AP Statistics 6.8 Confidence Intervals for the Difference of Two Proportions Study Notes
AP Statistics 6.8 Confidence Intervals for the Difference of Two Proportions Study Notes- New syllabus
AP Statistics 6.8 Confidence Intervals for the Difference of Two 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:
- Confidence Intervals for the Difference Between Two Population Proportions
- Verifying Conditions for a Two-Sample Proportion Confidence Interval
- Calculate an Appropriate Confidence Interval for a Comparison of Population Proportions
- Calculate an Interval Estimate Based on a Confidence Interval for a Difference of Proportions
Confidence Intervals for the Difference Between Two Population Proportions
Confidence Intervals for the Difference Between Two Population Proportions
Appropriate Procedure:
To estimate the difference between two population proportions (\( p_1 – p_2 \)), 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 standard Normal distribution (depends on confidence level)
Conditions to Verify:
- Random: Samples must come from independent random samples or randomized experiments.
- Independent Samples: Each sample must be independent of the other.
- Normal Approximation: For each sample, check:
- n₁ * p̂₁ ≥ 10 and n₁ * (1 – p̂₁) ≥ 10
- n₂ * p̂₂ ≥ 10 and n₂ * (1 – p̂₂) ≥ 10
Interpretation:
- The resulting interval estimates the true difference between population proportions \( p_1 – p_2 \).
- If 0 is not contained in the interval, there is evidence of a difference between the two populations.
- Always interpret the interval in context, including which population corresponds to \( p_1 \) and \( p_2 \).
Example:
A researcher wants to compare the proportion of male and female students who prefer online classes. A random sample finds:
- 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 \( p_1 – p_2 \).
▶️ Answer / Explanation
Step 1: Check conditions
- Random: samples are random
- Independent: samples are from different groups
- Normal approximation:
- Males: 60*0.667 = 40, 60*0.333 = 20
- Females: 80*0.625 = 50, 80*0.375 = 30
Step 2: Compute standard error
SE = √[0.667*0.333/60 + 0.625*0.375/80] ≈ √[0.0037 + 0.00293] ≈ √0.00663 ≈ 0.0814
Step 3: Critical value
95% CI → z* = 1.96
Step 4: Confidence interval
(0.667 – 0.625) ± 1.96 * 0.0814 → 0.042 ± 0.159 → [-0.117, 0.201]
Step 5: Interpretation
We are 95% confident that the true difference in preference (males – females) lies between -0.117 and 0.201. Since 0 is in the interval, there is no strong evidence of a difference in preference between male and female students.
Verifying Conditions for a Two-Sample Proportion Confidence Interval
Verifying Conditions for a Two-Sample Proportion Confidence Interval
Before constructing a confidence interval for the difference between two population proportions (\( p_1 – p_2 \)), the following conditions must be verified:
1. Randomness:
- Each sample must be collected randomly or come from a randomized experiment.
2. Independence:
- The two samples must be independent of each other.
- Within each sample, observations must be independent.
- If sampling without replacement, check the 10% condition: \( n_1 \le 0.1 N_1 \) and \( n_2 \le 0.1 N_2 \).
3. Normal Approximation (Large Counts):
- For each sample, both the number of successes and failures should be at least 10:
- Sample 1: \( n_1 \hat{p}_1 \ge 10 \) and \( n_1 (1 – \hat{p}_1) \ge 10 \)
- Sample 2: \( n_2 \hat{p}_2 \ge 10 \) and \( n_2 (1 – \hat{p}_2) \ge 10 \)
- This ensures the sampling distribution of \( \hat{p}_1 – \hat{p}_2 \) is approximately normal.
Summary:
- All three conditions (Randomness, Independence, Normal approximation) must be satisfied before constructing a two-sample proportion confidence interval.
- If any condition fails, the resulting confidence interval may not be reliable.
Example:
A researcher wants to compare 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 (\( \hat{p}_2 = 0.625 \))
Verify whether it is appropriate to construct a 95% confidence interval for \( p_1 – p_2 \).
