AP Statistics (AP Stats): Sampling Distribution of a Sample Proportion (p?) – Conditions (10% condition, np?10, n(1?p)?10)

AP Statistics – Sampling Distribution of a Sample Proportion (p?) – Conditions (10% condition, np?10, n(1?p)?10)

AP Statistics Study Guide: Sampling Distribution of a Sample Proportion (p?) – Conditions

What This Is

The sampling distribution of a sample proportion (p?) describes how the proportion of successes in a sample varies from sample to sample. It’s essential for constructing confidence intervals and performing hypothesis tests about a population proportion (p). The conditions (10% condition, np-10, n(1–p)-10) ensure that the sampling distribution is approximately Normal, allowing us to use the z-distribution for inference.

Real-world example: A pharmaceutical company tests whether a new vaccine is effective in at least 90% of cases. They take a random sample of 500 patients and find that 460 respond positively (p? = 0.92). Before making conclusions, they must verify that the sampling distribution of p? is approximately Normal by checking the conditions.

Key Terms & Formulas

• Sample proportion (p?): p? = X/n, where X = number of successes, n = sample size.
• Population proportion (p): The true proportion of successes in the entire population.
• Sampling distribution of p?: The distribution of all possible sample proportions (p?) from samples of size n.
• Mean of p?: ? = p (the sampling distribution is centered at the true population proportion).
• Standard deviation of p? (standard error): ? = ?(p(1–p)/n) (measures spread of the sampling distribution).
• Normal approximation conditions:
• 10% condition: n-0.10N (sample size-10% of the population)-Ensures independence when sampling without replacement.
• Large Counts condition: np-10 and n(1–p)-10-Ensures the sampling distribution is approximately Normal.
• One-proportion z-test: z = (p? – p?) / ?(p?(1–p?)/n)-Tests H?: p = p? vs. H?: p-p? (or one-sided).
• One-proportion z-interval: p? ± z ?(p?(1–p?)/n)-Confidence interval for p*.
• 1-PropZTest (TI-84): STAT-TESTS-5:1-PropZTest-Input p?, x (successes), n, and H?.
• 1-PropZInt (TI-84): STAT-TESTS-A:1-PropZInt-Input x, n, and confidence level.

Step-by-Step / Process Flow

For a Hypothesis Test (FRQ Example)

1. State hypotheses:
2. H?: p = p? (null hypothesis, e.g., p = 0.90 for the vaccine example).

3. H?: p-p? (or p > p? or p < p?, depending on the claim).

4. Check conditions:

5. Random: The sample must be randomly selected (stated in the problem).
6. 10% condition: n-0.10N (if sampling without replacement).

7. Large Counts: np?-10 and n(1–p?)-10 (use p? from H?, not p?).

8. Compute the test statistic:

9. z = (p? – p?) / ?(p?(1–p?)/n)

10. Calculator: Use 1-PropZTest (input p?, x, n, and H?).

11. Find the p-value:

12. Use normalcdf(lower, upper, 0, 1) for two-tailed tests (e.g., normalcdf(-1E99, -z, 0, 1) * 2).

13. Calculator: 1-PropZTest gives the p-value directly.

14. Make a conclusion in context:

15. If p-value < ? (e.g., 0.05), reject H? and conclude there is significant evidence for H?.
16. If p-value-?, fail to reject H? (do not say "accept H?").

For a Confidence Interval (FRQ Example)

1. State the parameter: "We want to estimate the true proportion p of [context]."

2. Check conditions:

3. Random: The sample must be randomly selected.
4. 10% condition: n-0.10N (if sampling without replacement).

5. Large Counts: np?-10 and n(1–p?)-10 (use p?, not p).

6. Compute the interval:

7. p? ± z ?(p?(1–p?)/n)*

8. Calculator: Use 1-PropZInt (input x, n, and confidence level).

9. Interpret the interval in context:

10. "We are [C%] confident that the true proportion of [context] is between [lower bound] and [upper bound]."

Common Mistakes

• Mistake: Forgetting to check the 10% condition when sampling without replacement.

• Correction: Always verify n-0.10N unless the problem states sampling with replacement (rare in AP Stats). The 10% condition ensures independence between observations.

• Mistake: Using p? instead of p? in the Large Counts condition for a hypothesis test.

