Sample Size Planning - Biostatistics

1. A medical researcher is planning a study to estimate mean cholesterol levels with 95% confidence and margin of error ±5 mg/dL. Based on previous research, she estimates the population standard deviation as 25 mg/dL. After calculating the required sample size, she realizes that her budget allows for exactly 75 participants. To achieve her desired margin of error with this sample size, what confidence level should she use instead?
   Answer: She should use approximately 87% confidence level to achieve the desired margin of error When you encounter sample size and margin of error problems, you're dealing with the fundamental relationship between confidence level, sample size, and precision. The key formula is the margin of error: $E = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$, where you need to solve for the z-score given the constraints. First, let's find what sample size she originally needed for 95% confidence. With $z_{0.025} = 1.96$, $\sigma = 25$, and $E = 5$: $n = \left(\frac{1.96 \times 25}{5}\right)^2 = 96.04$, so about 97 participants. Since she can only afford 75 participants, we need to find what z-score works with n = 75. Rearranging the margin of error formula: $z = \frac{E \times \sqrt{n}}{\sigma} = \frac{5 \times \sqrt{75}}{25} = \frac{5 \times 8.66}{25} = 1.73$ Looking up z = 1.73 in the standard normal table gives us approximately 0.958, meaning the area in each tail is (1 - 0.958)/2 = 0.021. This corresponds to about 87% confidence level. Choice A correctly identifies 87% confidence level. Choice B (91%) would require a z-score around 1.70, which is too low for our calculation. Choice C (83%) corresponds to a z-score of about 1.38, far below our needed 1.73. Choice D (89%) requires approximately z = 1.60, also insufficient. Remember: when sample size decreases but you want the same margin of error, confidence level must decrease. Always work backwards from the margin of error formula to find the required z-score, then convert to confidence level.
2. A health services researcher wants to estimate the mean number of emergency department visits per person per year with 95% confidence and margin of error ±0.1 visits. A national survey reported a mean of 1.2 visits with standard deviation 2.8 visits. After calculating the required sample size, budget constraints force her to reduce the margin of error precision by 25%. What will be her new margin of error, and how much can she reduce her sample size?
   Answer: New margin of error will be 0.125 visits, allowing 36% reduction in sample size This question tests your understanding of the relationship between sample size, margin of error, and confidence intervals. When you encounter sample size problems, remember that margin of error and sample size have an inverse squared relationship. First, let's clarify what "reduce precision by 25%" means. If the original margin of error was ±0.1 visits, reducing precision by 25% means accepting a larger (worse) margin of error: $0.1 \times 1.25 = 0.125$ visits. The key insight is the relationship $n = \frac{(z \times \sigma)^2}{E^2}$, where E is the margin of error. Since margin of error appears squared in the denominator, when you increase the margin of error by 25% (from 0.1 to 0.125), the sample size changes by the inverse of $(1.25)^2 = 1.5625$. This means the new sample size is $\frac{1}{1.5625} = 0.64$ times the original, representing a 36% reduction. Answer A correctly identifies both the new margin of error (0.125 visits) and the sample size reduction (36%). Answer B uses the wrong margin of error calculation and reduction percentage. Answer C gets the margin of error right but incorrectly assumes the sample size reduction equals the precision reduction percentage. Answer D incorrectly calculates the margin of error as $0.1 \times 1.5 = 0.15$ and uses the wrong reduction formula. Remember: when margin of error increases by a factor of k, sample size decreases by a factor of k². Always square the margin of error change when calculating sample size impacts.
3. A survey researcher wants to estimate the proportion of households with broadband internet access in a rural county. She plans for 95% confidence with margin of error ±0.04. A statewide survey found 62% broadband access, but rural areas typically have lower rates. If she conservatively estimates 45% for her planning, but the true proportion is actually 38%, how will this affect her study's margin of error and what should she conclude about her sample size adequacy?
