Math 115: Introduction to Statistics – 12. Inference for Two Independent Means

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	One numerical	$\bar{x}$	Cherry Blossom — mean finish time
12 (today)	Numerical + binary categorical	$\bar{x}_1 - \bar{x}_2$	Birth weights — by smoking habit
13	Numerical, paired	$\bar{x}_d$	Textbook prices — UCLA vs Amazon

Same framework throughout — only the statistic changes row to row.

Birth Weights and Smoking

Do infants born to smokers have different mean birth weights than those born to nonsmokers? Random sample of 981 U.S. births from 2014.

Response: birth weight (pounds) — numerical
Explanatory: maternal smoking habit (smoker / nonsmoker) — binary

Hypotheses (two-sided). Let $\mu_n, \mu_s$ = mean birth weight for all U.S. babies born to nonsmoking / smoking mothers.

$H_0: \mu_n - \mu_s = 0$ (no difference in mean birth weights)
$H_A: \mu_n - \mu_s \neq 0$ (a difference exists)

EDA: Birth Weights

Group	n	$\bar{x}$ (lbs)	$s$ (lbs)
Nonsmoker	867	7.27	1.23
Smoker	114	6.68	1.6

Observed difference: $\bar{x}_n - \bar{x}_s = 7.27 - 6.68 \approx 0.59$ lbs. Babies born to nonsmokers weigh a little over half a pound more on average. Real shift, or sampling noise?

A New CLT Result: Difference in Means

When conditions are met, the sampling distribution of $\bar{x}_1 - \bar{x}_2$ is approximately normal with:

Mean: $\mu_1 - \mu_2$ (the true difference)
Standard Error: depends on two sample standard deviations

Another CLT result. Just like $\bar{x}$ in topic 11 and $\hat{p}_1 - \hat{p}_2$ in topic 10, the difference of two sample means becomes approximately normal under independence + adequate sample sizes.

Same catch as topic 11: we don’t know the population SDs $\sigma_1, \sigma_2$. We estimate from the sample SDs $s_1, s_2$ — then use a t-distribution to account for the extra uncertainty.

Pooled SD and SE for the Difference

When the two groups have similar SDs, treat them as sharing a common population SD. Estimate it by pooling the two sample SDs:

\[s_p = \sqrt{\frac{(n_1 - 1) s_1^2 + (n_2 - 1) s_2^2}{n_1 + n_2 - 2}}\]

Then the standard error for the difference is:

\[SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}\]

And use a t-distribution with $df = n_1 + n_2 - 2$.

“Pooled” parallel. Same idea as the pooled $\hat{p}$ in topic 10 — combine information from both groups to estimate a common population value.

When the two groups have similar standard deviations, we treat them as sharing a common population standard deviation, and estimate it by pooling the two sample standard deviations. The pooled standard deviation s sub p is a weighted average of the two sample variances, weighted by group degrees of freedom — n sub one minus one and n sub two minus one — with the whole thing divided by n sub one plus n sub two minus two. Then the standard error for the difference is s sub p times the square root of one over n sub one plus one over n sub two. And the reference distribution is t with degrees of freedom equal to n sub one plus n sub two minus two — the total sample size minus two, one degree of freedom subtracted for each estimated group mean. The pooled parallel here matches the pooled p hat from topic ten: combine information from both groups to estimate a common population value.

Technical Conditions: Two Means

Three conditions for inference about $\mu_1 - \mu_2$ from two independent samples:

Condition	Rule
Independence	Observations independent within and across groups — typically from random samples (each $n < 10\%$ of population) or random assignment, otherwise justified from study context
Normality	Each group: $n \geq 30$ → sample data may show mild skew, but no extreme outliers. If $n < 30$, group data should look roughly normal.
Equal variance	Group SDs similar (within a factor of $\approx$ 2) — used for the pooled SE

Try it: do all three conditions hold for the Birth Weights sample?

Independence: random sample of 2014 U.S. births (well under 10% of all 2014 U.S. births) ✓. Normality: $n_n = 867$, $n_s = 114$ — both well above 30, no extreme outliers in the histograms ✓. Equal variance: $s_n \approx 1.23$, $s_s \approx 1.6$ — within a factor of 2 ✓. All three met.

