CI for a Proportion - Biostatistics

A pharmaceutical company tests a new drug on 500 patients and observes that 175 show improvement. The 95% confidence interval for the proportion showing improvement is calculated as (0.31, 0.39). What is the most likely explanation if a colleague calculated the interval as (0.29, 0.41)?
Answer: The colleague used a 99% confidence level instead of 95% When analyzing confidence intervals, remember that the width of the interval is directly related to the confidence level - higher confidence requires a wider interval to capture the true parameter with greater certainty. Let's examine what happened here. With 175 improvements out of 500 patients, the sample proportion is $\hat{p} = 175/500 = 0.35$ . Notice that both intervals are centered around this same value (0.35), so the sample proportion calculation wasn't the issue. The key difference is that your colleague's interval (0.29, 0.41) is wider than the original 95% interval (0.31, 0.39). Since a wider interval corresponds to higher confidence, your colleague most likely used a 99% confidence level instead of 95%. This makes option B correct. The 99% level requires a larger critical value (approximately 2.576 vs 1.96 for 95%), which creates the wider interval needed to be more confident about capturing the true proportion. Option A is wrong because a 90% confidence level would produce a narrower interval than 95%, not wider. Option C is incorrect because both intervals are centered at 0.35, indicating the same sample proportion was used. Option D doesn't fit because using the wrong sample size would likely shift the center point or create a dramatically different width, not just a modest symmetric expansion. Study tip: Remember the confidence level trade-off: higher confidence = wider intervals. When comparing two intervals with the same center, the wider one always represents higher confidence.
Two researchers independently survey patients about medication compliance. Researcher A surveys 100 patients and finds 30% compliant. Researcher B surveys 400 patients and finds 30% compliant. Both calculate 95% confidence intervals. How do their intervals compare?
Answer: Researcher A's interval is twice as wide as Researcher B's interval When you encounter questions about confidence interval width, remember that the key factor is sample size. The width of a confidence interval is inversely related to the square root of the sample size. For a proportion, the 95% confidence interval width is approximately $4 \times \sqrt{\frac{p(1-p)}{n}}$ , where p is the sample proportion and n is the sample size. Since both researchers found the same proportion (30%), the only difference comes from their sample sizes. Researcher A has n = 100, while Researcher B has n = 400. The ratio of their interval widths equals $\sqrt{\frac{400}{100}} = \sqrt{4} = 2$ . Therefore, Researcher A's interval is twice as wide as Researcher B's interval, making B correct. Let's examine why the other options are wrong. Choice A incorrectly assumes that only the sample proportion matters for interval width, ignoring the crucial role of sample size. Choice C suggests the width ratio equals the sample size ratio (400/100 = 4), but this ignores the square root relationship. Choice D proposes a factor of $\sqrt{2}$ , which would only be correct if one sample were twice the size of the other, not four times larger. Study tip: Remember the "square root rule" for confidence intervals: when sample size increases by a factor of k², the interval width decreases by a factor of k. This relationship appears frequently on biostatistics exams, so practice identifying how changes in sample size affect precision.
An epidemiologist studies vaccination rates in a city of 50,000 residents. She randomly samples 500 residents and finds that 350 are vaccinated. When calculating the 95% confidence interval (0.66, 0.74), should she apply a finite population correction?
Answer: No, because the sample size is less than 10% of the population size When you encounter confidence interval questions involving finite populations, you need to determine whether a finite population correction (FPC) is necessary. The FPC adjusts standard errors when sampling from smaller populations where the sample represents a substantial portion of the total population. The key rule is the 10% threshold: apply FPC when your sample size exceeds 10% of the population size. Here, you have 500 residents sampled from 50,000 total residents. Calculate: $\frac{500}{50,000} = 0.01 = 1\%$ . Since 1% is well below the 10% threshold, no finite population correction is needed. The population is large enough relative to the sample that you can treat it as essentially infinite. Let's examine why the other options are incorrect. Option A incorrectly links FPC to the sample size being over 30 - this confuses the central limit theorem's sample size requirement with finite population correction criteria. Option B suggests that exceeding 1% of the population triggers FPC, but this threshold is too low; the standard is 10%, not 1%. Option D mentions the population exceeding 10,000 as the determining factor, but FPC decisions depend on the ratio of sample to population size, not the absolute population size. Study tip: Remember the 10% rule for finite population correction. When sample size ÷ population size > 0.10, apply FPC. This is one of the most testable concepts in sampling theory, so always calculate this ratio when you see finite population scenarios.