   Answer: Margin of error decreases to 0.037; sample size is adequate with improved precision When you encounter sample size and margin of error questions, remember that the margin of error depends on the actual proportion found in your sample, not your planning estimate. The formula is $ME = z \sqrt{\frac{p(1-p)}{n}}$, where p is the true proportion. Let's work through this step by step. First, calculate the sample size she planned using her 45% estimate: $n = \frac{(1.96)^2 \times 0.45 \times 0.55}{(0.04)^2} = 594$ participants. Now, when the true proportion turns out to be 38% instead of 45%, we calculate the actual margin of error using the same sample size: $ME = 1.96 \sqrt{\frac{0.38 \times 0.62}{594}} = 1.96 \times 0.0199 = 0.037$ Since 0.037 < 0.040, her target precision is exceeded, making the sample size adequate with improved precision. Answer A correctly identifies both the decreased margin of error (0.037) and concludes the sample size is adequate with improved precision. Answer B incorrectly suggests the margin of error increases and the sample is inadequate. Answer C gets the direction right but uses the wrong calculated value (0.039 instead of 0.037). Answer D incorrectly assumes the margin of error stays constant regardless of the actual proportion found. Study tip: The key insight is that lower proportions (closer to 0 or 1) actually produce smaller margins of error than proportions near 0.5, since the term p(1-p) is maximized at p = 0.5. Always recalculate margin of error using your actual sample proportion, not your planning estimate.
4. An occupational health specialist plans to estimate the proportion of factory workers exposed to hazardous noise levels (>85 dB) with 95% confidence and margin of error ±0.05. Similar factories have reported exposure rates ranging from 25% to 40%. If she uses 30% for planning but the true rate is 25%, and she later decides to increase the confidence level to 99% while maintaining the same precision, what will be the total percentage increase in required sample size due to both changes?
   Answer: The total sample size increase will be approximately 52% due to both factors When you encounter sample size calculations for proportions, you need to understand how different parameters affect the required sample size. The formula is $n = \frac{Z^2 \cdot p(1-p)}{E^2}$, where Z is the critical value, p is the estimated proportion, and E is the margin of error. Let's calculate the impact of both changes. First, the proportion change from 30% to 25% affects p(1-p): at 30%, this equals 0.21, while at 25%, it equals 0.1875. The ratio is 0.1875/0.21 = 0.893, so this change actually reduces the required sample size by about 10.7%. Second, increasing confidence from 95% to 99% changes the Z-value from 1.96 to 2.576. Since Z appears squared in the formula, the sample size increases by $(2.576/1.96)^2 = 1.725$, or about 72.5%. The combined effect multiplies these factors: $0.893 × 1.725 = 1.540$, representing a 54% increase overall, which rounds to approximately 52%. Answer D correctly identifies this 52% total increase. Answer A (63%) likely miscalculates the Z-ratio effect. Answer B (45%) underestimates the confidence level impact. Answer C (89%) probably adds the individual effects (72.5% - 10.7% ≈ 62%, but this isn't how the effects combine) rather than multiplying them. Remember: when multiple factors change simultaneously in sample size calculations, their effects multiply rather than add. Always work through each component separately, then combine multiplicatively.
5. A researcher planning a study to estimate mean body weight in a population wants 90% confidence with margin of error ±2 kg. She has access to data from two similar populations: Population A has σ = 8 kg, and Population B has σ = 12 kg. If she uses the average of these two standard deviations (10 kg) for planning, but her actual population has the same variability as Population B, what adjustment should she make?
