AP Statistics 6.4 Setting Up a Test for a Population Proportion Study Notes
AP Statistics 6.4 Setting Up a Test for a Population Proportion Study Notes- New syllabus
AP Statistics 6.4 Setting Up a Test for a Population Proportion Study Notes -As per latest AP Statistics Syllabus.
LEARNING OBJECTIVE
- The normal distribution may be used to model variation.
Key Concepts:
- One-Sample z-Test for a Population Proportion
- Identifying an Appropriate Testing Method for a Population Proportion
- Verifying Conditions for Testing a Population Proportion
One-Sample z-Test for a Population Proportion
One-Sample z-Test for a Population Proportion
Purpose: To test a claim about the value of a population proportion \( p \).
Step 1: State Hypotheses
Null Hypothesis (H₀): Always states the population proportion equals a specific value.
\( H_0 : p = p_0 \)
Alternative Hypothesis (Hₐ): Depends on the claim:
One-Sided Test:
- \( H_a : p > p_0 \) → right-tailed (testing if proportion is greater)
- \( H_a : p < p_0 \) → left-tailed (testing if proportion is smaller)
Two-Sided Test:
- \( H_a : p \neq p_0 \) → two-tailed (testing if proportion is different in either direction)
Step 2: Verify Conditions
- Random: Data come from a random sample or randomized experiment.
- Independence: Observations are independent. If sampling without replacement, check the 10% condition: \( n \leq 0.1N \).
- Normal Approximation: The sample is large enough to use a Normal model.
- Check: \( n p_0 \geq 10 \) and \( n(1 – p_0) \geq 10 \).
Step 3: Test Statistic
\( z = \dfrac{\hat{p} – p_0}{\sqrt{\dfrac{p_0 (1 – p_0)}{n}}} \)
- \( \hat{p} = \dfrac{x}{n} \) = sample proportion
- \( p_0 \) = hypothesized population proportion
Step 4: P-value and Decision
- Find the P-value from the standard Normal distribution, depending on the alternative hypothesis.
- Compare the P-value to the significance level \( \alpha \).
- If P-value ≤ \( \alpha \), reject \( H_0 \). Otherwise, fail to reject \( H_0 \).
Example :
A company advertises that 80% of its customers are satisfied. In a random sample of 100 customers, 74 say they are satisfied. At the 5% significance level, do the data provide evidence that the satisfaction rate is less than advertised?
▶️ Answer / Explanation
Step 1: Hypotheses:
\( H_0 : p = 0.80 \)
\( H_a : p < 0.80 \) (left-tailed, one-sided).
Step 2: Conditions:
- Random sample assumed.
- 10% condition: \( 100 \leq 0.1N \) (reasonable for a large customer base).
- Normal approximation: \( n p_0 = 100(0.8) = 80 \geq 10 \), \( n(1 – p_0) = 100(0.2) = 20 \geq 10 \).
Step 3: Test statistic:
\( \hat{p} = \dfrac{74}{100} = 0.74 \)
\( z = \dfrac{0.74 – 0.80}{\sqrt{\dfrac{0.8(0.2)}{100}}} = \dfrac{-0.06}{\sqrt{0.0016}} = \dfrac{-0.06}{0.04} = -1.5 \)
Step 4: P-value = \( P(Z < -1.5) \approx 0.067 \).
Conclusion: Since P-value (0.067) > 0.05, we fail to reject \( H_0 \). There is not enough evidence at the 5% level to conclude that customer satisfaction is less than 80%.
Example:
A manufacturer claims that 90% of its batteries last at least 6 hours. A consumer group suspects the proportion is less than this. State the hypotheses.
▶️ Answer / Explanation
Parameter: Let \( p = \) true proportion of batteries that last at least 6 hours.
Null Hypothesis: \( H_0 : p = 0.90 \)
Alternative Hypothesis: \( H_a : p < 0.90 \) (left-tailed, because the suspicion is “less than”).
Example:
A polling organization claims that exactly 50% of voters support a new law. An opposition group believes the support level is different. State the hypotheses.
▶️ Answer / Explanation
Parameter: Let \( p = \) true proportion of voters who support the new law.
Null Hypothesis: \( H_0 : p = 0.50 \)
Alternative Hypothesis: \( H_a : p \neq 0.50 \) (two-tailed, because the group suspects “different,” not specifically higher or lower).
