Predict differences between sample means

Try

In a synthetic measurement model, population A has mean 72 units and standard deviation 12 units. Population B has mean 68 and standard deviation 9. Take independent random samples of 36 and 81 measurements. Predict the difference between the two sample means before any data arrive.

The population distributions describe individual measurements. One sample supplies a realized mean. The sampling distribution describes that mean, or a difference of means, over repeated samples. A population spread of 12 is not the spread of a mean of 36 measurements. Averaging changes the latter without changing the population itself.

Predict

Let D = X̄A−X̄B; the bar means an arithmetic average. Predict its center and decide whether its spread is 12−9, √(12²+9²), or something involving sample sizes. Then predict what happens when both sample sizes are quadrupled. Write a reason, not just a choice.

Experiment

Under independent observations within and between samples, E(D) = 72−68 = 4. A sample mean has variance σ²/n, where σ is the population standard deviation. Therefore

Var(D) = 12²/36 + 9²/81 = 4+1 = 5,

and SD(D) = √5 ≈ 2.2361 units. Variance has squared units; standard deviation returns to the measurement units. Quadrupling both sample sizes leaves the center at 4 and divides variance by 4. The new standard deviation is √5/2 ≈ 1.1180.

Independent sample sizes	Center	Variance	Standard deviation
36 and 81	4	5	√5
144 and 324	4	5/4	√5/2

For this drawing, additionally assume both individual populations are normal. The difference is then exactly normal under the model. Use the shared axis to compare how much probability lies between 2 and 6. Explain why the taller dashed curve represents the same total probability as the solid curve.

Understand

Linearity of expectation gives the center even without normality. Independence and finite population variances give the variance formula. Normality is a further statement about shape: independent linear combinations of normal variables are normal. If the populations are not normal, these moment calculations remain valid under the independent sampling assumptions, but a normal tail probability needs a central-limit approximation argument.

There is no universal rule that sample size 30 ensures accuracy. Strong skewness and rare extreme measurements can make convergence slow. Knowledge of two means and two variances alone cannot certify a particular finite-sample tail accuracy. Examine shape, tail behavior and the sampling design before choosing a reference distribution.

Under our stated normal model, find P(D > 4+√5). Subtract center 4 and divide by √5 to obtain z = 1. The supplied CDF Φ(1) ≈ 0.8413 gives upper-tail probability 1−0.8413 = 0.1587. Normal shape is exact here, but the displayed probability is rounded. Supplying this CDF value does not demonstrate calculator proficiency.

Simple random sampling without replacement requires another distinction. With population variance defined using denominator N, the exact mean variance is σ²/n times (N−n)/(N−1). A sample of 36 from 240 is 15%, so it fails the usual 10% approximation check. A sample of 81 from 2000 is 4.05% and passes. Passing is permission for a common approximation, not literal independence within a finite sample. Independence between the sampling processes still needs its own justification.

Apply

Suppose the realized sample means are 73 and 67, with sample standard deviations 15 and 8. The observed difference is 6. A Welch interval uses estimated standard error √(15²/36+8²/81) ≈ 2.6533 and a suitable t reference, rather than the stipulated model spread √5. The interval is centered at the observed 6; the known sampling distribution is centered at the population difference 4. Explain why these statements can both be true.

An equality null about population means does not imply equal variances. Pooling quantitative sample variances requires an additional equal-variance assumption. Do not copy the pooled Bernoulli null formula from the preceding lesson merely because both procedures compare groups.

Now change the design to paired measurements on 25 independent units. Suppose each pair has standard deviations 4 and 3 and covariance 10. The variance of one difference is 16+9−2·10 = 5. The mean of 25 independent paired differences therefore has variance 1/5 and standard deviation 1/√5. Treating columns as independent would instead give variance 1. Positive covariance here makes paired differences less variable because the two measurements tend to move together.

Covariance means the expected product of deviations from their means. Expanding the squared centered difference gives the minus twice covariance term. It is not legitimate to subtract the two variances, and pairing alone does not tell you the numerical covariance. An unfamiliar paired study needs information about the differences or their joint behavior.

Continue

In the practice cases, separate four claims: the center, the variance, the distributional shape, and the numerical probability. Each has its own assumptions. Check whether parameters are stipulated population facts or estimates from a sample. Finally, identify whether independent units are individual measurements, groups, or pairs. A correct arithmetic answer with the wrong independent unit is not a justified sampling model.