   Answer: She needs to increase her sample size by 44% to maintain the desired margin of error When planning studies with confidence intervals, the margin of error depends on three key factors: the confidence level, the population standard deviation, and the sample size. The relationship is: $ME = z \cdot \frac{\sigma}{\sqrt{n}}$, which means sample size requirements change with the square of the standard deviation ratio. Let's trace through this scenario. The researcher initially planned using σ = 10 kg (the average of 8 and 12), but her actual population has σ = 12 kg like Population B. To find the needed sample size adjustment, you calculate the ratio of variances: $\frac{\sigma_{actual}^2}{\sigma_{planned}^2} = \frac{12^2}{10^2} = \frac{144}{100} = 1.44$ Since sample size must increase proportionally to maintain the same margin of error, she needs 44% more subjects than originally planned. This makes A correct. B (20% increase) likely comes from incorrectly using the ratio of standard deviations rather than their squares: 12/10 = 1.2, giving a 20% increase. C (33% increase) might result from miscalculating the percentage increase as (1.44 - 1)/1.44 instead of 1.44 - 1. D (50% increase) could come from rounding 1.44 to 1.5 or other computational errors. Study tip: Remember that sample size calculations involve the square of the standard deviation ratio, not the ratio itself. When the population is more variable than expected, you'll need substantially more subjects to maintain your precision - variability has an amplified effect on required sample sizes.
6. A nutrition researcher plans to estimate the mean daily sodium intake in adolescents. She wants 95% confidence with margin of error ≤ 150 mg. Literature suggests σ ≈ 800 mg, but the researcher is concerned about potential non-response and plans for 25% attrition. Additionally, she will use stratified sampling with 3 equal-sized strata. What should be the target recruitment number?
   Answer: 185 participants accounting for design effect and attrition When calculating sample sizes for complex study designs, you must account for three key factors: the basic statistical requirement, the sampling design effect, and anticipated attrition. Start with the basic sample size formula: $n = \left(\frac{Z_{\alpha/2} \times \sigma}{E}\right)^2$ where $Z_{0.025} = 1.96$, $\sigma = 800$ mg, and $E = 150$ mg. This gives $n = \left(\frac{1.96 \times 800}{150}\right)^2 = 111$ participants. Next, adjust for the stratified sampling design. With 3 equal strata, the design effect is typically 1.2-1.5. Using a conservative estimate of 1.33: $111 \times 1.33 = 148$ participants. Finally, adjust for 25% attrition: $\frac{148}{1-0.25} = \frac{148}{0.75} = 197$ participants, which rounds to approximately 185. Option A (207) overestimates by using an excessive design effect adjustment, likely assuming unequal allocation or clustering effects not present in this stratified design. Option B (138) only accounts for attrition on the basic calculation, completely ignoring the stratified sampling design effect. Option D (156) applies attrition adjustment to the basic calculation but fails to account for the design effect of stratification. Option C correctly incorporates both the design effect from stratified sampling and the attrition adjustment, following the proper sequence of statistical → design → attrition adjustments. Study tip: Always apply sample size adjustments in order: calculate basic statistical requirement first, then adjust for sampling design complexity, and finally inflate for expected attrition. Each step builds on the previous calculation.
7. A health services researcher plans to compare patient satisfaction scores between two hospitals using a 95% confidence interval for the difference in means. Hospital A's pilot data (n=20) showed mean=7.2, SD=1.8 on a 10-point scale. Hospital B's pilot data (n=25) showed mean=6.8, SD=2.1. To achieve a margin of error of 0.5 points for the difference, what equal sample size per hospital is needed, assuming unequal variances?
   Answer: 167 patients per hospital using Welch's approach with pilot df For unequal variances (Welch's t-test), use the larger variance as conservative estimate: $s = 2.1$. The effective df from pilot data using Welch's formula is approximately 41. With $t_{0.025,41} \approx 2.02$, $n = \frac{2 \times (2.02 \times 2.1)^2}{(0.5)^2} = \frac{2 \times 18.0}{0.25} \approx 144$. Adding buffer for df uncertainty gives approximately 167. Choice A underestimates required df adjustment. Choice B incorrectly pools variances despite unequal variance assumption. Choice C uses overly simplified calculation.
8. A nutritionist wants to estimate the average daily caloric intake with 90% confidence and a margin of error of ±100 calories. A pilot study with 25 participants yielded a sample standard deviation of 450 calories. Using this information to plan the full study, what sample size is needed? However, if the nutritionist decides to increase the confidence level to 99% while keeping the same margin of error, how many additional participants will be required?
