Common Errors With Two-Proportion Intervals

Three Errors That Can Change the Conclusion

A two-proportion confidence interval estimates a difference between two population proportions. As covered in Constructing a Two-Proportion z-Interval by Hand, the interval uses the difference between the sample proportions and a standard error that accounts for uncertainty in both samples. Common errors can arise even when the arithmetic looks familiar: comparing two separate intervals as if that were a test, pooling proportions when calculating an interval, or losing track of which group is subtracted from which.

These errors matter because they can change what the evidence appears to say. A pair of overlapping individual intervals may still accompany a two-proportion interval that excludes zero. A pooled standard error may give a different interval from the correct one. And reversing the group order reverses the sign and the meaning of the difference.

Key idea: For a two-proportion confidence interval, estimate $p_1-p_2$ using $\hat{p}_1-\hat{p}_2$, and calculate the standard error with the two separate sample proportions. Keep the group order consistent from the parameter definition through the conclusion.

Error 1: Treating Overlapping Individual Intervals as a Test

An interval for $p_1$ estimates the population proportion in Group 1. An interval for $p_2$ estimates the population proportion in Group 2. Those are not the same parameter as $p_1-p_2$. Looking at whether the two individual intervals overlap is therefore not a formal confidence interval or test for the difference.

Overlap does not establish that the population proportions are equal or that their difference is zero. Separate intervals each reflect uncertainty in one group; a two-proportion interval directly estimates the difference and combines uncertainty from both groups. If the question concerns $p_1-p_2$, calculate and interpret an interval for $p_1-p_2$.

Important distinction: If a confidence interval for $p_1-p_2$ contains zero, zero is one plausible value for the difference at that confidence level; nonzero values inside the interval remain plausible too. If the entire interval is above or below zero, the interval supports a direction for the difference. Do not substitute overlap between individual intervals for this direct check.

Worked Example: Individual Intervals Overlap, but the Difference Interval Does Not Contain Zero

Setting: In a fictional survey, separate random samples of 1,000 customers are taken from two regions. In Region A, 550 sampled customers say they would recommend a local repair service; in Region B, 500 do. Compare the proportions, Region A minus Region B, using 95% confidence intervals.

State: Let $p_1$ be the true proportion of customers in Region A who would recommend the service, and $p_2$ the corresponding proportion in Region B. The target difference is $p_1-p_2$, Region A minus Region B.

Plan: The samples are separate random samples, so the Random condition and independence between groups are supported. Assume each region has at least 10,000 customers; then each sample is at most 10% of its region. For Large Counts, Region A has 550 successes and $1{,}000-550=450$ failures; Region B has 500 successes and $1{,}000-500=500$ failures. All four counts are at least 10. These conditions support the two-proportion interval and the individual proportion intervals.

Do: The sample proportions are $\hat{p}_1=550/1{,}000=0.55$ and $\hat{p}_2=500/1{,}000=0.50$. For the individual 95% intervals, use $z^*=1.96$. For Region A:

SE_{\hat{p}_1}=\sqrt{\frac{0.55(0.45)}{1{,}000}} =\sqrt{0.0002475}\approx0.015732

The margin of error is $1.96(0.015732)\approx0.030835$, so the 95% interval for $p_1$ is approximately $(0.5192,\ 0.5808)$. For Region B:

SE_{\hat{p}_2}=\sqrt{\frac{0.50(0.50)}{1{,}000}} =\sqrt{0.0002500}\approx0.015811

The margin of error is $1.96(0.015811)\approx0.030990$, giving an individual 95% interval of approximately $(0.4690,\ 0.5310)$. These individual intervals overlap from about $0.5192$ to $0.5310$.

Now calculate the interval directly for $p_1-p_2$. The point estimate is $0.55-0.50=0.05$. The unpooled standard error is:

\begin{aligned} SE_{\hat{p}_1-\hat{p}_2} &=\sqrt{\frac{0.55(0.45)}{1{,}000}+\frac{0.50(0.50)}{1{,}000}}\\ &=\sqrt{0.0002475+0.0002500}\\ &=\sqrt{0.0004975}\approx0.022305 \end{aligned}

The margin of error is $1.96(0.022305)\approx0.043717$. Thus, the 95% interval is $0.05\mathbin{\pm}0.043717=(0.0063,\ 0.0937)$, rounded to four decimal places.