Three conditions to use the t-distribution for two-mean inference. First, independence — observations must be independent within and across groups. The standard paths are random samples, with each sample less than ten percent of its population, or random assignment to the two groups; otherwise we justify independence from study context. Second, normality — for each group, if n is at least thirty, the Central Limit Theorem covers it as long as there are no extreme outliers. If n is less than thirty, the group’s data should look roughly normal. Third, equal variance — the two group standard deviations should be similar, within a factor of about two, since the pooled SE assumes a common population standard deviation. Pause now and check all three for the Birth Weights sample before advancing. For Birth Weights, all three check out. The data come from a random sample of twenty-fourteen U.S. births — well under ten percent of all U.S. births that year — so independence holds. Both group sizes — eight hundred sixty-seven nonsmokers and one hundred fourteen smokers — are well above thirty, with no extreme outliers in the histograms, so normality is satisfied through the Central Limit Theorem. And the two sample standard deviations, one point two three and one point six pounds, are within a factor of two of each other, so the equal-variance condition is reasonable. As with topic eleven, these are the same conditions for today’s hypothesis test and the confidence interval in twelve B.

Test Statistic for Two Means

The T-statistic measures how far the observed difference is from the null value, in SE units:

\[T = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{SE} = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{1/n_1 + 1/n_2}}\]

$\bar{x}_1 - \bar{x}_2$: observed difference
0: null value (from $H_0: \mu_1 - \mu_2 = 0$)
Compare to a t-distribution with $df = n_1 + n_2 - 2$

Same formula structure as Z and the one-mean T — observed minus null over SE. Because we estimate $\sigma_1, \sigma_2$ from samples, the reference distribution is a t-distribution (with $df = n_1 + n_2 - 2$ here) — coming up next.

The t-Distribution (Again)

Under $H_0$, $T = (\bar{x}_1 - \bar{x}_2)/SE$ follows a t-distribution with $df = n_1 + n_2 - 2$.

Bell-shaped and symmetric like the normal
Thicker tails (extra uncertainty from estimating $\sigma_1, \sigma_2$ from samples)
Smaller total $n$ → thicker tails
As $df$ grows, t approaches the normal

For Birth Weights, $df = 979$ — the t curve is essentially normal here.

Same reference distribution from topic eleven, just with new degrees of freedom. Under the null, T follows a t-distribution with df equal to n sub one plus n sub two minus two — the total sample size minus two, with one degree of freedom subtracted for each estimated group mean. Bell-shaped and symmetric like the normal, but with thicker tails to absorb the extra uncertainty from estimating two standard deviations from sample data. Smaller total sample sizes give thicker tails; as df grows, the t-distribution converges to the normal. 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. For Birth Weights, degrees of freedom is nine hundred seventy-nine, so our curve is effectively normal — but t is still the correct reference.

Statistic & Test Statistic

For each scenario, identify the statistic of interest and the test statistic ($Z$ or $T$).

A campus survey asks 600 students whether they have used a generative AI tool. Is the rate different from 50%? Statistic: $\hat{p}$ (single proportion). Test statistic: $Z$ — normal model (topic 6).
50 plots are randomly assigned to two fertilizer treatments. Do the mean yields differ? Statistic: $\bar{x}_1 - \bar{x}_2$ (difference of means). Test statistic: $T$ — t-distribution with $df = n_1 + n_2 - 2$ (today).
A nutrition study measures sodium in 40 fast-food burgers. Is the mean sodium different from 1200 mg? Statistic: $\bar{x}$ (single mean). Test statistic: $T$ — t-distribution with $df = n - 1$ (topic 11).

Time to recognize which test statistic goes with which scenario. For each of these three, name the statistic of interest and whether the test statistic is Z or T. Pause and think before advancing. First scenario — a campus survey about AI tool use. The variable is binary categorical: used or didn’t use. 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 — two fertilizer treatments on plots. The response is yield, a numerical measurement, and the explanatory is the treatment, binary categorical. The statistic of interest is the difference of two sample means, x-bar one minus x-bar two. Test statistic is T, because we estimate the standard deviations from sample data — that’s today’s framework. Third scenario — sodium content in burgers. One numerical variable, no grouping. The statistic of interest is x-bar, a single sample mean. Test statistic is T with df equal to n minus one — that’s the topic eleven framework.

Putting It Together for Birth Weights

Hypotheses (two-sided): $H_0: \mu_n - \mu_s = 0$ vs $H_A: \mu_n - \mu_s \neq 0$

Samples

Nonsmokers: $n_n = 867$, $\bar{x}_n = 7.27$, $s_n = 1.23$
Smokers: $n_s = 114$, $\bar{x}_s = 6.68$, $s_s = 1.6$
Observed difference: $\bar{x}_n - \bar{x}_s \approx 0.59$ lbs

Conditions met (random sample; both group sizes ≥ 30; similar SDs). Coming up: compute $s_p$ → $SE$ → $T$ → p-value → conclude.

