Business Statistics Quiz: Margin Of Error And Sample Size
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Margin Of Error And Sample SizeQuestion 1 of 20

An auditor wants to estimate the mean value of accounts receivable and needs a sample size that provides a margin of error of $25 with 95% confidence. A preliminary estimate for the population standard deviation is $180. However, the company's records are kept in two separate divisions, A and B. If the standard deviation of accounts in Division A is $150 and in Division B is $210, what is the consequence of using the pooled estimate of $180 for sample size planning?

The overall sample size will be correctly estimated only if the sample is drawn equally from both divisions.
The margin of error will be larger than $25 for Division B and smaller than $25 for Division A, assuming equal sample sizes are drawn from each.
The calculated sample size will be unnecessarily large for the required precision, wasting resources.
Using the pooled estimate is invalid; separate sample size calculations must be performed for each division.
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Business Statistics Quiz

Business Statistics Quiz: Margin Of Error And Sample Size

Practice Margin Of Error And Sample Size in Business Statistics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Margin Of Error And Sample Size, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Statistics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

An auditor wants to estimate the mean value of accounts receivable and needs a sample size that provides a margin of error of $25 with 95% confidence. A preliminary estimate for the population standard deviation is $180. However, the company's records are kept in two separate divisions, A and B. If the standard deviation of accounts in Division A is $150 and in Division B is $210, what is the consequence of using the pooled estimate of $180 for sample size planning?

  1. The overall sample size will be correctly estimated only if the sample is drawn equally from both divisions.
  2. The margin of error will be larger than $25 for Division B and smaller than $25 for Division A, assuming equal sample sizes are drawn from each. (correct answer)
  3. The calculated sample size will be unnecessarily large for the required precision, wasting resources.
  4. Using the pooled estimate is invalid; separate sample size calculations must be performed for each division.
Explanation: The sample size calculation depends directly on the standard deviation σ\sigma. The calculated sample size is n=(1.96×180/25)2≈199.1n = (1.96 \times 180 / 25)^2 \approx 199.1, so 200. If we allocate this sample (e.g., 100 to each division), the margin of error for each division can be checked. For Division A, ME_A = 1.96 \times 150 / \sqrt{100} = \29.4.ForDivisionB,. For Division B, ME_B = 1.96 \times 210 / \sqrt{100} = $41.16.TheMEisnotconstant.Letmere−readthis.Thisisaboutstratifiedsampling.Ifweusethepooledestimate. The ME is not constant. Let me re-read this. This is about stratified sampling. If we use the pooled estimate \sigma=180,wecalculate, we calculate n = (1.96 \times 180 / 25)^2 = 200.Ifwethensample100fromeachdivision,theMEweachieveforanoverallestimatedependsonthecombinedstandarderror.Let′sre−thinkthequestion.Thequestionissimpler.It′sabouttheconsequenceofusingone. If we then sample 100 from each division, the ME we achieve for an overall estimate depends on the combined standard error. Let's re-think the question. The question is simpler. It's about the consequence of using one \sigmawhentherearetwodifferentones.Ifweusewhen there are two different ones. If we usen=200andtake100fromeachdivision,theMEforasubsamplefromAwouldbeand take 100 from each division, the ME for a subsample from A would be1.96 \times 150 / \sqrt{100} = 29.4.ForB,itwouldbe. For B, it would be 1.96 \times 210 / \sqrt{100} = 41.16.Thequestionissubtle.Maybeit′snotaboutsubsamples,butaboutthewholesample.Ifwedrawasimplerandomsampleof200fromthewholepopulation,theactualMEwilldependontheproportionofaccountsfromeachdivisioninoursample.However,choiceBfocusesonthedivision−specificproperties.Ifweapplythesamplesizelogictoeachdivision,thesamplesizeneededforDivisionAwouldbesmaller,andforDivisionBwouldbelarger.Therefore,ifwedrawanequalnumberofsamplesbasedontheaverage. The question is subtle. Maybe it's not about subsamples, but about the whole sample. If we draw a simple random sample of 200 from the whole population, the actual ME will depend on the proportion of accounts from each division in our sample. However, choice B focuses on the division-specific properties. If we apply the sample size logic to each division, the sample size needed for Division A would be smaller, and for Division B would be larger. Therefore, if we draw an equal number of samples based on the average \sigma$, the estimate for B will be less precise (larger ME) than desired, and the estimate for A will be more precise (smaller ME) than needed. Choice B states this perfectly. Distractor A is incorrect because equal sampling doesn't guarantee the overall ME is correct. C is partially true for Division A but false for Division B, making it an incomplete answer. D is a possible strategy (stratified sampling) but doesn't describe the consequence of using the pooled estimate.