Conclude: We are 95% confident that the true difference in recommendation proportions, Region A minus Region B, is between about $0.0063$ and $0.0937$. The individual intervals overlap, but the interval for the difference is entirely above zero. The overlap between individual intervals is not a substitute for calculating the interval for the parameter of interest.

Error 2: Pooling When Calculating a Confidence Interval

The two-proportion confidence interval uses an unpooled standard error: it uses $\hat{p}_1$ for the first group’s contribution and $\hat{p}_2$ for the second group’s contribution. This matches the interval’s purpose of estimating two potentially different population proportions.

A pooled proportion combines the two groups’ successes and sample sizes into one estimate. Pooling is used for a two-proportion $z$-test when the null hypothesis says the population proportions are equal. Under that null, both groups are modeled as having one common proportion. A confidence interval does not assume the two proportions are equal, so inserting a pooled proportion into its standard-error formula is not the correct interval calculation.

Formula: For a two-proportion confidence interval, use $$ SE_{\hat{p}_1-\hat{p}_2} = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}. $$ Do not replace the two sample proportions with a pooled proportion when constructing this interval.

Worked Example: Use the Unpooled Standard Error for the Interval

Setting: In a fictional poll, 48 of 80 respondents in Group 1 favor a proposed community garden, compared with 40 of 100 respondents in Group 2. Find a 95% confidence interval for the difference, Group 1 minus Group 2, and compare the correct standard error with a pooled one.

State: Let $p_1$ and $p_2$ be the true proportions favoring the proposal in Groups 1 and 2. The parameter is $p_1-p_2$, Group 1 minus Group 2.

Plan: Assume the data come from separate random samples, and assume each sample is no more than 10% of its population. The samples are independent. For Large Counts, Group 1 has 48 successes and $80-48=32$ failures; Group 2 has 40 successes and $100-40=60$ failures. Every count is at least 10, so the conditions for a two-proportion interval are met.

Do: The sample proportions are $\hat{p}_1=48/80=0.60$ and $\hat{p}_2=40/100=0.40$. The point estimate is $0.60-0.40=0.20$. Use the unpooled standard error:

\begin{aligned} SE_{\hat{p}_1-\hat{p}_2} &=\sqrt{\frac{0.60(0.40)}{80}+\frac{0.40(0.60)}{100}}\\ &=\sqrt{0.003000+0.002400}\\ &=\sqrt{0.005400}\approx0.073485 \end{aligned}

The margin of error is $1.96(0.073485)\approx0.144030$. The 95% interval is $0.20\mathbin{\pm}0.144030=(0.0560,\ 0.3440)$, rounded to four decimal places.

For comparison, the pooled proportion would be $(48+40)/(80+100)=88/180\approx0.4889$. Using it in a pooled standard error would give:

\sqrt{0.4889(1-0.4889)\left(\frac{1}{80}+\frac{1}{100}\right)} \approx\sqrt{0.0056222}\approx0.074981

That is not the standard error used for this confidence interval. It comes from treating the groups as having a common proportion, the assumption used in a pooled two-proportion test under its null hypothesis. The similar numerical result here does not make the pooled calculation appropriate.

Conclude: We are 95% confident that the true proportion favoring the proposal in Group 1 is between about $0.0560$ and $0.3440$ higher than the true proportion in Group 2. The interval uses the separate sample proportions in its standard error; the pooled calculation is not the confidence-interval method.

Error 3: Reversing or Losing the Group Order

The expression $p_1-p_2$ always means the proportion in Group 1 minus the proportion in Group 2. A positive difference indicates a higher proportion in Group 1; a negative difference indicates a lower proportion in Group 1. The interpretation must match the order used in the parameter, point estimate, interval, and conclusion.

If the order is reversed, the sign of the estimate and the interval endpoints must also be reversed. For an interval $(L,U)$ for $p_1-p_2$, the corresponding interval for $p_2-p_1$ is $(-U,-L)$. Merely changing the group names in the conclusion without changing the signs is an error.