Computing T for Birth Weights

Pooled SD

\[s_p = \sqrt{\frac{(867 - 1)(1.23)^2 + (114 - 1)(1.6)^2}{867 + 114 - 2}} \approx 1.28\]

Standard error

\[SE = 1.28 \times \sqrt{\frac{1}{867} + \frac{1}{114}} \approx 0.128 \text{ lbs}\]

Try it: plug into $T = \dfrac{\bar{x}_n - \bar{x}_s}{SE}$.

$T = \dfrac{\bar{x}_n - \bar{x}_s}{SE} \approx 4.65$ — observed difference sits about 4.6 SEs from zero.

Calculating the P-value

Two-sided p-value = 2 × (smaller-tail area) ≈ < 0.0001 — the observed difference sits far out in the tail.

Conclusion for Birth Weights

Results: $T \approx 4.65$ ($df = 979$), p-value $< 0.0001$ — well below 0.05 → reject $H_0$.

Conclusion: Convincing evidence that mean birth weight differs between U.S. babies born to nonsmoking vs. smoking mothers.

In plain terms: nonsmokers’ babies averaged about 0.59 pounds heavier in the sample; a gap this large is extremely unlikely if the population means were equal.

Scope: random sample of 2014 U.S. births → conclusion generalizes to all 2014 U.S. births. Observational data → cannot conclude that smoking causes lower birth weight (third-variable explanations like prenatal care, nutrition, alcohol use, age remain possible).

Putting it together. T is about four point six five with nine hundred seventy-nine degrees of freedom, and the p-value is less than zero point zero zero zero one — far less than zero point zero five. We reject the null. The data provide convincing evidence that mean birth weight differs between U.S. babies born to nonsmoking and smoking mothers. In plain terms — nonsmokers’ babies averaged about zero point five nine pounds heavier in the sample, and that gap is much too large to attribute to sampling noise. On scope of inference: a random sample of twenty-fourteen U.S. births generalizes the conclusion to all twenty-fourteen U.S. births. But this is observational data — mothers chose to smoke or not — so we cannot conclude that smoking causes lower birth weight. Other factors that go with smoking — prenatal care, nutrition, alcohol use, maternal age — could explain some or all of the observed difference.

Mario Kart Auctions

You’ve been collecting copies of Mario Kart for the Nintendo Wii and decide to sell them on eBay. Should you bother photographing each disc yourself, or just use a stock photo of the box? Does using a stock photo affect the selling price?

Data (Diez et al., OpenIntro):

Photo	$n$	$\bar{x}$ ($)	$s$ ($)
Stock (yes)	105	48.50	9.47
No stock (no)	36	44.33	7.25

State $H_0$ and $H_A$ for $\mu_y - \mu_n$ (two-sided). $H_0: \mu_y - \mu_n = 0$ $H_A: \mu_y - \mu_n \neq 0$.
What’s the observed difference? $48.50 - 44.33 = \$4.17$ — stock-photo auctions averaged about four dollars more in the sample.

Now for your turn. You’ve been collecting copies of Mario Kart for the Nintendo Wee, and you’ve decided to sell them on eBay. Should you bother photographing each disc yourself, or just use a stock photo of the box? Here are the data — one hundred five auctions used a stock photo, with mean price forty-eight dollars fifty and standard deviation nine dollars forty-seven; thirty-six auctions did not use a stock photo, with mean price forty-four dollars thirty-three and standard deviation seven dollars twenty-five. Two short questions — advance when you are ready to see each answer. Hypotheses are two-sided on the difference between stock and no-stock means. Null: mu sub y minus mu sub n equals zero — same mean price regardless of stock photo. Alternative: mu sub y minus mu sub n is not equal to zero. Observed difference is forty-eight dollars fifty minus forty-four dollars thirty-three, about four dollars and seventeen cents — auctions with a stock photo averaged about four dollars more in the sample.

EDA: Mario Kart

Take a moment with the shapes and spreads of each group — you’ll use what you see on the next slide.

Mario Kart — Conditions

Recall: stock (yes): $n_y = 105$, $\bar{x}_y = 48.50$, $s_y = 9.47$. No stock: $n_n = 36$, $\bar{x}_n = 44.33$, $s_n = 7.25$.