Question 2

A research firm's initial plan for a survey requires a 95% confidence level and a 3% margin of error for a proportion. A new directive requires the confidence level to be increased to 99% and the margin of error to be decreased to 1.5%. The new required sample size will be approximately how many times larger than the originally planned sample size?

  1. 2.6 times
  2. 3.5 times
  3. 4.0 times
  4. 6.9 times (correct answer)
Explanation: The sample size is proportional to (z∗/ME)2(z^*/ME)^2. The ratio of the new sample size to the old one is calculated as (znewzold)2×(MEoldMEnew)2\left(\frac{z_{\text{new}}}{z_{\text{old}}}\right)^2 \times \left(\frac{ME_{\text{old}}}{ME_{\text{new}}}\right)^2. For a 95% confidence level, zold≈1.96z_{\text{old}} \approx 1.96. For a 99% confidence level, znew≈2.576z_{\text{new}} \approx 2.576. The ratio is (2.5761.96)2×(0.030.015)2≈(1.314)2×(2)2≈1.728×4≈6.91\left(\frac{2.576}{1.96}\right)^2 \times \left(\frac{0.03}{0.015}\right)^2 \approx (1.314)^2 \times (2)^2 \approx 1.728 \times 4 \approx 6.91. Therefore, the new sample size must be about 6.9 times larger. Distractor A results from incorrectly assuming a linear relationship (i.e., not squaring the ratios). Distractor C results from only accounting for the change in the margin of error. Distractor B results from squaring the z-score ratio but not the margin of error ratio.

Question 3

A beverage company wants to estimate the proportion of consumers who prefer their new soda flavor. They require a 95% confidence level and a margin of error of no more than 2.5%. A pilot study of 50 consumers found that 15 preferred the new flavor. What is the total additional number of consumers they must survey to meet their requirements?

  1. 1291
  2. 1241 (correct answer)
  3. 1487
  4. 1537
Explanation: First, calculate the total required sample size using the information from the pilot study. The pilot study's sample proportion is p^=15/50=0.30\hat{p} = 15/50 = 0.30. This is the best estimate to use for planning. For a 95% confidence level, z∗≈1.96z^* \approx 1.96, and the desired margin of error is ME=0.025ME = 0.025. The required sample size nn is n=(z∗ME)2p^(1−p^)=(1.960.025)2(0.30)(0.70)=(78.4)2(0.21)≈1290.78n = \left(\frac{z^*}{ME}\right)^2 \hat{p}(1-\hat{p}) = \left(\frac{1.96}{0.025}\right)^2 (0.30)(0.70) = (78.4)^2 (0.21) \approx 1290.78. Since the sample size must be a whole number, we round up to 1291. The question asks for the additional number of consumers needed. Since 50 were already surveyed, the additional number is 1291−50=12411291 - 50 = 1241. Distractor A is the total sample size required. Distractor D is the total sample size if the conservative estimate p=0.5p=0.5 were used. Distractor C is the additional sample size if the conservative estimate were used (1537−501537 - 50).

Question 4

A political campaign wants to estimate the proportion of voters in a district who favor their candidate. They need a 95% confidence interval with a margin of error of at most 3%. In this district, 45% of voters are registered as Democrats, 35% as Republicans, and 20% as Independents. The campaign expects strong support (around 80%) from Democrats, weak support (around 20%) from Republicans, and uncertain support (around 50%) from Independents. To minimize the total sample size, which group's expected proportion should they use to plan their study?