A hospital surveys 160 patients about satisfaction with nursing care. Among these, 128 patients report being satisfied. The hospital administration wants to use this data to construct a confidence interval, but they're concerned about the validity of the normal approximation. What should they check first?
Answer: Whether both $160 \times 0.8 = 128$ and $160 \times 0.2 = 32$ are at least 5 When constructing confidence intervals for proportions, you must verify that the normal approximation to the binomial distribution is valid. This requires checking specific conditions about your sample data. The key condition is that both $np$ and $n(1-p)$ must be at least 5, where $n$ is your sample size and $p$ is your sample proportion. Here, with 128 satisfied patients out of 160, your sample proportion is $p = 128/160 = 0.8$ . So you need to check: $np = 160 \times 0.8 = 128 \geq 5$ ✓ and $n(1-p) = 160 \times 0.2 = 32 \geq 5$ ✓. This is exactly what option B describes. Option A incorrectly applies the $n > 30$ rule, which applies to means and the central limit theorem, not proportion confidence intervals. The sample size requirement for proportions depends on the success-failure condition, not a fixed threshold. Option C is wrong because you don't need the population to be normally distributed. The normal approximation applies to the sampling distribution of the sample proportion, not the original population distribution. Option D misunderstands the requirements. While proportions closer to 0.5 do have better normal approximations, there's no requirement that $p$ be "sufficiently close" to 0.5. The success-failure condition ( $np \geq 5$ and $n(1-p) \geq 5$ ) handles this automatically. Study tip: For proportion confidence intervals, always check the success-failure condition first: both $np$ and $n(1-p)$ must be at least 5. This single check validates your normal approximation.
A researcher surveys 180 patients and finds that 72 experienced side effects from a new medication. When calculating a 95% confidence interval for the proportion of patients experiencing side effects, what is the most appropriate reason for using the normal approximation?
Answer: Both $np = 72$ and $n(1-p) = 108$ are greater than 5, satisfying the success-failure condition When you encounter questions about confidence intervals for proportions, you need to determine when the normal approximation to the binomial distribution is valid. This is crucial because using an inappropriate approximation can lead to inaccurate confidence intervals. The correct criterion is the success-failure condition: both $np$ and $n(1-p)$ must be at least 5. With $n = 180$ patients and $\hat{p} = 72/180 = 0.4$ , we get $np = 180 \times 0.4 = 72$ and $n(1-p) = 180 \times 0.6 = 108$ . Since both values are much greater than 5, the normal approximation is appropriate. Answer B correctly identifies this fundamental requirement. Answer A incorrectly applies the central limit theorem's $n > 30$ rule, which applies to sample means, not proportions. For proportions, sample size alone isn't sufficient—you need adequate numbers of both successes and failures. Answer C confuses concepts from confidence intervals for means. When estimating population means with unknown standard deviations, you use the t-distribution, not the normal distribution for proportions. Answer D suggests that $\hat{p} = 0.4$ being "close to 0.5" ensures normality, but this isn't the correct criterion. While proportions near 0.5 do approach normality faster, the formal requirement is still the success-failure condition, regardless of how close $\hat{p}$ is to 0.5. Study tip: For proportion confidence intervals, always check $np \geq 5$ and $n(1-p) \geq 5$ before using the normal approximation. Don't rely on sample size rules from other contexts.
A study reports that among 200 patients, 74 showed improvement, yielding a 90% confidence interval of (0.31, 0.43) for the improvement rate. The study director claims this interval is too wide and wants to reduce its width by half. Assuming the same improvement rate in future studies, approximately how many total patients would be needed?
Answer: 800 patients, since the margin of error is proportional to $1/\sqrt{n}$ , so reducing it by half requires quadrupling $n$ The margin of error is $z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ , which is proportional to $1/\sqrt{n}$ . To reduce the margin of error (and thus interval width) by half, we need $\frac{1}{\sqrt{n_{new}}} = \frac{1}{2} \cdot \frac{1}{\sqrt{n_{old}}}$ , which gives $n_{new} = 4n_{old} = 4 \times 200 = 800$ . Choice A incorrectly assumes a linear relationship. Choice C uses an incorrect multiplier. Choice D incorrectly suggests the critical value changes (it stays the same for the same confidence level).