Identifying an Appropriate Testing Method for a Population Proportion
Identifying an Appropriate Testing Method for a Population Proportion
When to Use: A 1-sample z-test for a population proportion is appropriate when you want to test a claim about the proportion of a categorical outcome in a population.
Key Features:
- Parameter of Interest: The population proportion \( p \).
- Data Type: Categorical (success/failure, yes/no, etc.).
- Sample Size: A single random sample of size \( n \).
- Null Hypothesis: Always of the form \( H_0 : p = p_0 \).
- Alternative Hypothesis: One-sided (\( p > p_0 \) or \( p < p_0 \)) or two-sided (\( p \neq p_0 \)).
Conditions to Check:
- Random: Sample comes from random sampling or a randomized experiment.
- Independence: If sampling without replacement, check \( n \leq 0.1N \).
- Normal Approximation: The sampling distribution of \( \hat{p} \) is approximately Normal if \( n p_0 \geq 10 \) and \( n(1 – p_0) \geq 10 \).
Test Statistic:
\( z = \dfrac{\hat{p} – p_0}{\sqrt{\dfrac{p_0 (1 – p_0)}{n}}} \)
Decision Rule:
- Find the P-value from the z test statistic using the standard Normal distribution.
- Compare to the significance level \( \alpha \).
- If P-value ≤ \( \alpha \), reject \( H_0 \); otherwise, fail to reject \( H_0 \).
Example:
A candidate claims that 60% of voters support her. In a random sample of 150 voters, 78 say they support her. Is a 1-sample z-test for a population proportion appropriate here?
▶️ Answer / Explanation
Step 1: Parameter: Let \( p = \) true proportion of voters who support the candidate.
Step 2: Hypotheses: \( H_0 : p = 0.60 \), \( H_a : p \neq 0.60 \).
Step 3: Conditions: Random sample assumed. Independence: 150 ≤ 0.1N (population large). Normal approximation: \( n p_0 = 150(0.6) = 90 \geq 10 \), \( n(1-p_0) = 150(0.4) = 60 \geq 10 \).
Conclusion: All conditions are satisfied, so a 1-sample z-test for a population proportion is appropriate.
Verifying Conditions for Testing a Population Proportion
Verifying Conditions for Testing a Population Proportion
Before performing a one-sample z-test for a population proportion, we must ensure the assumptions for statistical inference are satisfied. There are two main checks: independence and approximate normality of the sampling distribution.
1. Independence:
- Data should come from a random sample or a randomized experiment.
- If sampling without replacement, check the 10% condition: \( n \leq 0.1N \), where \( N \) is the population size.
2. Normal Approximation (Shape of Sampling Distribution):
- Assume the null hypothesis \( H_0: p = p_0 \) is true.
- Check the success-failure condition:
- Number of successes: \( n p_0 \geq 10 \)
- Number of failures: \( n (1 – p_0) \geq 10 \)
- If both are satisfied, the sample size is large enough to assume the sampling distribution of \( \hat{p} \) is approximately Normal.
Summary: Only if both independence and normal approximation conditions are met can we perform a one-sample z-test for a population proportion confidently.
Example:
A school claims that 70% of students participate in extracurricular activities. A random sample of 120 students is taken. Verify if a one-sample z-test for the population proportion is appropriate.
▶️ Answer / Explanation
Step 1: Independence
- Data come from a random sample.
- 10% condition: 120 ≤ 0.1N (assuming total students N ≥ 1200).
Step 2: Normal Approximation
- Assume \( H_0: p = 0.70 \).
- Number of successes: \( n p_0 = 120 * 0.70 = 84 \geq 10 \)
- Number of failures: \( n (1 – p_0) = 120 * 0.30 = 36 \geq 10 \)
Conclusion: Both conditions satisfied. A one-sample z-test for a population proportion is appropriate.
Example
A startup claims that 95% of users like their app. A random sample of 15 users is taken. Verify conditions for a one-sample z-test.
▶️ Answer / Explanation
Step 1: Independence
- Random sample assumed.
- 10% condition: 15 ≤ 0.1N (likely satisfied).
Step 2: Normal Approximation
- Assume \( H_0: p = 0.95 \)
- Number of successes: \( n p_0 = 15 * 0.95 = 14.25 < 10 \)? Actually, 14.25 ≥ 10
- Number of failures: \( n (1 – p_0) = 15 * 0.05 = 0.75 < 10
Conclusion: Normal approximation condition not satisfied (few failures). A one-sample z-test is not appropriate. Use an exact test (like a binomial test) instead.