  1. The expected proportion from Democrats (80%) because they are the largest group.
  2. The expected proportion from Republicans (20%) because it is the furthest from 50%.
  3. The expected proportion from Independents (50%) because it yields the largest variance. (correct answer)
  4. A weighted average of the expected proportions based on the registration breakdown.
Explanation: The sample size required for a proportion is largest when the population proportion pp is closest to 0.5, because this maximizes the variance term p(1−p)p(1-p). The campaign wants to ensure their sample is large enough regardless of the overall outcome. To be safe, they should use the value of pp that requires the largest sample size. Comparing the variance terms: for Democrats, p(1−p)=0.8(0.2)=0.16p(1-p) = 0.8(0.2) = 0.16; for Republicans, p(1−p)=0.2(0.8)=0.16p(1-p) = 0.2(0.8) = 0.16; for Independents, p(1−p)=0.5(0.5)=0.25p(1-p) = 0.5(0.5) = 0.25. The Independent group's expected proportion of 50% yields the highest variance and thus requires the largest sample size for a given margin of error. Using this value provides the most conservative estimate among the known subgroups. Distractor D is incorrect because a simple weighted average of proportions does not correctly account for the variance calculation needed for sample size planning.

Question 5

A consulting firm reported a 95% confidence interval for the average employee satisfaction score (on a 1-100 scale) as [72.5, 77.5]. The report, based on a sample of 150 employees, omitted the sample standard deviation. What was the approximate sample standard deviation of the satisfaction scores?

  1. 2.55
  2. 15.47 (correct answer)
  3. 30.92
  4. 5.00
Explanation: This requires working backward from the confidence interval. First, calculate the margin of error (ME). The width of the interval is 77.5−72.5=5.077.5 - 72.5 = 5.0. The margin of error is half the width, so ME=5.0/2=2.5ME = 5.0 / 2 = 2.5. The formula for the margin of error for a mean is ME=z∗snME = z^* \frac{s}{\sqrt{n}}. For a 95% confidence interval, z∗≈1.96z^* \approx 1.96. We are given ME=2.5ME = 2.5 and n=150n = 150. We can rearrange the formula to solve for the sample standard deviation, ss: s=ME×nz∗=2.5×1501.96≈2.5×12.2471.96≈30.61751.96≈15.62s = \frac{ME \times \sqrt{n}}{z^*} = \frac{2.5 \times \sqrt{150}}{1.96} \approx \frac{2.5 \times 12.247}{1.96} \approx \frac{30.6175}{1.96} \approx 15.62. The closest answer is 15.47. Distractor D is the width of the interval. Distractor C incorrectly multiplies the margin of error by n\sqrt{n} but forgets to divide by z∗z^*. Distractor A is the margin of error itself.

Question 6

A large corporation with 50,000 employees wants to survey a sample of them to estimate the mean daily commute time. The desired margin of error is 2 minutes with 95% confidence, and the standard deviation of commute times is estimated to be 15 minutes. The calculated required sample size is 217. A manager argues that since the company has only 1,000 employees at its headquarters, a separate survey there should use the finite population correction (FPC) factor. If the FPC is applied for the headquarters survey (N=1000), what is the adjusted sample size?

  1. 178 (correct answer)
  2. 212
  3. 46
  4. 217
Explanation: First, let's verify the initial sample size calculation: n=(z∗σME)2=(1.96×152)2=(14.7)2=216.09n = \left(\frac{z^* \sigma}{ME}\right)^2 = \left(\frac{1.96 \times 15}{2}\right)^2 = (14.7)^2 = 216.09, which rounds up to 217. This is the sample size assuming an infinite population, let's call it n0n_0. The finite population correction formula for adjusting the sample size is n=n01+n0−1Nn = \frac{n_0}{1 + \frac{n_0 - 1}{N}}. Here, n0=217n_0 = 217 and the relevant population size is N=1000N = 1000. So, n=2171+217−11000=2171+2161000=2171.216≈178.45n = \frac{217}{1 + \frac{217 - 1}{1000}} = \frac{217}{1 + \frac{216}{1000}} = \frac{217}{1.216} \approx 178.45. The adjusted required sample size would be 179 (rounding up), or approximately 178. Distractor D suggests the FPC has no effect. Distractor B makes a small, insignificant adjustment. Distractor C results from a significant miscalculation of the formula.

Question 7

A hospital administrator wants to estimate the proportion of patients who would rate their satisfaction as 'excellent'. A previous, large-scale study at a different hospital found this proportion to be 30%. The administrator wants to achieve a margin of error of 4% or less with 95% confidence. However, the budget only allows for a sample of 350 patients. Which of the following is the most accurate conclusion?