A clinical trial enrolled 240 patients to test a new treatment. Of these, 156 patients showed improvement. When calculating a 95% confidence interval for the true proportion of patients who would show improvement, which of the following represents the correct margin of error?
Answer: $1.96 \times \sqrt{\frac{0.65 \times 0.35}{240}}$ The margin of error for a 95% CI for a proportion is $z_{\alpha/2} \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ . Here, $\hat{p} = 156/240 = 0.65$ , so the margin of error is $1.96 \times \sqrt{\frac{0.65 \times 0.35}{240}}$ . Choice B uses raw counts instead of proportions. Choice C incorrectly uses 156 as the denominator instead of the total sample size. Choice D has an extra factor of 240 in the denominator.
A pharmaceutical company tests a new drug on 180 patients and finds that 63 patients show significant improvement. When constructing a 99% confidence interval for the true proportion of patients who would show improvement, which condition for using the z-interval is most likely to be violated?
Answer: The random sampling condition, since clinical trial participants are typically volunteers who may not represent the general population For a valid z-interval, we need: (1) random sampling, (2) independence, (3) success-failure condition ( $n\hat{p} \geq 10$ and $n(1-\hat{p}) \geq 10$ ). Here, $n\hat{p} = 63 \geq 10$ and $n(1-\hat{p}) = 117 \geq 10$ , so the success-failure condition is satisfied. Clinical trial volunteers often differ systematically from the general population, violating the random sampling assumption. Choice A incorrectly suggests we need $n\hat{p} \geq 100$ . Choice B overstates patient interaction effects. Choice C misunderstands the success-failure condition.
A clinical trial with 400 participants shows that 120 patients responded positively to treatment. If the 90% confidence interval for the true response rate is calculated as (0.26, 0.34), what can be concluded about a 95% confidence interval for the same data?
Answer: The 95% confidence interval will be wider than (0.26, 0.34) with the same center point of 0.30 When you encounter confidence interval questions, remember that confidence level and interval width are directly related - higher confidence requires wider intervals to maintain the same probability of capturing the true parameter. The sample proportion here is $\hat{p} = 120/400 = 0.30$ , which serves as the center point for any confidence interval. The key insight is understanding how confidence level affects interval width. A 95% confidence interval requires capturing the true parameter 95% of the time versus 90% for the given interval. To achieve this higher confidence, you need a wider interval that casts a "broader net" around the estimate. Since both intervals use the same sample data, they share the same center point (0.30) and standard error. The only difference is the critical value (z-score): 1.645 for 90% confidence versus 1.96 for 95% confidence. The larger critical value creates a larger margin of error, making the 95% interval wider than (0.26, 0.34) while keeping the same center. Option A incorrectly states the 95% interval will be (0.25, 0.35) - while this is wider, you can't determine exact bounds without calculation. Option B wrongly suggests the 95% interval will be narrower, confusing confidence level with precision. Option D is incorrect because the standard error depends only on the sample data, which remains unchanged. Study tip: Remember the confidence level trade-off: higher confidence = wider intervals. When comparing intervals from the same data, the more confident interval is always wider, centered at the same point.
A medical researcher surveys 150 doctors and finds that 45 prefer electronic health records over paper records. When calculating a confidence interval for the true proportion, the researcher gets a margin of error of 0.074 using a 95% confidence level. What sample size would be needed to reduce this margin of error to 0.037?