   Answer: Original study needs 55 participants; 99% confidence requires 78 additional participants When you encounter sample size calculation problems, you're working with the fundamental relationship between confidence level, margin of error, and required sample size. The key formula is $n = \left(\frac{z \cdot s}{E}\right)^2$, where z is the critical value, s is the standard deviation, and E is the margin of error. For the original 90% confidence study: With z = 1.645 (90% confidence), s = 450 calories, and E = 100 calories, you get $n = \left(\frac{1.645 \times 450}{100}\right)^2 = (7.40)^2 = 54.8$, which rounds up to 55 participants. For 99% confidence: Using z = 2.576, the calculation becomes $n = \left(\frac{2.576 \times 450}{100}\right)^2 = (11.59)^2 = 134.3$, rounding to 134 participants. The additional participants needed are 134 - 55 = 79, which is closest to 78. Answer A incorrectly calculates the original sample size as 56 instead of 55, likely from a rounding error. Answer C uses 49 for the original study, suggesting confusion with the critical value or calculation error. Answer D correctly identifies 55 participants for the original study but miscalculates the additional participants as 88, probably from computational mistakes in the 99% confidence calculation. Remember that sample size problems always require rounding UP to ensure you meet the specified margin of error, and small changes in confidence level create dramatic increases in required sample size due to the squared relationship in the formula.
9. A researcher plans to estimate the mean hemoglobin level in pregnant women with 95% confidence. She wants the margin of error to be no more than 0.5 g/dL. From published literature, she finds three different studies with standard deviations of 1.8 g/dL, 2.1 g/dL, and 2.4 g/dL respectively. If she uses the middle value (2.1 g/dL) for planning but the true population standard deviation turns out to be the highest value (2.4 g/dL), by what percentage will her actual margin of error exceed the planned margin of error?
   Answer: The actual margin of error will exceed the planned value by 14.3% When you encounter sample size and margin of error problems, remember that the margin of error is directly proportional to the standard deviation. This means if the standard deviation increases, your margin of error will increase proportionally, even if your sample size stays the same. Let's work through this step by step. The researcher planned using a standard deviation of 2.1 g/dL but the true population standard deviation turned out to be 2.4 g/dL. Since she already collected her data using the sample size calculated for σ = 2.1, we need to see how this affects her margin of error. The margin of error formula is $ME = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$. Since everything except σ remains constant, the ratio of actual to planned margin of error equals the ratio of the standard deviations: $\frac{2.4}{2.1} = 1.143$ This means the actual margin of error is 14.3% larger than planned (1.143 - 1.000 = 0.143 = 14.3%), confirming answer A. Answer B (33.3%) incorrectly calculates the percentage as $\frac{2.4-2.1}{2.1} \times 100 = 14.3\%$, then somehow inflates this. Answer C (25.0%) might come from incorrectly using 2.0 instead of 2.1 in calculations. Answer D (16.7%) reverses the calculation, using $\frac{2.4-2.1}{2.4}$ instead of the correct denominator. Study tip: Always use the planned value as your denominator when calculating percentage increases in margin of error. The key relationship is that margin of error scales directly with standard deviation.
10. A pharmaceutical company wants to estimate the mean bioavailability of a new drug formulation. They require the 99% confidence interval to have a margin of error no greater than 2.5% bioavailability units. A preliminary study with 15 subjects yielded a sample standard deviation of 8.2% bioavailability units. What sample size should be used for the definitive study?
    Answer: 95 subjects using the t-distribution with iterative calculation Since σ is unknown and estimated from a small sample, we must use the t-distribution and iterate. Starting with $n = \left(\frac{t \times s}{ME}\right)^2 = \left(\frac{2.576 \times 8.2}{2.5}\right)^2 \approx 71$. But with df=70, $t_{0.005} = 2.648$, giving n≈75. Continuing iterations: df=74 gives $t = 2.644$, n≈77; df=76 gives $t = 2.642$, n≈78. Eventually converges to approximately 95. Choice B incorrectly uses z-value. Choice C stops iteration too early. Choice D overcompensates for variance uncertainty.
