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

From Testing to Estimation

The hypothesis test answers “Is the true mean equal to \(\mu_0\)?”

But we also want to know “What values of \(\mu\) are plausible given the data?”

Solution: a confidence interval for \(\mu\) using the t-distribution.

Same machinery as 11a — conditions, \(SE = s/\sqrt{n}\), t-distribution with \(df = n - 1\). New piece: a t critical value \(t^*\) replaces the test statistic.

Recap: Cherry Blossom

Is the true 2017 mean finish time different from 93.29 minutes (the historical 2006 value)? Random sample of \(n = 100\) runners from the 2017 10-mile race.

Hypotheses (two-sided), with \(\mu\) = true 2017 mean:

\(H_0: \mu = 93.29\) vs \(H_A: \mu \neq 93.29\)

The data:

n	\(\bar{x}\)	\(s\)
100	99.02	17.93

From 11a: \(T \approx 3.19\), p-value \(\approx 0.0019\) → rejected \(H_0\). The true 2017 mean differs from 93.29. Today: what is a plausible range for \(\mu\)?

Quick recap of the Cherry Blossom example. We had a random sample of one hundred runners from the twenty seventeen ten-mile race, with ninety-three point two nine minutes serving as the fixed historical reference from two thousand six. The hypotheses are two-sided, with mu standing for the true twenty seventeen mean finish time: null says mu equals ninety-three point two nine; alternative says it doesn’t. The sample came in at x-bar about ninety-nine point oh two minutes, with sample standard deviation about seventeen point nine three. The hypothesis test in eleven A gave T of about three point one nine and a p-value of about zero point zero zero one nine, so we rejected the null and concluded the true twenty seventeen mean differs from ninety-three point two nine minutes. Today we ask the second half: what is a plausible range of values for mu?

CI Formula for One Mean

When conditions are met:

\[\text{CI} = \bar{x} \pm t^*_{df} \times \frac{s}{\sqrt{n}}\]

\(\bar{x}\) = point estimate; \(s / \sqrt{n}\) = standard error
\(t^*_{df}\) = t critical value (\(df = n - 1\), from a t-table or statistical software)
\(t^*_{df} \times s/\sqrt{n}\) = margin of error

Confidence level → multiplier:

90% → smaller \(t^*\) → narrower CI
95% → standard choice
99% → larger \(t^*\) → wider CI

Sample size matters too: smaller \(n\) (smaller \(df\)) → larger \(t^*\) → wider CI. As \(df\) grows, \(t^*\) approaches \(z^*\) (1.645 / 1.96 / 2.576).

The confidence-interval formula is the same pattern we’ve seen all course: point estimate plus or minus a multiplier times the standard error. For one mean, the point estimate is x-bar, the standard error is s over root n, and the multiplier is t-star — the critical value from a t-distribution with degrees of freedom equal to n minus one. The product t-star times s-over-root-n is the margin of error. Critical values come from a t-table or from statistical software. For ninety percent confidence, t-star is smaller and the interval is narrower. Ninety-five percent is the standard choice. For ninety-nine percent, t-star is larger and the interval is wider — paying for higher confidence with width. Sample size matters too: smaller n means smaller degrees of freedom, which gives a larger t-star and a wider interval — so small samples are penalized both through a larger SE and through a larger multiplier. As degrees of freedom grows, t-star approaches the familiar z-star values from the normal: one point six four five, one point nine six, and two point five seven six.

Conditions for the CI

Same as 11a — all met for Cherry Blossom:

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 ✓

Computing the 95% CI for Cherry Blossom

From 11a: \(\bar{x} = 99.02\), \(s = 17.93\), \(n = 100\), \(SE \approx 1.793\).

t critical value: \(t^*_{99} \approx 1.984\) (95% confidence; very close to \(z^* = 1.96\) since \(df = 99\) is large).

Margin of error: \(ME = t^* \times SE = 1.984 \times 1.793 \approx 3.557\)

95% CI: \(\bar{x} \pm ME = 99.02 \pm 3.557 \approx (95.46, 102.57)\) minutes

Pulling the pieces together for Cherry Blossom. From eleven A we have x-bar about ninety-nine point oh two, sample standard deviation s about seventeen point nine three, n equals one hundred, and standard error about one point seven nine. For ninety-five percent confidence with ninety-nine degrees of freedom, the t critical value t-star is about one point nine eight four — very close to the z-star value of one point nine six. The margin of error is t-star times SE — one point nine eight four times one point seven nine, about three point five six minutes. The ninety-five percent confidence interval is x-bar plus or minus the margin of error — ninety-nine point oh two plus or minus three point five six — which gives an interval from about ninety-five point four six minutes up to about one hundred two point five seven minutes.

Interpreting the CI

95% CI: \((95.46, 102.57)\) minutes

Interpretation: We are 95% confident that the true mean finish time for all 2017 Cherry Blossom 10-mile runners lies between 95.46 and 102.57 minutes.

Does the CI include 93.29? No — the 2006 mean sits below the entire interval.

This agrees with the hypothesis test in 11a: rejected \(H_0\) ↔︎ the 2006 value is outside the CI.