Check independence, normality, and equal variance. Independence: auctions are separate listings, reasonably independent ✓. Normality: both group sizes $\geq 30$, so CLT applies; histograms showed no extreme outliers ✓. Equal variance: $s_y \approx 9.47$ vs. $s_n \approx 7.25$ — within a factor of two, consistent with the similar spreads in the histograms ✓. Conditions met.

Mario Kart — Compute T

Recall: stock (yes): $n_y = 105$, $\bar{x}_y = 48.50$, $s_y = 9.47$. No stock: $n_n = 36$, $\bar{x}_n = 44.33$, $s_n = 7.25$. Observed diff $\bar{x}_y - \bar{x}_n = \$4.17$; conditions met.

Compute the degrees of freedom. $df = n_y + n_n - 2 = 105 + 36 - 2 = 139$
Compute the pooled SD $s_p$. $s_p = \sqrt{\dfrac{104 \cdot 9.47^2 + 35 \cdot 7.25^2}{139}} \approx 8.96$
Compute the standard error $SE$. $SE = s_p \sqrt{\tfrac{1}{n_y} + \tfrac{1}{n_n}} = 8.96 \times \sqrt{\tfrac{1}{105} + \tfrac{1}{36}} \approx 1.73$
Compute $T$. $T = \dfrac{\bar{x}_y - \bar{x}_n}{SE} = \dfrac{4.17}{1.73} \approx 2.41$

With conditions met, work through the test statistic in four short steps. Pause and try each one — advance when you are ready to see each answer. Degrees of freedom is total sample size minus two — one hundred five plus thirty-six minus two — equals one hundred thirty-nine. Pooled standard deviation is the square root of one hundred four times nine point four seven squared, plus thirty-five times seven point two five squared, all divided by one hundred thirty-nine — about eight dollars ninety-six. Standard error is s sub p times the square root of one over n sub y plus one over n sub n — eight dollars ninety-six times the square root of one over one hundred five plus one over thirty-six, which is about one dollar seventy-three. T is the observed difference of four dollars seventeen divided by the standard error of one dollar seventy-three, about two point four one.

Mario Kart — Conclusion

Recall: $T \approx 2.41$, $df = 139$. Observed diff $\bar{x}_y - \bar{x}_n = \$4.17$ (stock-photo higher); hypotheses two-sided.

Two-sided p-value $\approx$ 0.017. State your conclusion at $\alpha = 0.05$, including scope of inference. p-value $< 0.05$ → reject $H_0$. Convincing evidence that stock-photo and no-stock auctions have different mean prices (and stock-photo auctions sell for more on average). Scope: sample of eBay Mario Kart Wii auctions from a specific window — not a random sample of any wider population — so the conclusion is most defensible for similar listings. Observational (sellers chose how to photograph), so we can’t claim using a stock photo causes higher prices — it may reflect product condition, bundling, or listing style.

With T and the p-value in hand, write the conclusion in context. Advance when you are ready to see the answer. The two-sided p-value of about zero point zero one seven is below zero point zero five, so we reject the null. The data provide convincing evidence that stock-photo and no-stock auctions have different mean prices, with stock-photo auctions averaging higher. On scope: this is a sample of eBay Mario Kart Wii auctions from a specific window, not a random sample of any wider population, so the conclusion is most defensible for similar eBay listings. And because it’s observational data — sellers chose how to photograph their listings — we cannot claim that using a stock photo causes higher prices. Sellers who used stock photos may have systematically different products, condition, or listing styles.

Looking Ahead

We’ve tested “Are the two means different?” using the t-distribution with $df = n_1 + n_2 - 2$.

Next time: a confidence interval for $\mu_1 - \mu_2$ — “By how much do the two means differ?” — using the same machinery.

References

Introduction to Modern Statistics (2e) textbook by Mine Çetinkaya-Rundel and Johanna Hardin
Sections 20.3, 20.4
Diez, D. M., Çetinkaya-Rundel, M., & Barr, C. D. OpenIntro Statistics — Mario Kart auction data.

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	One numerical	\(\bar{x}\)	Cherry Blossom — mean finish time
12 (today)	Numerical + binary categorical	\(\bar{x}_1 - \bar{x}_2\)	Birth weights — by smoking habit
13	Numerical, paired	\(\bar{x}_d\)	Textbook prices — UCLA vs Amazon