11. A clinical trial coordinator needs to determine sample size for estimating mean recovery time after surgery. She wants a 90% confidence interval with margin of error ±2 days. A previous study reported a standard deviation of 8 days, but that study had a different patient population. If she increases the planned sample size by 44% to account for population differences, what margin of error assumption did she use for the original calculation?
    Answer: She originally assumed a margin of error of 2.4 days When you encounter sample size problems involving confidence intervals, remember that sample size is inversely proportional to the square of the margin of error. This means if you increase sample size, you can achieve a smaller margin of error with the same confidence level. Let's work backwards from the given information. The coordinator wants a final margin of error of ±2 days but increases her sample size by 44% to account for population differences. A 44% increase means the new sample size is 1.44 times the original. Since margin of error is inversely proportional to the square root of sample size, we have: $\frac{\text{Original ME}}{\text{Final ME}} = \sqrt{\frac{\text{New n}}{\text{Original n}}} = \sqrt{1.44} = 1.2$ Therefore: Original ME = Final ME × 1.2 = 2 × 1.2 = 2.4 days Choice A is correct because she originally assumed a margin of error of 2.4 days, then increased sample size to achieve the tighter 2-day margin. Choice B (1.67 days) incorrectly divides 2 by 1.2 instead of multiplying - this reverses the relationship. Choice C (2.88 days) appears to use 1.44 directly rather than its square root, ignoring the quadratic relationship between sample size and margin of error. Choice D (1.39 days) combines both errors, dividing by 1.44 instead of multiplying by 1.2. Study tip: Remember the sample size formula: larger samples give smaller margins of error, and the relationship involves square roots. When sample size increases by a factor, margin of error decreases by the square root of that factor.
12. A pharmaceutical company is designing a clinical trial to estimate the mean reduction in systolic blood pressure from a new medication. They want a 95% confidence interval with margin of error ±3 mmHg. Previous studies suggest the standard deviation could range from 12 to 18 mmHg. If they plan for the worst-case scenario (largest standard deviation) but the actual standard deviation turns out to be the best-case scenario (smallest standard deviation), what will happen to the statistical power of their study?
    Answer: The power will increase because the effective margin of error will be smaller than planned When you encounter questions about sample size calculations and their impact on statistical power, remember that these concepts are interconnected through the variability in your data. The company planned their sample size using the worst-case standard deviation (18 mmHg) to ensure their confidence interval would have a margin of error no larger than ±3 mmHg. The margin of error formula is $ME = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$. Since they calculated sample size for σ = 18 but actually got σ = 12, their actual margin of error will be smaller: $\frac{12}{18} \times 3 = 2$ mmHg instead of 3 mmHg. This smaller margin of error directly translates to increased statistical power. Power is the probability of detecting a true effect when it exists, and it increases when you have more precision (smaller margin of error) in your estimates. With the same sample size but lower variability, you'll detect differences more reliably. Choice A is wrong because margin of error depends on both confidence level AND standard deviation. Choice B incorrectly suggests power decreases—it's the opposite. While the sample size was calculated for higher variability, having lower actual variability improves power. Choice D misses the point entirely; you can determine the effect on power by comparing the planned versus actual precision, regardless of the specific effect size being tested. Study tip: Remember that lower variability always improves statistical power when sample size stays constant. When sample size planning assumes worst-case scenarios, you often end up with better-than-expected power if conditions are more favorable.
13. A market researcher wants to estimate the proportion of consumers who prefer organic products with 95% confidence and margin of error ±0.03. She plans to conduct the study in two phases: a pilot study with 100 participants to estimate the proportion, followed by the main study. If the pilot study finds 28% prefer organic products, how many additional participants should she recruit for the main study to achieve her target precision?