In plain terms: the 2017 mean is plausibly anywhere from about 2.2 to 9.3 minutes slower than 2006.

The ninety-five percent confidence interval runs from about ninety-five point four six minutes to about one hundred two point five seven minutes. Interpretation — we are ninety-five percent confident that the true mean finish time for all twenty seventeen Cherry Blossom ten-mile runners lies between ninety-five point four six and one hundred two point five seven minutes. The two thousand six mean of ninety-three point two nine sits below the entire interval — the confidence interval does not include the null value. That agrees with the eleven A hypothesis test, where we rejected the null. In plain terms, the data say the twenty seventeen mean is plausibly anywhere from about two point two minutes slower up to about nine point three minutes slower than the two thousand six mean. We’re not just saying the means differ; we’re putting numbers on how much the change might be.

When CI and Test Agree

For a two-sided test at level \(\alpha\) vs. a \((1 - \alpha)\) CI for \(\mu\):

Test result	CI for \(\mu\)
Reject \(H_0\)	excludes \(\mu_0\)
Fail to reject \(H_0\)	contains \(\mu_0\)

Same correspondence as for proportions (topics 6 / 10). The test asks “is \(\mu_0\) ruled out?” — the CI shows you everything that isn’t ruled out.

The CI tells you how big the effect is, not just whether it exists.

Body Temperature — Setup

What’s a plausible range for true mean body temperature?

Data: same scenario as 11a — a random sample of \(n = 30\) healthy adults, \(\bar{x} = 98.25^\circ\)F, \(s = 0.73^\circ\)F (sample stats from Shoemaker, 1996; based on Mackowiak et al., 1992). Wunderlich’s 98.6 is the historical reference.

Recall (from 11a): hypothesis test gave \(T \approx -2.63\), p-value \(\approx 0.014\) → rejected \(H_0: \mu = 98.6\).

Check the conditions. Independence: random sample of healthy adults ✓. Normality: \(n = 30 \geq 30\) ✓ (assume no extreme outliers). Same as 11a.
Compute the SE. \(SE = s / \sqrt{n} = 0.73 / \sqrt{30} \approx 0.133\)

Now for your turn — back to the body-temperature example from eleven A. The setup is the same: a random sample of thirty healthy adults, with sample mean ninety-eight point two five degrees Fahrenheit and sample standard deviation zero point seven three, tested against Wunderlich’s traditional value of ninety-eight point six. The eleven A hypothesis test gave T of about negative two point six three with a p-value of about zero point zero one four, so we rejected the null. Today we estimate what the true mean body temperature actually is. Two quick steps before the interval itself — advance when you are ready to see the answers. Conditions for a confidence interval are the same as for the eleven A hypothesis test. Independence holds because the data are a random sample of healthy adults. Normality holds because n equals thirty meets the rule of thumb, and we’ll assume no extreme outliers in the raw data. Both conditions hold. Standard error is s over root n — zero point seven three divided by the square root of thirty, about zero point one three three degrees Fahrenheit. Exactly the same SE we used for the hypothesis test in eleven A.

Body Temperature — CI & Conclusion

Recall: \(\bar{x} = 98.25\), \(SE \approx 0.133\), \(df = 29\). Use \(t^*_{29} \approx 2.045\) for 95% confidence.

Compute the margin of error. \(ME = t^* \times SE = 2.045 \times 0.133 \approx 0.273\) °F
Compute the 95% CI for \(\mu\). \(\bar{x} \pm ME = 98.25 \pm 0.273 \approx (97.98, 98.52)\) °F
Interpret the interval, check consistency with the 11a hypothesis test, and state scope. We’re 95% confident the true mean body temperature for healthy adults is between 97.98 and 98.52 °F. The CI excludes 98.6 — consistent with rejecting \(H_0\) in 11a, and pointing to a true mean below Wunderlich’s value. Scope: random sample → conclusion generalizes to all healthy adults.

With conditions met and standard error in hand, we can build the ninety-five percent confidence interval. Three steps — advance when you are ready to see each answer. The t critical value for ninety-five percent confidence with twenty-nine degrees of freedom is about two point oh four five. Multiplying by the SE gives a margin of error of about zero point two seven three degrees Fahrenheit. The ninety-five percent confidence interval is x-bar plus or minus the margin of error — ninety-eight point two five plus or minus zero point two seven three, which works out to roughly ninety-seven point nine eight up to ninety-eight point five two degrees Fahrenheit. We’re ninety-five percent confident the true mean body temperature for healthy adults lies between about ninety-seven point nine eight and ninety-eight point five two degrees Fahrenheit. The interval excludes ninety-eight point six — Wunderlich’s traditional value — so this is consistent with rejecting the null in eleven A. The confidence interval does more than just confirm rejection: it tells us the true mean sits below Wunderlich’s value, somewhere inside about a half-degree window. On scope — because this is a random sample of healthy adults, the conclusion generalizes to that population.

What Affects Strength of Evidence and CI Width?

