AP Statistics Unit 6 Inference for Proportions

Topic 6.5: Interpreting p-Values

Calculating Test Statistics, Calculating & Interpreting p-Values
Video Follow-Along Worksheet
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Learning Objectives: Essential Knowledge:

Key Vocabulary

Standardized Test Statistic (statistic − null value) / standard deviation of statistic. Measures how far the sample result is from the null value in standard-deviation units.
p-value The probability of obtaining evidence for Ha as strong as or stronger than the observed evidence, by chance alone, when H0 is true.
Standard Error An estimate of the standard deviation of a statistic. For a significance test, we use p0 (not p) since we assume H0 is true.
One-Sided p-value Area in one tail of the standard normal distribution (Ha uses < or >).
Two-Sided p-value Area in both tails of the standard normal distribution (Ha uses ≠). Calculated as 2 × (one-tail area).

Video 1: Calculating a Test Statistic & p-Value

[0:00 – 9:24]
Context — "Does Green = More Natural?" (continued) [0:34] Same study from 6.4: 30 randomly selected students, 18 of 30 chose the green cup. From previous videos: H0: p = 0.50, Ha: p > 0.50. Conditions were met.
1. [1:33] The sample proportion is p = 18/30 = . This gives evidence for Ha because p is the null value of 0.50.
2. [2:00] We need to know how likely it is to get evidence this strong or just by when the null hypothesis is .

Standardized Test Statistic for a One-Sample z-Test

z = (p − p0) / √(p0(1 − p0) / n)
[2:22] General form: (statistic − null value) / standard deviation of statistic
3. [2:46] In the formula, we use (not p) in the denominator because we are .
4. [3:21] Calculate the test statistic for the lemonade study:
(a) Numerator: p − p0 = 0.60 − 0.50 =
(b) Denominator: √(0.50(1 − 0.50) / 30) = √(0.50 × 0.50 / 30) = (round to 4 decimal places)
(c) z = 0.10 / 0.0913 =
5. [3:21] A p of 0.60 is only about standard deviations above the mean. Josh Tabor says this is "."
Formula Sheet Reminder [3:56] The standardized test statistic formula and the standard deviation/standard error of p are both on the AP formula sheet. For a significance test, use the formula (with p0), not the standard error formula (with p), because we assume H0 is true.
6. [5:23] The p-value is the proportion of values in the that are as or more extreme than the observed test statistic, in the direction of the .
7. [5:42] Because Ha: p > 0.50, we find the probability of getting a z-score 1.10. Using Table A or technology, the p-value = .
Practice — Football Favorites: [6:23] 40% of U.S. adults say football is their favorite sport. A mayor wonders if the proportion in his town differs from the national proportion. Random sample of 100 adults: 29 named football. From previous videos: H0: p = 0.40, Ha: p ≠ 0.40. Conditions were met.
8. [6:58] Calculate the test statistic for the football study:
(a) p =
(b) z = (0.29 − 0.40) / √(0.40(1 − 0.40) / 100) =
9. [7:19] The value of p = 0.29 is more than standard deviations the mean. Josh Tabor says this is "pretty ."
10. [7:44] Because Ha is p ≠ 0.40 (a test), we must find area in of the standard normal curve.
11. [8:21] For the football study:
(a) P(z ≤ −2.25) =
(b) P(z ≥ 2.25) =
(c) p-value = 0.0122 + 0.0122 =
p-Value Calculation Summary: [8:37]

Video 2: Interpreting p-Values

[0:00 – 5:40]
12. [1:38] A p-value measures how likely it is to get evidence for Ha as strong as or even than the observed evidence by when .
Lemonade Study Interpretation: [2:06] "Assuming that 50% of all students at this school would choose the green cup, there is a 0.1357 probability of getting a sample proportion of 0.60 or greater by chance alone in a random sample of 30 students from this school."
13. [2:26] A p-value interpretation must include THREE elements:
(a) A statement that the is true (the "assuming" clause).
(b) The probability of getting a result as extreme as or than the observed value.
(c) The phrase "by ."
14. [2:26] You can think of a p-value as a , where the condition is that the is true.
Practice — Football Study Interpretation: [3:20] H0: p = 0.40, Ha: p ≠ 0.40, z = −2.25, p-value = 0.0244.
15. [4:30] Interpret the p-value of 0.0244 for the football study:
"Assuming that % of all adults in the town would say football is their favorite sport, there is a probability of getting a sample proportion as extreme as or more extreme than in by chance alone in a random sample of adults in this town."
16. [4:52] For the two-sided test, we say "in either direction" because the alternative is p ≠ 0.40, meaning evidence the null value supports Ha.

Key Takeaways

[Summary]
  1. Test statistic formula: z = (p − p0) / √(p0(1 − p0) / n). Uses in the denominator, not p.
  2. The p-value is found using the distribution.
  3. For a one-sided test, find area in . For a two-sided test, find area in .
  4. Interpretation must state H0 is true, describe evidence "as extreme or more extreme," and include "by ."

Reflection Questions

[AI-Graded]
R1. In the lemonade study, the p-value was 0.1357. A student says: "There is a 13.57% probability that H0 is true." Explain what is wrong with this interpretation and provide a correct interpretation of the p-value.
R2. Compare the p-values for the lemonade study (0.1357, one-sided) and the football study (0.0244, two-sided). Without yet learning a formal decision rule, which study provides stronger evidence against H0? Explain your reasoning using the meaning of a p-value.

Exit Ticket

A researcher believes that the proportion of left-handed students at a large university is less than 0.10. She takes a random sample of 150 students and finds that 9 are left-handed (p = 0.06).

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