• Correction: For hypothesis tests, use p? (from H?) to check np?-10 and n(1–p?)-10. For confidence intervals, use p?.

• Mistake: Rounding p? too early (e.g., using p? = 0.4 instead of p? = 0.405 in calculations).

• Correction: Keep p? in decimal form (e.g., 405/1000 = 0.405) until the final answer to avoid rounding errors.

• Mistake: Misinterpreting the confidence level as the probability that p is in the interval.

• Correction: The confidence level (e.g., 95%) is the long-run success rate of the method, not the probability for this specific interval. Say: "We are 95% confident that the true proportion is between X and Y."

• Mistake: Using the t-distribution instead of the z-distribution for proportions.

• Correction: Proportions always use the z-distribution (Normal). The t-distribution is for means when ? is unknown.

AP Exam Insights

• FRQs often combine conditions with inference: Expect a question like:

  "A researcher claims that 60% of high school students support a later start time. In a random sample of 200 students, 130 support the change. Do these data provide convincing evidence against the researcher’s claim at the-= 0.05 level?"

• You must check conditions, perform a z-test, and interpret the p-value.

• Tricky distinction: p vs. p? vs. p?:

• p = true population proportion (unknown).
• p? = sample proportion (observed).

• p? = hypothesized proportion (from H?).

• Calculator pitfall: 1-PropZTest and 1-PropZInt do not check conditions for you—you must verify them manually.

• Common FRQ setup: A problem gives a sample proportion and asks for a confidence interval or hypothesis test, requiring you to:

• State the parameter.
• Check conditions.
• Perform calculations.
• Interpret results in context.

Quick Check Questions

Question 1 (Multiple Choice)

A factory claims that 5% of its light bulbs are defective. In a random sample of 200 bulbs, 18 are defective. Which condition is not satisfied for performing a one-proportion z-test? (A) The sample is random. (B) n-0.10N (C) np?-10 and n(1–p?)-10 (D) The sample size is large enough for the Central Limit Theorem.

Answer: (C) Explanation: np? = 200(0.05) = 10 (barely meets the condition), but n(1–p?) = 200(0.95) = 190-10 is satisfied. However, np? = 10 is the minimum requirement, and the problem may expect you to recognize that np? = 10 is technically acceptable but borderline. The 10% condition (B) is likely satisfied (unless the population is very small), and (A) and (D) are correct. The most questionable condition is (C).

Question 2 (FRQ Part)

A school newspaper reports that 70% of students favor a new dress code. A student disagrees and surveys a random sample of 50 students, finding that 30 favor the dress code. (a) Check the conditions for constructing a 95% confidence interval for the true proportion of students who favor the dress code. (b) Construct and interpret the 95% confidence interval.

Answer: (a) Conditions: - Random: The sample is stated to be random. - 10% condition: 50-0.10N (assuming the school has at least 500 students, which is reasonable). - Large Counts: np? = 50(0.6) = 30-10 and n(1–p?) = 50(0.4) = 20-10. All conditions are satisfied.

(b) Interval: - p? = 30/50 = 0.6 - z = 1.96 (for 95% confidence) - SE = ?(0.6(0.4)/50)-0.0693 - Margin of Error = 1.96(0.0693)-0.136 - Interval: 0.6 ± 0.136? (0.464, 0.736) - Interpretation: We are 95% confident that the true proportion of students who favor the new dress code is between 46.4% and 73.6%.

Last-Minute Cram Sheet

1. Sampling distribution of p?: ? = p, ? = ?(p(1–p)/n).
2. Conditions for Normal approximation:
3. 10% condition: n-0.10N (sampling without replacement).
4. Large Counts: np-10 and n(1–p)-10 (use p? for tests, p? for intervals).
5. One-proportion z-test: z = (p? – p?) / ?(p?(1–p?)/n).
6. One-proportion z-interval: p? ± z ?(p?(1–p?)/n)*.
7. TI-84 commands:
8. 1-PropZTest (hypothesis test).
9. 1-PropZInt (confidence interval).
10. Always check conditions before calculating!
11. For hypothesis tests, use p? in Large Counts; for intervals, use p?.
12. Never say "the probability that p is in the interval is 95%."
13. Round p? to 3 decimal places in calculations.
14. If np < 10 or n(1–p) < 10, the Normal approximation is invalid—use binomial methods (rare on AP Exam).