  1. The budget is insufficient, as the required sample size is approximately 505 patients. (correct answer)
  2. The budget is sufficient, as the required sample size is approximately 323 patients.
  3. The budget is insufficient; the smallest margin of error achievable with 350 patients is approximately 5.2%.
  4. The budget is sufficient, but only if the true proportion is closer to 15% than to 30%.
Explanation: When estimating proportions with confidence intervals, you need to determine the required sample size using the formula n=z2⋅p(1−p)E2n = \frac{z^2 \cdot p(1-p)}{E^2}, where z is the critical value, p is the estimated proportion, and E is the desired margin of error. For this problem, you have a 95% confidence level (z = 1.96), an estimated proportion of 30% (p = 0.30) from the previous study, and a desired margin of error of 4% (E = 0.04). Plugging these values in: n=(1.96)2⋅0.30(0.70)(0.04)2=3.84⋅0.210.0016=0.80640.0016=504n = \frac{(1.96)^2 \cdot 0.30(0.70)}{(0.04)^2} = \frac{3.84 \cdot 0.21}{0.0016} = \frac{0.8064}{0.0016} = 504 This means approximately 505 patients are needed, making the 350-patient budget insufficient. Answer A correctly identifies this insufficient budget and the proper required sample size of approximately 505 patients. Answer B incorrectly suggests the budget is sufficient with only 323 patients needed—this would result from using an incorrect margin of error or confidence level. Answer C reverses the calculation by finding what margin of error 350 patients would yield, rather than what sample size is needed for 4% margin of error. Answer D incorrectly suggests changing the proportion estimate would help significantly, but even at 15%, you'd still need about 490 patients. Remember: always calculate the required sample size first before determining budget adequacy. Don't get trapped by answer choices that work backwards from the available budget—the statistical requirement drives the sample size, not the budget constraints.

Question 8

A sample size calculation for a proportion results in n=600.27n = 600.27. An analyst reports that the minimum required sample size is 600. A senior statistician reviews the report and flags this as an error. Why is reporting 600 as the sample size an error?

  1. The sample size must always be rounded up to ensure the margin of error is not greater than the desired amount. (correct answer)
  2. The calculation should have been rounded to the nearest whole number, which is 600.
  3. The fractional result indicates that the confidence level used was not a standard one like 90%, 95%, or 99%.
  4. For sample sizes this large, the number should be reported to the nearest hundred, so 600 is correct.
Explanation: When you encounter sample size calculations in business statistics, remember that these calculations determine the minimum number of observations needed to achieve your desired precision. The key principle is that you must meet or exceed the calculated requirement. Since sample size calculations often produce non-whole numbers (like 600.27), you face a rounding decision that directly impacts your statistical precision. If you round down from 600.27 to 600, you're using fewer observations than the calculation determined necessary. This means your actual margin of error will be slightly larger than what you intended, potentially compromising the reliability of your results. The correct approach is always to round up to the next whole number (601 in this case) to ensure you meet or exceed your precision requirements. Looking at the wrong answers: Option B incorrectly suggests rounding to the nearest whole number, but "nearest" could mean rounding down, which reduces precision below acceptable levels. Option C misunderstands what causes fractional results - any confidence level can produce fractional sample sizes depending on the other parameters in the calculation. Option D suggests an arbitrary rounding rule based on magnitude, but statistical precision requirements don't change based on whether your sample is large or small. Study tip: Always remember "round up for sample size" - this ensures you never fall short of your required statistical precision. This principle applies whether you're calculating sample sizes for proportions, means, or any other parameter estimation problem.

Question 9

A research report states that based on a simple random sample of 120 customers, the 95% confidence interval for the mean monthly expenditure is [$210, $250]. The report also claims that the population standard deviation is known to be $80. Which of the following statements is the most likely conclusion about the report's findings?