Answer: 600 doctors When you encounter questions about sample size and margin of error, remember that these have an inverse square root relationship - a fundamental principle in confidence interval calculations. The margin of error formula for proportions is $E = z \sqrt{\frac{p(1-p)}{n}}$ . Since everything else remains constant (confidence level, proportion estimate), the relationship between sample sizes and margins of error follows: $\frac{n_2}{n_1} = \left(\frac{E_1}{E_2}\right)^2$ Starting with 150 doctors and a margin of error of 0.074, you want to reduce this to 0.037. Notice that 0.037 is exactly half of 0.074. When you halve the margin of error, you must quadruple the sample size: $n_2 = 150 \times \left(\frac{0.074}{0.037}\right)^2 = 150 \times 2^2 = 150 \times 4 = 600$ Option A (300 doctors) would only double the sample size, which reduces margin of error by a factor of $\sqrt{2} \approx 1.41$ , not the needed factor of 2. Option B (450 doctors) triples the sample size, reducing margin of error by $\sqrt{3} \approx 1.73$ . Option D (750 doctors) represents a 5-fold increase, which would reduce margin of error by $\sqrt{5} \approx 2.24$ - more than necessary. The correct answer is C (600 doctors). Study tip: Remember the inverse square root relationship: to cut margin of error in half, you need four times the sample size. To reduce it by one-third, you need nine times the sample size. This pattern appears frequently in biostatistics problems.
A medical researcher calculates a 95% confidence interval for the proportion of patients who respond to a new therapy. The interval is (0.45, 0.65). A colleague argues that this interval is too wide to be useful and suggests increasing the sample size. To cut the interval width in half while maintaining 95% confidence, by what factor must the sample size be increased?
Answer: 4 When you encounter confidence interval width problems, remember that the width depends on the standard error, which is inversely related to the square root of sample size. The margin of error for a proportion confidence interval is $z \times \sqrt{\frac{p(1-p)}{n}}$ , where n is the sample size. Since the interval width equals twice the margin of error, to halve the width, you must halve the margin of error. To cut the margin of error in half while keeping the confidence level (and thus z-value) constant, you need: $\frac{1}{2}\sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{p(1-p)}{n_{new}}}$ Squaring both sides: $\frac{1}{4} \times \frac{p(1-p)}{n} = \frac{p(1-p)}{n_{new}}$ Solving for the new sample size: $n_{new} = 4n$ Therefore, you must increase the sample size by a factor of 4, making B correct. Choice A (factor of 2) represents the common misconception that doubling sample size halves the interval width - this ignores the square root relationship. Choice C (factor of 8) might arise from incorrectly cubing the factor of 2. Choice D (factor of 16) could result from squaring the factor of 4, perhaps confusing this with variance calculations. Study tip: Remember the "square root rule" - because standard error involves $\sqrt{n}$ , you must quadruple the sample size to halve the margin of error. This relationship appears frequently in sample size calculations across different statistical contexts.
A researcher calculates a 95% confidence interval for the proportion of patients responding to treatment as (0.42, 0.68). If the same data were used to construct a 90% confidence interval instead, which statement about the new interval is most accurate?
Answer: The new interval will have the same center but will be narrower, though the exact endpoints require recalculation using the original sample statistics Both intervals will have the same center (sample proportion), but the 90% CI will be narrower because $z_{0.05} = 1.645 < 1.96 = z_{0.025}$ . The exact endpoints require recalculating with $\hat{p} \pm 1.645\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ , which needs the original sample data. From the given CI, $\hat{p} = 0.55$ , but we need $n$ for exact calculation. Choice A provides specific values without justification. Choice B attempts calculation without sufficient information. Choice D incorrectly suggests new data collection is needed.
Two studies examined the same treatment. Study A: 40 successes out of 100 patients, 95% CI = (0.304, 0.496). Study B: 80 successes out of 200 patients, 95% CI = (0.329, 0.471). Both studies have the same sample proportion (0.40), but Study B has a narrower confidence interval. What is the primary reason for this difference?
Answer: Study B has reduced sampling variability due to the larger sample size, resulting in a smaller standard error The margin of error for a proportion CI is $z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ . With the same $\hat{p}$ and confidence level, the margin of error decreases as $n$ increases because the standard error $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ decreases. Study B (n=200) has a smaller standard error than Study A (n=100). Choice A incorrectly suggests different methods were used. Choice C confuses precision with bias reduction. Choice D incorrectly focuses on absolute number of successes rather than sample size.
A medical device company tests their product on 320 patients and observes 96 successful outcomes. They calculate a 95% confidence interval for the success rate. Due to a data entry error, they initially used $n = 300$ instead of $n = 320$ in their calculations. How did this error affect their reported confidence interval?
