AP Statistics Unit 6 Inference for Proportions

Topics 6.1–6.2: Why Be Normal? & Constructing a Confidence Interval for p

The Logic of Significance Testing, CI Procedures, Conditions & Calculations
Video Follow-Along Worksheet
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Learning Objectives:
🖩 Practice on the TI-84 →

Key Vocabulary

Significance Testing Determining if evidence is convincing by evaluating whether results could plausibly occur by random chance alone
Confidence Interval An interval of values used to estimate a population parameter, accounting for sampling variability
Point Estimate A single-value estimate of a population parameter (e.g., p̂ estimates p)
Margin of Error How much a sample statistic is likely to vary from the population parameter; ME = z* × SE
Standard Error An estimate of the standard deviation of the sampling distribution: SEp̂ = √(p̂(1 − p̂) / n)
Critical Value (z*) A multiplier that sets the confidence level (e.g., z* = 1.96 for 95% confidence)
One-Sample z-Interval The specific procedure for constructing a CI for a single population proportion

Video 1: Why Be Normal? (Topic 6.1)

[Video: ~7 min]
1. [0:13] This video introduces two key questions: How do we identify for a claim? And how do we determine if the evidence for a claim is ?
Green Cup Example: [0:43] Two student researchers randomly selected 30 students. Each tasted lemonade from a green cup and a white cup (same lemonade) and said which tasted “more natural.”
2. [1:11] Of the 30 students, stated the lemonade in the green cup tasted more natural.
3. [1:56] If the color had no effect, we’d expect % of students to pick the green cup. The sample proportion was p̂ = 18/30 = .
4. [2:27] Explanation #1: The color had no effect, and the result of 60% occurred by alone.
5. [2:55] If it’s pretty to get p̂ ≥ 0.60 by chance alone, we can rule out Explanation #1.
6. [3:45] To simulate random choice (50/50), the researchers flipped a times per trial.
7. [5:22] In 100 simulated trials, p̂ ≥ 0.60 occurred times. Estimated probability ≈ .
8. [5:57] Because getting p̂ ≥ 0.60 is somewhat likely by chance, the data provide convincing evidence that students associate green with being more natural.
Key Takeaway: [6:26] To determine if evidence is convincing: (1) Consider two explanations — random chance or a real effect. (2) Estimate the probability of the evidence occurring by chance alone. (3) If you can eliminate random chance as a plausible explanation, the evidence is convincing.
9. [6:26] If you can eliminate as a plausible explanation for the evidence, the evidence for that claim is convincing.

Video 2: Identifying the Procedure & Checking Conditions (Topic 6.2)

[Video: ~10 min]
10. [0:52] The two major types of inference are: (used to estimate a parameter) and (used to test a claim about a parameter).
Verifying Signatures Example: [2:28] Supporters of Proposition 100 submitted 9,388 signatures. City administrators randomly selected 500 to verify. Of those, 364 were from registered voters. Goal: construct a 95% confidence interval for the proportion of all valid signatures.
11. [4:10] When estimating the proportion of successes in a single population, we use a .
Conditions for a One-Sample z-Interval for p: [5:11]
  1. Random: Data collected using a random sample from the population.
  2. 10% Condition: When sampling without replacement, n ≤ 10% of N.
  3. Large Counts: np̂ ≥ 10 AND n(1 − p̂) ≥ 10.
12. [5:18] Condition 1 (Independence): Data must be collected using a from the population.
13. [5:31] Condition 2: When sampling without replacement, the sample size must be at most % of the population size.
14. [5:58] Condition 3 (Large Counts): Both np̂ and n(1 − p̂) must be at least . (Number of successes AND failures both ≥ 10)
15. [6:47] Applying to the signature example: p̂ = 364 / 500 = .
16. [7:07] 10% check: 10% of 9,388 = . Since 500 ≤ 938.8, the condition is .
17. [7:22] Large Counts check: Successes = ≥ 10 ✓.   Failures = ≥ 10 ✓.
18. [8:44] Practice (Part 2): If they only checked the first 40 signatures (38 verified), the Random condition fails because they used a sample.
19. [9:03] The Large Counts condition also fails because the number of failures is only , which is less than 10.

Video 3: Calculating the Confidence Interval (Topic 6.2)

[Video: ~7 min]

Confidence Interval Formula

p̂ ± z* × √(p̂(1 − p̂) / n)
point estimate ± margin of error
20. [1:46] A confidence interval has the form: ± .
21. [2:42] The margin of error is determined by two factors: the and the .
22. [3:33] Standard error of p̂: SE = √(p̂(1 − p̂) / ). (We use p̂ instead of p because we don’t know the true population proportion.)
23. [4:03] For a 95% confidence interval, the critical value z* = .
24. [4:26] Full calculation: With p̂ = 0.728 and n = 500:
ME = 1.96 × √(0.728 × 0.272 / 500) ≈ .
25. [4:26] 95% CI: (, ).
26. [5:51] Converting to counts: Estimated valid signatures: 9,388 × (0.689 to 0.767) ≈ (, ).
Formula Sheet Tip: [5:20] You don’t need to memorize the full CI formula! The AP formula sheet provides: (1) the generic CI form: statistic ± z* × SE, and (2) the standard error: SEp̂ = √(p̂(1 − p̂) / n).

Video 4: Determining Minimum Sample Size (Topic 6.2)

[Video: ~7 min]
27. [0:29] In sample-size problems, we are in the stage — we have not collected data yet.
28. [1:59] If no prior estimate of p̂ exists, use p̂ = because it produces the possible margin of error (an upper bound for n).
Example 1: [0:07] A food services director wants to estimate the proportion of students who always bring lunch. What is the smallest sample size for ME ≤ 0.05 with 95% confidence?
29. [3:28] For 95% confidence with ME ≤ 0.05 and p̂ = 0.5: n ≥ . (Always round to the next whole number!)
Example 2: [4:09] A newspaper editor wants ME ≤ 0.03 with 90% confidence for the proportion of subscribers who read the comics.
30. [4:48] For 90% confidence, the critical value z* = .
31. [5:40] With ME ≤ 0.03 and 90% confidence: n ≥ .

Reflection Questions

[AI-Graded]
R1. Explain the logic of significance testing using the green cup example from Topic 6.1. In your explanation, address: (a) What is the evidence in this study? (b) What are the two possible explanations for the evidence? (c) How did the simulation help evaluate these explanations? (d) Why did we conclude that the evidence was NOT convincing?
R2. A school nurse randomly selects 150 students from a school of 2,400 to survey about flu vaccination. She finds that 96 of the 150 have received the flu vaccine. (a) Verify all three conditions for constructing a confidence interval. (b) Construct a 95% confidence interval for the proportion of all students who have received the flu vaccine. (c) Interpret your interval in context using: “We are 95% confident that…”

Exit Ticket

A city council member claims that more than half of residents support a new park project. A research firm randomly selects 400 residents from a city of 50,000 and finds that 224 support the project (p̂ = 0.56).

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