AP Statistics 6.9 Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions Study Notes
AP Statistics 6.9 Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions Study Notes- New syllabus
AP Statistics 6.9 Justifying a Claim Based on a Confidence Interval for a Difference of Population Proportions Study Notes -As per latest AP Statistics Syllabus.
LEARNING OBJECTIVE
- An interval of values should be used to estimate parameters, in order to account for uncertainty.
Key Concepts:
- Justifying Confidence Interval for a Difference Between Two Population Proportions
Justifying Confidence Interval for a Difference Between Two Population Proportions
Confidence Interval for a Difference Between Two Population Proportions
A confidence interval (CI) for a difference in population proportions provides a range of plausible values for the true difference \( p_1 – p_2 \). It can be used to estimate or justify claims about the populations.
Procedure:
Use a two-sample z-interval for proportions:
\( (\hat{p}_1 – \hat{p}_2) \pm z^* \sqrt{\dfrac{\hat{p}_1 (1 – \hat{p}_1)}{n_1} + \dfrac{\hat{p}_2 (1 – \hat{p}_2)}{n_2}} \)
- \( \hat{p}_1, \hat{p}_2 \) = sample proportions
- \( n_1, n_2 \) = sample sizes
- \( z^* \) = critical value from the standard Normal distribution for the confidence level
Conditions to Verify:
- Random samples from independent populations.
- Independent samples: observations in one sample do not affect the other.
- Normal approximation: \( n_1 \hat{p}_1 \ge 10, n_1 (1-\hat{p}_1) \ge 10, n_2 \hat{p}_2 \ge 10, n_2 (1-\hat{p}_2) \ge 10 \).
Stepwise Calculation:
- Compute the point estimate: \( \hat{p}_1 – \hat{p}_2 \).
- Compute the standard error: \( SE = \sqrt{\dfrac{\hat{p}_1 (1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2 (1-\hat{p}_2)}{n_2}} \).
- Determine the critical value \( z^* \) for the desired confidence level.
- Compute the margin of error: \( ME = z^* \cdot SE \).
- Construct the confidence interval: \( (\hat{p}_1 – \hat{p}_2) \pm ME \).
- Interpret the interval in context, including reference to the samples taken and the populations they represent.
Interpretation Concepts:
- In repeated random sampling with the same sample sizes, approximately C% of the confidence intervals created will capture the true difference in population proportions \( p_1 – p_2 \).
- An interpretation of a CI should include:
- Reference to the specific sample(s) used
- Details about the populations the sample(s) represent
- Contextual meaning of the interval (plausible range for the difference)
Example:
A researcher compares the proportion of male and female students who prefer online learning. Sample data:
- Males: 40 out of 60 prefer online classes (\( \hat{p}_1 = 0.667 \))
- Females: 50 out of 80 prefer online classes (\( \hat{p}_2 = 0.625 \))
Construct a 95% confidence interval for \( p_1 – p_2 \) and interpret it.
▶️ Answer / Explanation
Step 1: Verify conditions
- Random samples
- Independent samples
- Normal approximation: Males: \( n_1 \hat{p}_1 = 40 \), \( n_1 (1-\hat{p}_1) = 20 \); Females: \( n_2 \hat{p}_2 = 50 \), \( n_2 (1-\hat{p}_2) = 30 \)
Step 2: Point estimate
\( \hat{p}_1 – \hat{p}_2 = 0.667 – 0.625 = 0.042 \)
Step 3: Standard error
\( SE = \sqrt{\dfrac{0.667*0.333}{60} + \dfrac{0.625*0.375}{80}} \approx 0.0814 \)
Step 4: Critical value
95% CI → \( z^* = 1.96 \)
Step 5: Margin of error
\( ME = 1.96 * 0.0814 \approx 0.159 \)
Step 6: Confidence interval
\( (\hat{p}_1 – \hat{p}_2) \pm ME = 0.042 \pm 0.159 \Rightarrow [-0.117, 0.201] \)
Step 7: Interpretation
- We are 95% confident that the true difference in preference (males – females) lies between -0.117 and 0.201.
- This interval references the sample data and the populations they represent.
- Since 0 is in the interval, there is insufficient evidence of a difference in preference between male and female students.
- In repeated sampling with the same sample sizes, approximately 95% of such confidence intervals would capture the true difference \( p_1 – p_2 \).
Example :
A school wants to compare the proportion of male and female students who prefer online learning. Sample data:
- Males: 45 out of 75 prefer online classes (\( \hat{p}_1 = 0.6 \))
- Females: 50 out of 80 prefer online classes (\( \hat{p}_2 = 0.625 \))
A 95% confidence interval for \( p_1 – p_2 \) is constructed. Which of the following statements is correct?
- The interval suggests that male students are significantly more likely to prefer online learning than female students.
- The interval suggests that female students are significantly more likely to prefer online learning than male students.
- The interval likely contains 0, so there is insufficient evidence of a difference in preference.
- The interval can prove which gender prefers online learning.
▶️ Answer / Explanation
Step 1: Point estimate
\( \hat{p}_1 – \hat{p}_2 = 0.6 – 0.625 = -0.025 \)
Step 2: Standard error
\( SE = \sqrt{\dfrac{0.6 * 0.4}{75} + \dfrac{0.625 * 0.375}{80}} \approx \sqrt{0.0032 + 0.00293} \approx 0.077 \)
Step 3: Margin of error
95% CI → \( z^* = 1.96 \), so \( ME = 1.96*0.077 \approx 0.151 \)
Step 4: Confidence interval
\( -0.025 \pm 0.151 \Rightarrow [-0.176, 0.126] \)
Step 5: Interpretation
- The interval contains 0, so there is insufficient evidence of a difference in preference between male and female students.
- Conclusion: Option C is correct.