Math 115: Introduction to Statistics – 11. Inference for One Mean

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.

Let’s step back and orient by the type of data we’re working with. So far the variables have been categorical. Topic six was one binary categorical variable — the payday-loan study, looking at the fraction of borrowers supporting tighter regulation — and the statistic was a single proportion p hat. Topic ten was two binary categorical variables — the CPR study comparing treatment and control survival rates — and the statistic was the difference of two proportions, p hat one minus p hat two. Today we switch to numerical data. Topic eleven, our running example, is one numerical variable — Cherry Blossom finish times — and the statistic is x-bar, the sample mean. Topic twelve next time will add a binary categorical variable for grouping — birth weights by maternal smoking — where the statistic is the difference of two sample means. The framework stays the same 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.

Two conditions for inference about mu from a single sample. The first is independence — knowing the value of one observation should give us no information about any other observation in the sample. The standard path is random sampling, as long as our sample is less than ten percent of the population; otherwise we justify independence from study context. The second condition is normality. If n is at least thirty, the Central Limit Theorem makes the sampling distribution of x-bar approximately normal — even when the sample data shows some skew — as long as there are no extreme outliers. If n is less than thirty, we need the sample data itself to look roughly normal: no strong skew and no obvious outliers. Pause now and check both conditions for the Cherry Blossom sample before advancing. Note that the full 2017 ten-mile field had seventeen thousand four hundred forty-two runners. For Cherry Blossom, both conditions check out. The data come from a random sample of twenty seventeen runners, and our sample of one hundred is well under ten percent of the seventeen thousand four hundred forty-two runners in the full ten-mile field — so independence is satisfied. n equals one hundred — comfortably above thirty — and the histogram is roughly symmetric with no extreme outliers, so normality is satisfied through the Central Limit Theorem.

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.

Because T uses s in place of sigma, its sampling distribution under the null isn’t quite the standard normal — it’s a t-distribution, with degrees of freedom equal to n minus one. The shape looks a lot like the normal: bell-shaped, symmetric, centered at zero. The difference is in the tails, which are thicker — that’s how the distribution absorbs the extra uncertainty from estimating sigma. The plot shows three curves: solid black is the standard normal, dashed red is t with five degrees of freedom — clearly heavier in the tails — and dotted blue is t with thirty degrees of freedom, already nearly indistinguishable from the normal. Smaller samples mean thicker tails; as n grows, the t-distribution converges to the normal. For our Cherry Blossom sample, degrees of freedom is ninety-nine, so our curve is effectively normal — but t is still the correct reference. We’ll get a p-value from an area under this curve 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).

Time to put together what we’ve covered. For each of these three scenarios, identify the statistic of interest, and which test statistic — Z or T — we’d use. Pause and think about each one before advancing. First scenario — a pre-election poll about a ballot measure. The variable is whether a voter supports the measure: binary categorical. The statistic of interest is p hat, a single proportion. Test statistic is Z, the standard normal — that’s the topic six framework. Second scenario — drug versus placebo in a clinical trial. Two binary categorical groups, and we want to compare their recovery rates. The statistic of interest is the difference of two sample proportions, p hat one minus p hat two. Test statistic is Z again — the topic ten framework. Third scenario — sodium content in fast-food burgers. The variable is sodium content, a numerical measurement. The statistic of interest is x-bar, a single sample mean. Test statistic is T, because we have to estimate sigma from the sample — that’s today’s framework.

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 \(\bar{x} = 98.25^\circ\)F, \(s = 0.73^\circ\)F (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? \(98.25 - 98.6 = -0.35^\circ\)F. Sample mean is lower than the 98.6 reference — consistent with the modern challenge to Wunderlich’s value.

Now for your turn. Is the average human body temperature really ninety-eight point six degrees Fahrenheit? That value comes from nineteenth-century work by Wunderlich — but more recent work has challenged it. In a random sample of thirty healthy adults, the sample mean was ninety-eight point two five degrees Fahrenheit, with sample standard deviation of zero point seven three degrees. Two quick steps before the calculations — advance when you are ready to see the answers. For step one, we set up a two-sided test on the true mean body temperature mu. Null: mu equals ninety-eight point six — Wunderlich’s traditional value. Alternative: mu is not equal to ninety-eight point six — the true mean could differ in either direction. For step two, x-bar minus mu naught is ninety-eight point two five minus ninety-eight point six, which equals negative zero point three five degrees Fahrenheit. The sample mean is lower than the ninety-eight point six 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.

Conditions are met, so we can compute T and conclude. Four short steps — advance when you are ready to see each answer. Degrees of freedom is n minus one — thirty minus one is twenty-nine. Standard error is s over root n — zero point seven three divided by the square root of thirty, about zero point one three three. T is x-bar minus mu naught divided by SE — ninety-eight point two five minus ninety-eight point six is negative zero point three five, divided by zero point one three three, about negative two point six three. The p-value comes from a t-distribution with twenty-nine degrees of freedom — find the area past T equal to negative two point six three, then double it for two-sided. That works out to about zero point zero one four. Since that’s below zero point zero five, we reject the null. The data provide convincing evidence that the true mean body temperature for healthy adults differs from ninety-eight point six degrees Fahrenheit, and the sample suggests it’s lower than Wunderlich’s value. On scope — because this is a random sample of healthy adults, the conclusion generalizes to that population.

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.