Answer: The interval was too wide because the incorrect smaller sample size led to an overestimated standard error Using $n = 300$ instead of $n = 320$ affects both the sample proportion and standard error. The incorrect $\hat{p} = 96/300 = 0.32$ vs. correct $\hat{p} = 96/320 = 0.30$ . However, the primary effect on interval width comes from the standard error $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ . Using the smaller $n = 300$ gives a larger standard error and thus a wider interval. Choice B incorrectly focuses on bias rather than precision. Choice C incorrectly suggests width is unaffected. Choice D correctly identifies the proportion difference but incorrectly concludes this makes the interval wider.
A nurse practitioner finds that 48 out of 120 patients in a diabetes prevention program successfully lost weight. She calculates a 95% confidence interval for the success rate. If she had instead used a sample proportion of 0.41 (keeping the same sample size), how would the margin of error change?
Answer: The margin of error would increase because 0.41 is closer to 0.5 than the original proportion When you encounter confidence interval questions involving proportions, remember that the margin of error depends on how much variability exists in your sample proportion. The key insight is that proportions create maximum variability when they're closest to 0.5. Let's calculate the original proportion: 48/120 = 0.40. The margin of error for a proportion confidence interval is $ME = z_{\alpha/2}\sqrt{\frac{p(1-p)}{n}}$ . Notice that the term $p(1-p)$ determines how the margin of error changes with different proportions. To see which proportion creates more variability, compare $p(1-p)$ values: - Original: $0.40 \times 0.60 = 0.24$ - Alternative: $0.41 \times 0.59 = 0.2419$ Since 0.2419 > 0.24, the proportion of 0.41 produces a larger margin of error. This happens because 0.41 is closer to 0.5 than 0.40, and proportions near 0.5 create maximum uncertainty. Looking at the wrong answers: B incorrectly states the margin of error would decrease and wrongly claims 0.41 is farther from 0.5. C correctly identifies that 0.41 is closer to 0.5 but incorrectly predicts the margin of error would decrease. D gets both facts wrong—claiming 0.41 is farther from 0.5 and that this increases the margin of error. Study tip: Remember that proportion confidence intervals are widest (largest margin of error) when p = 0.5 and get narrower as proportions approach 0 or 1. Always check which proportion is closer to 0.5 to determine which has the larger margin of error.
A quality control manager at a pharmaceutical company needs to estimate the proportion of defective pills in a batch. A sample of 400 pills reveals 24 defective ones. If the manager constructs a 95% confidence interval, what assumption is most critical for the validity of this interval?
Answer: The sample of 400 pills was selected using simple random sampling When constructing confidence intervals for proportions, you're making statistical inferences about a population based on sample data. The validity of these inferences depends critically on how representative your sample is of the entire population. The most essential assumption for a valid confidence interval is that your sample was selected using simple random sampling (C). This means every pill in the batch had an equal chance of being selected, ensuring the sample is representative and unbiased. Without random sampling, your confidence interval becomes meaningless because systematic bias could skew your results in unknown ways. Let's examine why the other options are less critical. Option (A) about batch size being 10 times larger than sample size relates to the finite population correction factor, but this only affects the precision of your estimate slightly—it doesn't invalidate the interval entirely. Option (B) assumes defective pills are randomly distributed, but this is actually what we're trying to determine through our sampling process, not an assumption we need to make beforehand. Option (D) suggests the proportion must be less than 10%, but confidence intervals work regardless of the true proportion value, though they're most precise when the sample size satisfies $np \geq 5$ and $n(1-p) \geq 5$ . Study tip: When evaluating confidence interval validity, always prioritize sampling method over population characteristics. Random sampling is the foundation that makes statistical inference possible—without it, even perfect calculations on biased data will give you unreliable conclusions.
In a study of 400 patients, 92 developed complications. The 95% confidence interval for the proportion of patients who develop complications is calculated to be (0.185, 0.275). A colleague claims that because 0.25 falls within this interval, we can be 95% confident that exactly 25% of all patients will develop complications. Which statement best describes this interpretation?
Answer: The interpretation is incorrect; the confidence interval suggests that 25% is a plausible value, but we cannot assign 95% confidence to any specific value within the interval A confidence interval provides a range of plausible values for the population parameter, but the 95% confidence level applies to the interval construction method, not to individual values within the interval. Any specific value (like exactly 25%) within the interval is plausible but cannot be assigned the 95% confidence level. Choice A misunderstands confidence intervals. Choice C incorrectly suggests we can only be confident about the sample proportion. Choice D incorrectly validates the interpretation based on proximity to the sample proportion.