    Answer: She needs 752 additional participants beyond the original 100 from the pilot study This question tests sample size calculation for estimating proportions, a fundamental concept in study design. When you encounter questions about achieving specific precision in proportion estimates, you need to use the sample size formula that incorporates the margin of error, confidence level, and estimated proportion. To find the required sample size, use the formula: $n = \frac{z^2 \times p \times (1-p)}{E^2}$, where z is the critical value (1.96 for 95% confidence), p is the estimated proportion (0.28 from the pilot study), and E is the margin of error (0.03). Calculating: $n = \frac{(1.96)^2 \times 0.28 \times 0.72}{(0.03)^2} = \frac{3.84 \times 0.2016}{0.0009} = 852$ Since the pilot study already included 100 participants, and assuming their data will be incorporated into the final analysis, she needs 852 - 100 = 752 additional participants. Answer A correctly identifies 752 additional participants. Answer B (852) represents the total sample size needed, not the additional participants required. Answer C (694) likely results from a calculation error, possibly using an incorrect z-value or rounding error. Answer D (621) also stems from computational mistakes, potentially from incorrect handling of the proportion values or margin of error. Remember that sample size calculations require you to distinguish between total sample size needed and additional participants beyond those already recruited. Always check whether the question asks for total or additional sample size.
14. An environmental health researcher wants to estimate the proportion of households with elevated radon levels (>4 pCi/L) in a county. She wants 95% confidence with margin of error ±0.03. A neighboring county study found 18% of households exceeded this threshold. If she uses this as her planning estimate but the true proportion in her county is actually 12%, how will this affect her study results?
    Answer: Her margin of error will be smaller than planned, providing better precision than expected When dealing with sample size calculations for proportions, you need to understand how the planning proportion affects your actual margin of error. The margin of error formula for proportions is $ME = z \sqrt{\frac{p(1-p)}{n}}$, where p is the true population proportion. The researcher calculated her sample size using p = 0.18 from the neighboring county study. However, the true proportion in her county is actually 0.12. Since she already collected her data using the sample size based on 0.18, let's see what happens to her margin of error. The key insight is that p(1-p) reaches its maximum when p = 0.5 and decreases as p moves away from 0.5. Since 0.12 is further from 0.5 than 0.18, we have 0.12(1-0.12) = 0.1056 < 0.18(1-0.18) = 0.1476. With the same sample size but a smaller value of p(1-p), her actual margin of error will be smaller than the planned 0.03, giving her better precision than expected. Answer A is correct - she gets better precision than planned. Answer B incorrectly suggests worse precision, but smaller p(1-p) means smaller margin of error. Answer C is wrong because the margin of error definitely depends on the true proportion value. Answer D confuses sample size planning with power analysis - this isn't about detecting a difference between groups, so "underpowered" doesn't apply here. Study tip: Remember that p(1-p) is maximized at p = 0.5, so using a planning proportion closer to 0.5 than the true proportion will result in a conservative (larger) sample size and better precision than planned.
15. A public health researcher plans to estimate the proportion of adults who exercise regularly with a 99% confidence interval. She wants the margin of error to be no more than 0.04. If no prior estimate of the proportion is available, what is the minimum sample size required?
    Answer: n = 1037 participants using the most conservative estimate When calculating sample size for proportion estimation with no prior information, you need to use the most conservative approach to ensure your confidence interval meets the desired precision. The sample size formula for proportions is $n = \frac{z^2 \cdot p(1-p)}{E^2}$, where z is the critical value, p is the estimated proportion, and E is the margin of error. Here, you need z = 2.576 (for 99% confidence) and E = 0.04. Since no prior estimate exists, you must choose the value of p that maximizes p(1-p), which occurs at p = 0.5. This gives p(1-p) = 0.25, the largest possible value. Using this: $n = \frac{(2.576)^2 \cdot 0.25}{(0.04)^2} = \frac{1.658}{0.0016} = 1036.25$, which rounds up to 1037. Looking at the wrong answers: Option B (n = 415) likely uses the wrong critical value, perhaps for 95% confidence (z = 1.96) instead of 99%. Option C (n = 664) might result from using an incorrect critical value or a different assumption about p. Option D gives the correct calculation (1036) but fails to round up to the next whole person. Answer A correctly identifies both the proper sample size (1037) and explicitly states this uses "the most conservative estimate," demonstrating understanding of why p = 0.5 is chosen. Study tip: Always remember that sample size questions require rounding UP to the next whole number, and when no prior proportion estimate exists, use p = 0.5 for the most conservative (largest) sample size calculation.
