approx 0.133
Compute T. T = (\bar{x} - \mu_0) / SE = (98.25 - 98.6) / 0.133 \approx -2.63
The two-sided p-value is about 0.014. Conclusion at \alpha = 0.05, including scope? p-value < 0.05 → reject H_0. Convincing evidence the true mean body temperature differs from 98.6°F (and the sample suggests it’s lower). Scope: random sample → conclusion generalizes to all healthy adults.
Looking Ahead
We’ve tested “Is the true mean equal to \mu_0?” with the t-distribution.
Next time: a confidence interval for \mu — “What’s a plausible range for the true mean?” — using the same t-distribution machinery.
What about randomization? Simulation-based methods (like topics 3 and 9 for proportions) also exist for means and work in a similar way — useful when normality is in question (though they carry their own assumptions). We won’t cover them in the slides; you’ll meet them in a later lab.
References
- Introduction to Modern Statistics (2e) textbook by Mine Çetinkaya-Rundel and Johanna Hardin
- Section 19.2
- Shoemaker, A. L. (1996). What’s normal? — Temperature, gender, and heart rate. Journal of Statistics Education, 4(2). link
- Mackowiak, P. A., Wasserman, S. S., & Levine, M. M. (1992). A critical appraisal of 98.6°F, the upper limit of the normal body temperature, and other legacies of Carl Reinhold August Wunderlich. JAMA, 268(12), 1578–1580.