In a clinical study, 45 out of 150 patients experienced adverse events. A 95% confidence interval for the proportion is calculated. Later, it's discovered that 5 of the patients who were recorded as having adverse events actually did not. After correcting this error, how will the corrected confidence interval compare to the original?
Answer: The corrected interval will be shifted left and will be wider, since the corrected proportion is further from 0.5 where variance is minimized Originally: $\hat{p}_1 = 45/150 = 0.30$ . Corrected: $\hat{p}_2 = 40/150 = 0.267$ . The interval center shifts left. For width, the standard error is $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ . Since $\hat{p}(1-\hat{p})$ is maximized at $\hat{p} = 0.5$ , moving from 0.30 to 0.267 (further from 0.5) increases the variance $\hat{p}(1-\hat{p})$ , making the interval wider. Choice A ignores the change in variance. Choice B incorrectly states that 0.267 is closer to 0.5. Choice D correctly notes the direction but incorrectly states variance is maximized away from 0.5.
A clinical study reports that among 280 patients, 84 experienced adverse reactions. The researchers claim the 95% confidence interval for the adverse reaction rate is (0.24, 0.36). Which statement best evaluates this claim?
Answer: The interval is incorrect because it's too wide for the given sample size and proportion When you encounter confidence interval problems, you need to verify whether the reported interval matches what the data actually supports by calculating the correct interval yourself. Let's check this claim step by step. The sample proportion is $\hat{p} = 84/280 = 0.3$ . For a 95% confidence interval, we use $z = 1.96$ . The standard error is $SE = \sqrt{\frac{0.3(1-0.3)}{280}} = \sqrt{\frac{0.21}{280}} = 0.0274$ . The margin of error equals $1.96 \times 0.0274 = 0.0537$ . Therefore, the correct 95% confidence interval is $0.3 ± 0.0537$ , which gives us (0.246, 0.354) or approximately (0.25, 0.35). The researchers reported (0.24, 0.36), which has a margin of error of 0.06. This is noticeably wider than the correct margin of error of 0.054. Answer A is wrong because while 0.3 is indeed centered in the reported interval, this doesn't validate the interval's width. Answer C incorrectly suggests the interval is too narrow when it's actually too wide. Answer D is wrong because the margin of error should be about 0.054, not 0.06 as the reported interval suggests. Answer B correctly identifies that the interval is too wide for the given sample size and proportion. Study tip: Always calculate the margin of error yourself rather than trusting reported intervals. The formula $ME = z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ is essential for verifying confidence interval claims in biostatistics problems.
In a study of 250 medical students, 85 plan to specialize in primary care. A researcher calculates the margin of error for a 95% confidence interval as 0.058. If the same study were repeated with 1000 medical students and the same sample proportion, what would happen to the margin of error?
Answer: It would decrease to approximately 0.029 because the margin of error is inversely proportional to the square root of sample size When you encounter questions about margin of error and sample size, remember that this relationship follows a specific mathematical pattern based on the standard error formula. The margin of error for a proportion is calculated as $ME = z \times \sqrt{\frac{p(1-p)}{n}}$ , where z is the critical value, p is the sample proportion, and n is the sample size. The key insight is that sample size appears in the denominator under a square root, making the margin of error inversely proportional to $\sqrt{n}$ . Starting with n = 250 and ME = 0.058, when the sample size increases to 1000 (a 4-fold increase), the margin of error becomes $0.058 \times \sqrt{\frac{250}{1000}} = 0.058 \times \sqrt{0.25} = 0.058 \times 0.5 = 0.029$ . Option A incorrectly states that margin of error is inversely proportional to sample size itself, rather than its square root. This would give an incorrect calculation of 0.058 ÷ 4 = 0.0145. Option C makes a similar error by claiming that quadrupling sample size halves the margin of error, which would be true only if the relationship were inverse rather than inverse square root. Option D incorrectly assumes that margin of error depends only on the sample proportion, ignoring the crucial role of sample size in determining precision. Remember this pattern: when sample size increases by a factor of k, the margin of error decreases by a factor of $\sqrt{k}$ . Quadruple the sample size, and you halve the margin of error.

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