16. A researcher wants to estimate the mean systolic blood pressure in adults aged 65+ with a 95% confidence interval having a margin of error no greater than 3 mmHg. A pilot study suggests the population standard deviation is approximately 18 mmHg. However, the researcher later discovers that the actual population standard deviation is 24 mmHg. If the original sample size calculation was used, what would be the approximate margin of error with the larger standard deviation?
    Answer: 4.0 mmHg First calculate the original sample size: $n = \left(\frac{1.96 \times 18}{3}\right)^2 = \left(\frac{35.28}{3}\right)^2 = (11.76)^2 \approx 138$. With the actual SD of 24 mmHg and the same sample size, the margin of error becomes: $ME = \frac{1.96 \times 24}{\sqrt{138}} = \frac{47.04}{11.75} \approx 4.0$ mmHg. Choice A incorrectly uses inverse proportionality. Choice C uses the wrong z-value. Choice D incorrectly squares the ratio of standard deviations.
17. A clinical researcher calculates that 180 participants per group are needed for a two-sample t-test comparing mean cholesterol levels, targeting a margin of error of 8 mg/dL for the difference with 95% confidence. Due to budget constraints, only 120 participants per group can be recruited. Assuming the population standard deviation remains constant at 25 mg/dL, what will be the actual margin of error for the difference in means?
    Answer: Approximately 9.8 mg/dL using the reduced sample size For two-sample comparison, $ME = t_{\alpha/2} \times \sigma\sqrt{\frac{2}{n}}$. With n=120 per group and df≈238, $t_{0.025} \approx 1.97$. So $ME = 1.97 \times 25\sqrt{\frac{2}{120}} = 1.97 \times 25 \times 0.129 \approx 9.8$ mg/dL. Choice B incorrectly uses the ratio $8 \times \sqrt{180/120} = 8 \times 1.22 = 9.8$ but rounds to 12. Choice C uses incorrect degrees of freedom. Choice D underestimates the impact of sample size reduction.
18. A researcher wants to estimate the mean systolic blood pressure in a population with a 95% confidence interval having a margin of error no greater than 3 mmHg. From a pilot study, the estimated standard deviation is 15 mmHg. However, the researcher realizes that if the true population standard deviation is actually 18 mmHg instead of 15 mmHg, what will happen to the actual margin of error if the same sample size is used?
    Answer: The margin of error will increase to approximately 3.6 mmHg When you encounter confidence interval questions involving changes in population parameters, focus on how each component of the margin of error formula responds to those changes. The margin of error formula is $ME = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$. Initially, the researcher calculated the needed sample size using σ = 15 mmHg to achieve ME = 3 mmHg with 95% confidence (z = 1.96). If the true population standard deviation is actually 18 mmHg instead of 15 mmHg, the margin of error increases proportionally since σ appears in the numerator. The ratio of new to old standard deviation is 18/15 = 1.2. Therefore, the actual margin of error becomes: 3.0 × 1.2 = 3.6 mmHg. This confirms that answer A is correct. Answer B (2.5 mmHg) incorrectly assumes the margin of error decreases, which would only happen if the true standard deviation were smaller than estimated. Answer C (3.0 mmHg) mistakenly suggests the margin of error stays constant regardless of the true population variance—this ignores the direct relationship between σ and margin of error. Answer D (4.2 mmHg) represents an incorrect calculation, possibly from misapplying the ratio or using wrong values. Remember this key principle: margin of error is directly proportional to the population standard deviation. When the true σ is larger than your estimate, your confidence interval will be wider than planned. Always check whether parameter changes increase or decrease the components in your formulas—don't just memorize calculations.

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