▶️ Answer / Explanation
- Randomness: Samples are random
- Independence: Samples are from different groups, and n₁ ≤ 0.1N₁, n₂ ≤ 0.1N₂
- Normal approximation:
- Males: n₁ * p̂₁ = 60*0.667 = 40 ≥ 10, n₁*(1-p̂₁) = 60*0.333 = 20 ≥ 10
- Females: n₂ * p̂₂ = 80*0.625 = 50 ≥ 10, n₂*(1-p̂₂) = 80*0.375 = 30 ≥ 10
All conditions are satisfied, so constructing a two-sample proportion confidence interval is appropriate.
Calculate an Appropriate Confidence Interval for a Comparison of Population Proportions
Calculate an Appropriate Confidence Interval for a Comparison of Population Proportions
For comparing two population proportions, we construct a confidence interval for \( p_1 – p_2 \) to estimate the range of plausible values for the true difference between the two 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 (depends on confidence level)
Stepwise Approach:
- Verify conditions: Randomness, Independence, and Normal approximation.
- Compute the point estimate: \( \hat{p}_1 – \hat{p}_2 \).
- Calculate the standard error: \( SE = \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \).
- Find \( z^* \) for the desired confidence level.
- Determine the margin of error: \( ME = z^* \times SE \).
- Construct the confidence interval: \( (\hat{p}_1 – \hat{p}_2) \pm ME \).
- Interpret the interval in context.
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 \( p_1 – p_2 \).
▶️ Answer / Explanation
Step 1: Check conditions
- Random samples from independent groups.
- Normal approximation:
- Males: \( 60 \times 0.667 = 40 \ge 10 \), \( 60 \times 0.333 = 20 \ge 10 \)
- Females: \( 80 \times 0.625 = 50 \ge 10 \), \( 80 \times 0.375 = 30 \ge 10 \)
Step 2: Point estimate
\( \hat{p}_1 – \hat{p}_2 = 0.667 – 0.625 = 0.042 \)
Step 3: Standard error
\( SE = \sqrt{(0.667 \times 0.333 / 60) + (0.625 \times 0.375 / 80)} \approx 0.0814 \)
Step 4: Critical value and margin of error
95% CI → \( z^* = 1.96 \); \( ME = 1.96 \times 0.0814 = 0.159 \)
Step 5: Confidence interval
\( 0.042 \pm 0.159 \Rightarrow [-0.117,\; 0.201] \)
Step 6: Interpretation
We are 95% confident that the true difference in proportions (males – females) lies between -0.117 and 0.201. Since 0 is in the interval, there is no strong evidence of a difference in preference between genders.
Calculate an Interval Estimate Based on a Confidence Interval for a Difference of Proportions
Calculate an Interval Estimate Based on a Confidence Interval for a Difference of Proportions
A confidence interval for \( p_1 – p_2 \) can be expressed as an interval estimate with specified units (such as percentage points). This helps interpret the difference between two proportions in a more meaningful context.
Formula Reminder:
\( (\hat{p}_1 – \hat{p}_2) \pm z^* \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \)
- The resulting confidence interval represents the range of plausible differences in proportions.
- To convert the result into percentage points, multiply the proportions by 100.
Example:
From the previous example, the 95% confidence interval for \( p_1 – p_2 \) was \([-0.117, 0.201]\).
Convert this to percentage points and interpret.
▶️ Answer / Explanation
Step 1: Convert to percentage points
\([-0.117, 0.201] \times 100 = [-11.7,\; 20.1]\) percentage points.
Step 2: Interpret the interval estimate
- We are 95% confident that the proportion of males preferring online classes is between 11.7 percentage points lower and 20.1 percentage points higher than the proportion of females who prefer online classes.
- Since the interval includes 0, there is insufficient evidence to conclude a real difference between the two groups.
Final Interpretation (in context): The estimated difference between the two population proportions lies between −11.7 and +20.1 percentage points, showing that the true difference could be positive, negative, or zero.