AP Statistics Unit 5 Sampling Distributions

Topic 5.8: Sampling Distributions for Differences in Sample Means

From One Mean to Two — Comparing Populations with x̄1 − x̄2
Video Follow-Along Worksheet
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Learning Objectives: Essential Knowledge:

Key Vocabulary

Difference in Sample Means x̄1 − x̄2 — the statistic computed by subtracting the sample mean of one group from the sample mean of another
Sampling Distribution The distribution of x̄1 − x̄2 across all possible pairs of random samples of the given sizes from two populations
10% Condition Both sample sizes must be less than 10% of their respective population sizes for the SD formula to be valid
Independence of Samples The two samples must be selected independently of each other (knowing one sample tells you nothing about the other)

Part 1: Building the Sampling Distribution of x̄L − x̄O

[Video 1: 0:00 – 3:47]
Setup — Sampling Citrus: [0:51] A tree produces lemons with μL = 4 oz, σL = 0.5 oz. A different tree produces oranges with μO = 3 oz, σO = 0.4 oz. Both distributions are approximately normal. We select n = 6 fruit from each tree, find each sample mean, and compute x̄L − x̄O.
1. [1:55] Each dot on the sampling distribution represents the between the mean weight of 6 lemons and the mean weight of 6 oranges — not an individual fruit’s weight.
2. [3:28] After 500 simulated samples, the distribution of x̄L − x̄O is approximately , centered at ounce(s).
3. The center of the sampling distribution (1 oz) equals the difference in the two means (μL − μO = 4 − 3 = 1).

Part 2: Formulas & Conditions

[Video 1: 4:08 – 6:44]

Formulas (on your AP formula sheet)

Mean: μx̄1 − x̄2 = μ1 − μ2
SD: σx̄1 − x̄2 = √(σ1²/n1 + σ2²/n2)
4. [5:22] Apply the mean formula: μx̄L − x̄O = 4 − 3 = ounce
5. [5:48] Apply the SD formula: σx̄L − x̄O = √(0.5²/6 + 0.4²/6) = ounces
6. [5:06] Two conditions for the SD formula to be valid: both sample sizes must be less than of their population sizes, and the two samples must be of each other.

Pause & Think

7. Notice the SD formula adds the two fractions under the square root, even though we subtract the means. This is because variability always when you combine independent random variables — uncertainty from both samples contributes to the total spread.

Part 3: Shape — When Is the Distribution Approximately Normal?

[Video 1: 6:44 – 8:28]
Two Paths to Normality (only one needs to hold):
[6:44] Path 1: Both population distributions are approximately normal.
[7:06] Path 2: Both sample sizes are large enough (n1 ≥ 30 and n2 ≥ 30).
8. In the citrus example, both population distributions are approximately normal. This means the sampling distribution of x̄L − x̄O is approximately normal even though n = , which is less than 30.
9. If neither population were normal, we would need both sample sizes to be at least for the sampling distribution to be approximately normal.

Comparing Normality Conditions Across All Sampling Distributions

p̂1 − p̂2 x̄ (single mean) x̄1 − x̄2
Normality Condition Large Counts
(4 checks of np ≥ 10)		 Pop. normal
OR n ≥ 30		 Both pops. normal
OR both n ≥
SD Formula Condition Both n < 10% of N n < 10% of N Both n < 10% of N

Part 4: Interpreting Parameters in Context

[Video 2: 0:00 – 3:33]
10. [1:54] Interpreting the mean: A correct interpretation of μx̄L − x̄O = 1 must reference " random samples" of 6 lemons and 6 oranges and include .
11. [2:27] The order of subtraction matters: if we had subtracted O − L instead, the mean difference would be ounce.
12. [2:43] Interpreting the SD: σx̄L − x̄O = 0.26 means the differences in sample mean weight will vary by about 0.26 ounces from the true mean difference of 1 ounce. (Use "typically" or "on average.")

Part 5: Calculating & Interpreting Probabilities

[Video 2: 3:33 – 6:23]
Probability Question: [3:33] Would it be unusual to get a sample mean weight for the lemons that was more than 1.2 ounces greater than the sample mean weight for the oranges?
13. [4:19] Calculate the z-score: z = (1.2 − 1) / 0.26 =
14. [4:37] Using Table A or technology: P(x̄L − x̄O > 1.2) ≈
15. [4:45] Getting a difference of 1.2+ ounces happens in about % of all possible samples, so this result unusual.

Why Study Sampling Distributions?

16. [5:11] Knowing which values of a statistic are unusual is one of the most important reasons we study sampling distributions. This directly connects to in Units 6–9.

Reflection Questions

[AI-Graded]
R1. Compare the sampling distribution of x̄1 − x̄2 (difference in sample means) with the sampling distribution of p̂1 − p̂2 (difference in sample proportions) from Topic 5.6. How are the formulas for the mean and standard deviation similar? How are the normality conditions different?
R2. In the citrus example, the z-score for a difference of 1.2 ounces was only 0.77 (not unusual). Suppose we increased both sample sizes from 6 to 60. Without calculating the exact probability, explain what would happen to the standard deviation of the sampling distribution, the z-score, and the probability. Would a difference of 1.2 ounces become unusual? Explain your reasoning.

Exit Ticket

A species of oak tree produces acorns with lengths that are approximately normally distributed with μ1 = 50 mm and σ1 = 4 mm. A different species has acorns with μ2 = 45 mm and σ2 = 3 mm (also approximately normal). A botanist selects 16 acorns from each species.

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