  1. The sample size was too small to make a valid conclusion about the mean expenditure.
  2. The reported sample mean of $230 is a biased estimate of the population mean.
  3. The population of expenditures is likely not normally distributed, which invalidates the interval.
  4. The reported confidence interval is inconsistent with the other information provided. (correct answer)
Explanation: When you encounter confidence interval problems, always verify that the given information is mathematically consistent. This question tests whether you can spot when reported statistics don't align with the confidence interval formula. The formula for a 95% confidence interval with known population standard deviation is: xˉ±1.96×σn\bar{x} \pm 1.96 \times \frac{\sigma}{\sqrt{n}} Let's check the margin of error. With σ = $80 and n = 120, the margin of error should be: $1.96×80120=1.96×7.30=14.311.96 \times \frac{80}{\sqrt{120}} = 1.96 \times 7.30 = 14.31 $ The reported interval [$210, $250] has a range of $40, giving a margin of error of $20 (half the range). Since $20 ≠ $14.31, the confidence interval is inconsistent with the stated sample size and population standard deviation. This makes answer D correct. Answer A is wrong because n = 120 is plenty large for valid confidence intervals. Answer B incorrectly suggests the sample mean is biased—sample means from random samples are unbiased estimators regardless. Answer C misunderstands the Central Limit Theorem: with n = 120, the sampling distribution of the mean will be approximately normal even if the population isn't, making the confidence interval valid. Study tip: Always verify confidence interval calculations by plugging the given values into the formula. Look for inconsistencies between the reported interval width and what the margin of error should actually be—this is a common way exam writers test your understanding of the underlying mathematics.

Question 10

An operations manager wants to estimate the mean time to assemble a product. She wants to be 90% confident that the sample mean is within 1 minute of the true population mean. A small initial study suggests the standard deviation of assembly time is 5 minutes. After calculating the required sample size, she learns that the last hour of the workday has a much higher variability in assembly times. How should she adjust her sample size calculation?

  1. No adjustment is needed; the initial sample size calculation based on the average standard deviation is sufficient.
  2. She should use stratified sampling and calculate separate smaller sample sizes for the regular hours and the last hour.
  3. She should increase the confidence level to 99% to account for the increased variability, which will increase the sample size.
  4. She should use a larger estimate for the standard deviation in her calculation, which will increase the required sample size. (correct answer)
Explanation: When determining sample size for estimating a population mean, you need to consider how variability affects your precision. The sample size formula n=(zα/2⋅σE)2n = \left(\frac{z_{\alpha/2} \cdot \sigma}{E}\right)^2 shows that sample size increases with the square of the standard deviation (σ). If the operations manager discovers that assembly times have much higher variability during the last hour, this means her initial standard deviation estimate of 5 minutes is too low for the overall population. To maintain her desired precision (within 1 minute with 90% confidence), she must use a larger, more realistic estimate of the standard deviation in her calculation. This will increase the required sample size to account for the greater variability she'll encounter when sampling throughout the entire workday. Option A is wrong because using an underestimated standard deviation will result in too small a sample size, failing to achieve the desired precision. Option B misapplies stratified sampling - while stratification can be useful, the goal here is estimating the overall population mean, not comparing subgroups, and smaller sample sizes would reduce precision. Option C incorrectly suggests changing the confidence level, but the manager specifically wants 90% confidence; increasing to 99% doesn't address the variability issue and changes her original requirements. Study tip: Remember that in sample size calculations, the standard deviation estimate should reflect the full variability you expect to encounter in your actual sampling. Always use the most conservative (largest) reasonable estimate of σ to ensure adequate precision.

Question 11

A market research firm wants to estimate the proportion of consumers who prefer organic food products. They plan to use a 95% confidence interval with a margin of error no greater than 0.04. If a pilot study suggests the proportion is approximately 0.35, but the firm wants to be conservative and assumes no prior information about the proportion, what is the minimum sample size required?

  1. 425 consumers, since using p=0.35p = 0.35 from the pilot study provides the most accurate estimate
  2. 601 consumers, since assuming p=0.5p = 0.5 without prior information maximizes the required sample size (correct answer)
  3. 372 consumers, since the pilot study proportion of 0.35 should be used for planning purposes
  4. 545 consumers, since the average of 0.35 and 0.5 provides a reasonable compromise estimate
Explanation: When no prior information is assumed (conservative approach), we use p = 0.5 to maximize the sample size calculation. Using the formula n = (z²pq)/E², where z = 1.96, p = 0.5, q = 0.5, and E = 0.04: n = (1.96² × 0.5 × 0.5)/(0.04²) = 3.8416 × 0.25/0.0016 = 600.25, rounded up to 601. Choice A uses p = 0.35 incorrectly when being conservative. Choice C also incorrectly uses the pilot study value. Choice D incorrectly averages the proportions.

Question 12

A polling organization reports a margin of error of +/- 3% for a survey. A critic claims this margin of error is only valid if the sample proportion is 50%, and the true margin of error is likely smaller. Under which of the following circumstances is the critic's claim correct?