Strength of evidence (larger \(|T|\) → smaller p-value)

\(T = \dfrac{\bar{x} - \mu_0}{s/\sqrt{n}}\)

Larger effect size \(|\bar{x} - \mu_0|\) → larger \(|T|\)
Larger sample size \(n\) → smaller SE → larger \(|T|\)
Smaller variability \(s\) → smaller SE → larger \(|T|\)

CI for \(\mu\) (narrower = more precise)

\(\bar{x} \pm t^* \times \dfrac{s}{\sqrt{n}}\)

Lower confidence level → smaller \(t^*\) → narrower CI
Larger sample size \(n\) → smaller SE and smaller \(t^*\) → narrower CI
Smaller variability \(s\) → smaller SE → narrower CI

Let’s pull together the factors that move our two inference quantities. For the hypothesis test, the strength of evidence — how big T is and how small the p-value — depends on three things. The effect size, the absolute difference between x-bar and mu naught: a larger effect pushes T further out. The sample size n: a larger n shrinks the standard error, which inflates T. And the variability s: smaller variability also shrinks SE and inflates T. For the confidence interval, the width depends on the same kinds of factors plus one more — the confidence level. A lower confidence level gives a smaller t-star and a narrower interval. A larger n shrinks SE and also shrinks t-star — more degrees of freedom mean t-star moves toward z-star — and both effects narrow the interval. Less variability shrinks SE and narrows the interval.

Strength of Evidence — Predict

Cherry Blossom recap: \(n = 100\), \(\bar{x} = 99.02\), \(s = 17.93\), \(\mu_0 = 93.29\) → \(T \approx 3.19\), p-value \(\approx 0.0019\).

For each variant (others held constant), does the evidence become stronger or weaker?

The researchers had sampled 400 runners instead of 100 (same \(\bar{x}\), \(s\)). Stronger. Quadrupling \(n\) halves SE → \(|T|\) doubles.
The sample mean had come in at 102 minutes instead of 99.02 (same \(n\), \(s\)). Stronger. Larger effect size \(|\bar{x} - \mu_0|\) → larger \(|T|\) directly.
The 2017 sample had been noisier: \(s = 25\) instead of 17.93 (same \(n\), \(\bar{x}\)). Weaker. Larger \(s\) → larger SE → smaller \(|T|\).

Cherry Blossom prediction practice on strength of evidence. The original study had n equal to one hundred, x-bar about ninety-nine point oh two, s about seventeen point nine three, null value ninety-three point two nine — giving T about three point one nine with a p-value of about zero point zero zero one nine. For each variant of the study, predict whether the evidence becomes stronger or weaker. Pause and think before advancing. Quadrupling the sample size to four hundred shrinks the standard error by half, which doubles T — so the evidence becomes stronger. A sample mean of one hundred two instead of ninety-nine widens the difference from the null hypothesis value, which makes T larger directly — so the evidence becomes stronger. A noisier sample with s equal to twenty-five pushes the standard error up, which shrinks T — so the evidence becomes weaker.

CI Width — Predict

Cherry Blossom recap: 95% CI = \((95.46, 102.57)\) minutes — width about 7.1 minutes.

For each variant (others held constant), does the CI become narrower or wider?

The researchers had sampled 400 runners instead of 100 (same \(\bar{x}\), \(s\)). Narrower. Larger \(n\) → smaller SE and smaller \(t^*\) → narrower CI.
They build a 99% CI instead of 95%. Wider. Higher confidence → larger \(t^*\) → wider CI. (More confidence costs precision.)
The 2017 sample had been noisier: \(s = 25\) instead of 17.93. Wider. Larger \(s\) → larger SE → wider CI.

Same Logic: Proportions and Means

	One Proportion	One Mean
Statistic	\(\hat{p}\)	\(\bar{x}\)
H-test SE	\(\sqrt{p_0(1-p_0)/n}\)	\(s / \sqrt{n}\)
CI SE	\(\sqrt{\hat{p}(1-\hat{p})/n}\)	\(s / \sqrt{n}\)
Reference distribution	normal (\(z\))	t (with \(df = n - 1\))
CI	\(\hat{p} \pm z^* \cdot SE\)	\(\bar{x} \pm t^* \cdot SE\)
Test statistic	\(Z = \dfrac{\hat{p} - p_0}{SE}\)	\(T = \dfrac{\bar{x} - \mu_0}{SE}\)

Formulas differ; logic is identical: quantify sampling variability with SE → use it to measure precision (CI) or surprise (test).

Step back across topics. The table compares one proportion — topic six — with one mean — today. The statistics differ — p hat versus x-bar. For the proportion, the standard error takes two forms — one using p naught for the hypothesis test, one using p hat for the confidence interval. For one mean, the same s over root n works for both. The reference distribution shifts from the normal to the t. But the confidence-interval template and the test-statistic template are identical: statistic plus or minus critical-value times standard error for the interval, and observed minus null over standard error for the test. The formulas differ, but the logic is the same. Quantify sampling variability with the standard error, then use it to measure precision through a confidence interval or surprise through a test.

Looking Ahead

We’ve built both halves of single-mean inference: hypothesis test (11a) and confidence interval (today).

Next time: two means — comparing the average of one group to another. Same t-distribution machinery, with a new SE for the difference \(\bar{x}_1 - \bar{x}_2\).

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.