Worked Example: Interpret a Negative Difference in the Stated Order

Setting: In a fictional survey, 42 of 120 sampled residents in North District report using a bike lane weekly. In South District, 70 of 140 sampled residents do. Find and interpret a 95% confidence interval for North District minus South District.

State: Let $p_1$ be the true proportion of North District residents who use a bike lane weekly, and $p_2$ the true proportion of South District residents who do. The parameter is $p_1-p_2$, North minus South.

Plan: Assume the districts were sampled separately using random samples, so the Random condition and independence between groups are supported. Assume North District has at least 1,200 residents and South District at least 1,400; then $120\leq0.10(1{,}200)$ and $140\leq0.10(1{,}400)$, so the 10% condition holds. For Large Counts, North has 42 successes and $120-42=78$ failures; South has 70 successes and $140-70=70$ failures. Each count is at least 10. The conditions support a two-proportion interval.

Do: The sample proportions are $\hat{p}_1=42/120=0.35$ and $\hat{p}_2=70/140=0.50$. The point estimate is $0.35-0.50=-0.15$. Calculate the unpooled standard error:

\begin{aligned} SE_{\hat{p}_1-\hat{p}_2} &=\sqrt{\frac{0.35(0.65)}{120}+\frac{0.50(0.50)}{140}}\\ &=\sqrt{0.0018958+0.0017857}\\ &=\sqrt{0.0036815}\approx0.060676 \end{aligned}

The 95% margin of error is $1.96(0.060676)\approx0.118925$. The interval is $-0.15\mathbin{\pm}0.118925=(-0.2689,\ -0.0311)$, rounded to four decimal places.

Conclude: We are 95% confident that the true difference in weekly bike-lane use proportions, North District minus South District, is between about $-0.2689$ and $-0.0311$. Since both endpoints are negative, the interval indicates that the true proportion in North District is about 3.11 to 26.89 percentage points lower than in South District. The point estimate is 15 percentage points lower; that estimate is not the interval’s full range of plausible population differences.

Common Mistakes and AP Exam Tips

Using overlap as the decision rule: Do not conclude that there is no difference just because two separate intervals overlap. Calculate an interval for $p_1-p_2$ when that is the parameter the question asks about.
Misreading an interval that contains zero: Zero is then one plausible value of the difference, but nonzero values in the interval remain plausible too. Do not say that the proportions are proven equal.
Pooling for an interval: Use the separate sample proportions in the two-proportion interval’s standard error. A pooled proportion belongs to the test calculation when the null hypothesis assumes equal population proportions.
Switching order halfway through: Write “Group 1 minus Group 2” beside $p_1-p_2$. Check that the sample estimate and contextual conclusion use the same order.
Calling the point estimate a range: The point estimate is one number, such as $-0.15$. The interval gives a range of plausible values for the population difference.
Reporting a negative interval without translating it: State which group has the lower proportion and, when useful, express the endpoints as percentage points lower. Keep the estimate separate from the interval’s range.

AP Exam Tip: A complete response identifies the parameter and group order, checks the design and Large Counts conditions, shows the unpooled standard error, and interprets both endpoints in context. For a direction claim, inspect the signs of both endpoints—not just the sign of the point estimate.

Key Takeaway

A two-proportion interval directly estimates a population difference; two individual intervals do not. Use the unpooled standard error for the interval, preserve the subtraction order, and translate the signs into a contextual comparison. If zero is in the interval, zero and any nonzero values within the interval are plausible at that confidence level.

Key takeaway: Match the method to the parameter: estimate $p_1-p_2$ with a two-proportion interval, use separate sample proportions in its standard error, and interpret the result in the stated group order.

Check Your Understanding

Answer each question using the distinction between individual intervals and an interval for a difference.

Two separate 95% confidence intervals for $p_1$ and $p_2$ overlap. What can you conclude—and what can you not conclude—from that overlap alone?
For a confidence interval estimating $p_1-p_2$, why does the standard error use $\hat{p}_1$ and $\hat{p}_2$ separately?
A student uses the pooled sample proportion in a two-proportion confidence interval. Explain why this is not the correct interval calculation.
An interval for $p_1-p_2$ is $(-0.12,\ 0.04)$. What does zero’s presence mean, and what nonzero values remain plausible?
If an interval for North minus South is entirely negative, which group has the higher estimated population proportion according to the interval?