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Reading Time: 6 min
Last Updated: March 5, 2026
Main Ideas: 5
Reading Time: 6 min
Last Updated: March 5, 2026
Main Ideas: 5

Topic 5.8 Notes – Sampling Distributions for Differences in Sample Means

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Instead of looking at one sample from each population, we imagine taking many pairs of random samples and tracking how their mean differences behave. This lets us predict variability and calculate probabilities about differences between groups.

1. The Sampling Distribution of xˉ1−xˉ2 \bar{x}_1 - \bar{x}_2

Picture this: you randomly sample students from School A and School B, compute each sample’s mean test score, and subtract. If you repeated that process over and over, each time calculating

xˉ1−xˉ2, \bar{x}_1 - \bar{x}_2,

those values would form a sampling distribution.

Assume:

  • Population 1: mean μ1 \mu_1 , standard deviation σ1 \sigma_1 , sample size n1 n_1
  • Population 2: mean μ2 \mu_2 , standard deviation σ2 \sigma_2 , sample size n2 n_2
  • Samples are random and independent

Mean and Standard Deviation

The sampling distribution has predictable parameters:

μxˉ1−xˉ2=μ1−μ2 \mu_{\bar{x}_1 - \bar{x}_2} = \mu_1 - \mu_2

The average of all possible sample differences equals the true population difference. That’s why xˉ1−xˉ2 \bar{x}_1 - \bar{x}_2 is an unbiased estimator of μ1−μ2 \mu_1 - \mu_2 .

σxˉ1−xˉ2=σ12n1+σ22n2 \sigma_{\bar{x}_1 - \bar{x}_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}

Important details:

  • You add variances, not standard deviations.
  • Larger sample sizes make the distribution less spread out.
  • More variability in either population increases the spread.

Students often forget to square the standard deviations before dividing by sample size. That mistake costs easy points.

2. Shape of the Sampling Distribution

Now we care about when this distribution is normal (or approximately normal), because that’s what lets us calculate probabilities using z-scores.

If Both Populations Are Normal

If each population distribution is normal, then xˉ1−xˉ2 \bar{x}_1 - \bar{x}_2 is exactly normal, no matter the sample sizes.

If Populations Are Not Normal

This is where the Central Limit Theorem kicks in.

If both sample sizes are at least 30:

  • Each sample mean is approximately normal.
  • The difference of two approximately normal distributions is also approximately normal.

So we can model xˉ1−xˉ2 \bar{x}_1 - \bar{x}_2 as normal.

Here’s the idea visually. The left panel shows a heavily skewed population. As the sample size increases (middle and right panels), the sampling distribution of the mean becomes more bell-shaped.

Study guide illustration

Central Limit Theorem illustration

On tests, you must justify normality for both groups. If one sample is large and the other is small from a skewed population, you cannot assume normality.

3. Sampling With and Without Replacement

The standard deviation formula technically assumes sampling with replacement.

If sampling without replacement:

  • The true standard deviation is slightly smaller.
  • If each sample size is less than 10% of its population, the difference is negligible.

This is the familiar 10% condition. On the AP exam, you state it to justify independence when sampling from a large population without replacement.

If the two samples come from completely separate populations, they’re automatically independent.

4. Using the Distribution to Find Probabilities

Suppose:

  • μ1=82 \mu_1 = 82 , σ1=10 \sigma_1 = 10 , n1=40 n_1 = 40
  • μ2=78 \mu_2 = 78 , σ2=12 \sigma_2 = 12 , n2=50 n_2 = 50

First find the parameters:

μxˉ1−xˉ2=82−78=4 \mu_{\bar{x}_1 - \bar{x}_2} = 82 - 78 = 4

σxˉ1−xˉ2=10240+12250=10040+14450=2.5+2.88=5.38≈2.32 \sigma_{\bar{x}_1 - \bar{x}_2} = \sqrt{\frac{10^2}{40} + \frac{12^2}{50}} = \sqrt{\frac{100}{40} + \frac{144}{50}} = \sqrt{2.5 + 2.88} = \sqrt{5.38} \approx 2.32

If we want P(xˉ1−xˉ2>6) P(\bar{x}_1 - \bar{x}_2 > 6) :

z=6−42.32≈0.86 z = \frac{6 - 4}{2.32} \approx 0.86

Then use the normal table or calculator.

Interpretation in context matters. For example:

The mean of the sampling distribution is 4 points, meaning that in repeated sampling, the average difference in sample mean scores (School A minus School B) would be 4 points. The standard deviation is about 2.32 points, meaning the difference in sample means typically varies by about 2.32 points from sample to sample.

Units always stay attached.

5. Common Mistakes

  • Subtracting standard deviations instead of adding variances.
  • Checking the CLT condition for only one sample.
  • Forgetting to mention independence or the 10% condition.
  • Using this formula for paired data. Paired means use a different approach.
  • Giving probability answers without context or units.

Key Takeaways

The center of the sampling distribution is μ1−μ2 \mu_1 - \mu_2 , not xˉ1−xˉ2 \bar{x}_1 - \bar{x}_2 .
The spread is σ12/n1+σ22/n2 \sqrt{\sigma_1^2/n_1 + \sigma_2^2/n_2} ; always square before dividing.
For a normal model, each population must be normal or its sample size must be at least 30.
If sampling without replacement, check that each sample is under 10% of its population.
Interpret the mean and standard deviation of the sampling distribution in real-world units and in context.

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Notes

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