  1. The claim is always correct because the formula for margin of error does not depend on the sample proportion.
  2. The claim is correct if the true population proportion is very close to 0% or 100%. (correct answer)
  3. The claim is incorrect; the reported margin of error is an average over all possible sample proportions.
  4. The claim is only correct if the confidence level is less than 95%.
Explanation: The margin of error for a proportion is given by ME=z∗p(1−p)nME = z^* \sqrt{\frac{p(1-p)}{n}}. The term p(1−p)p(1-p) is maximized when p=0.5p=0.5. Polling organizations often report the most conservative (largest) margin of error, which is calculated using p=0.5p=0.5. If the true population proportion is far from 0.5 (i.e., close to 0 or 1), the term p(1−p)p(1-p) will be smaller than 0.5(1−0.5)=0.250.5(1-0.5)=0.25. For example, if p=0.1p=0.1, then p(1−p)=0.09p(1-p)=0.09. This would result in a smaller margin of error for a given sample size nn and confidence level z∗z^*. Therefore, the critic's claim that the true margin of error is likely smaller is correct if the actual proportion being measured is close to 0 or 1. Distractor A is incorrect because the formula explicitly depends on the proportion. C is incorrect as the ME is not an average. D is incorrect as the relationship holds for any confidence level.

Question 13

A marketing agency conducts a survey of 400 potential customers and finds that 120 are aware of a new product. They construct a 90% confidence interval. For a new study, they want to achieve the same margin of error but with 99% confidence. Using the most conservative estimate for the unknown proportion in the new study, what is the minimum required sample size?

  1. 663
  2. 980
  3. 1168 (correct answer)
  4. 400
Explanation: This is a two-step problem. First, calculate the margin of error (ME) from the original survey. The sample proportion is p^=120/400=0.30\hat{p} = 120/400 = 0.30. For 90% confidence, z∗≈1.645z^* \approx 1.645. The ME is ME=1.6450.30(1−0.30)400≈0.0377ME = 1.645 \sqrt{\frac{0.30(1-0.30)}{400}} \approx 0.0377. Second, use this ME to calculate the sample size for the new study. The new study requires 99% confidence (z∗≈2.576z^* \approx 2.576) and uses the conservative estimate p=0.5p=0.5. The formula is n=(z∗ME)2p(1−p)n = \left(\frac{z^*}{ME}\right)^2 p(1-p). So, n=(2.5760.0377)2(0.5)(0.5)≈(68.33)2(0.25)≈4668.7×0.25≈1167.18n = \left(\frac{2.576}{0.0377}\right)^2 (0.5)(0.5) \approx (68.33)^2 (0.25) \approx 4668.7 \times 0.25 \approx 1167.18. We must round up to the next integer, which is 1168. Distractor B arises from using the old proportion (0.3) instead of the conservative 0.5 for the new study. Distractor A arises from using the z-score for 95% confidence (1.96) instead of 99%. Distractor D incorrectly suggests no change in sample size is needed.

Question 14

A company wants to estimate the difference in the proportion of customers satisfied with two different service centers, A and B. They want the 95% confidence interval for the difference, pA−pBp_A - p_B, to have a margin of error no greater than 5%. Assuming they will sample an equal number of customers (nn) from each center, what is the minimum sample size nn required for each service center? Use the most conservative estimate for the proportions.

  1. 385
  2. 544
  3. 769 (correct answer)
  4. 1538
Explanation: The formula for the sample size required per group (n) when estimating the difference between two proportions is n=(z∗ME)2(pA(1−pA)+pB(1−pB))n = \left(\frac{z^*}{ME}\right)^2 (p_A(1-p_A) + p_B(1-p_B)). For a 95% confidence interval, z∗≈1.96z^* \approx 1.96. The desired margin of error is ME=0.05ME = 0.05. Using the most conservative estimate means we set pA=0.5p_A = 0.5 and pB=0.5p_B = 0.5. Plugging these values into the formula: n=(1.960.05)2(0.5(1−0.5)+0.5(1−0.5))=(39.2)2(0.25+0.25)=1536.64×0.5=768.32n = \left(\frac{1.96}{0.05}\right)^2 (0.5(1-0.5) + 0.5(1-0.5)) = (39.2)^2 (0.25 + 0.25) = 1536.64 \times 0.5 = 768.32. Since the sample size must be an integer, we round up to 769 for each center. Distractor A is the sample size required for estimating a single proportion, a common error. Distractor D is the total sample size for both groups (2×7692 \times 769). Distractor B is the result of using the z-score for 90% confidence (1.645) by mistake.

Question 15

A manufacturer is planning a study to estimate the proportion of defective products to within 1.5% with 90% confidence. The lead engineer insists on using a conservative estimate for the population proportion. The statistician argues that based on historical data, the defect rate has never exceeded 4%, and using this information would be more efficient. What is the approximate reduction in required sample size if the statistician's recommendation is followed instead of the engineer's?

  1. The sample size is reduced by about 83%. (correct answer)
  2. The sample size is reduced by about 15%.
  3. The sample size is reduced by about 96%.
  4. The sample size is reduced by about 4%.
Explanation: This question compares the sample size needed under two assumptions for the population proportion, pp. ME = 0.015 and confidence is 90% (z∗≈1.645z^* \approx 1.645). The engineer's conservative approach uses p=0.5p=0.5. The required sample size is neng=(1.6450.015)2(0.5)(0.5)≈(109.67)2(0.25)≈12027×0.25≈3007n_{eng} = \left(\frac{1.645}{0.015}\right)^2 (0.5)(0.5) \approx (109.67)^2(0.25) \approx 12027 \times 0.25 \approx 3007. The statistician's approach uses p=0.04p=0.04. The required sample size is nstat=(1.6450.015)2(0.04)(0.96)≈(109.67)2(0.0384)≈12027×0.0384≈462n_{stat} = \left(\frac{1.645}{0.015}\right)^2 (0.04)(0.96) \approx (109.67)^2(0.0384) \approx 12027 \times 0.0384 \approx 462. The reduction in sample size is 3007−462=25453007 - 462 = 2545. The percent reduction is 25453007×100%≈84.6%\frac{2545}{3007} \times 100\% \approx 84.6\%. The closest answer is 83%. The calculation can be simplified by looking at the ratio of the p(1−p)p(1-p) terms: p2(1−p2)p1(1−p1)=0.04(0.96)0.5(0.5)=0.03840.25=0.1536\frac{p_2(1-p_2)}{p_1(1-p_1)} = \frac{0.04(0.96)}{0.5(0.5)} = \frac{0.0384}{0.25} = 0.1536. This means the new sample size is only 15.36% of the original, so the reduction is 1−0.1536=0.84641 - 0.1536 = 0.8464 or 84.64%.

Question 16

A financial services company wants to estimate the mean household income of its clients. They want the margin of error for a 95% confidence interval to be no more than 2% of the estimated mean. A pilot study suggests the mean income is approximately $120,000 with a standard deviation of $30,000. What is the minimum sample size required?

  1. 25
  2. 2401
  3. 9604
  4. 601 (correct answer)
Explanation: When determining sample size for estimating a population mean with a specified margin of error, you need to work with the relationship between confidence level, standard deviation, and desired precision. The key formula is: n=(z⋅σE)2n = \left(\frac{z \cdot \sigma}{E}\right)^2, where z is the critical value, σ is the standard deviation, and E is the margin of error in absolute terms. First, convert the relative margin of error to absolute terms: 2% of $120,000 = $2,400. For a 95% confidence interval, z = 1.96. Substituting: $n=(1.96×30,0002,400)2=(58,8002,400)2=(24.5)2=600.25n = \left(\frac{1.96 \times 30,000}{2,400}\right)^2 = \left(\frac{58,800}{2,400}\right)^2 = (24.5)^2 = 600.25 $ Rounding up to ensure the margin of error doesn't exceed 2%, you need n = 601, making D correct. Choice A (25) represents a sample far too small for this precision requirement—it would yield a margin of error around 12% of the mean. Choice B (2401) appears to use z = 1.96 incorrectly, possibly squaring it in the calculation. Choice C (9604) likely results from using the margin of error as $600 instead of $2,400, or from other computational errors with the formula components. Remember that sample size calculations always round up to ensure you meet or exceed your precision requirements. Also, when margin of error is given as a percentage of the mean, always convert to absolute dollars first before applying the formula.

Question 17

A quality control department took a sample of 250 smart bulbs and constructed a 95% confidence interval for the mean lifespan: [8050 hours, 8150 hours]. Assuming the sample standard deviation is a reasonable estimate for the population standard deviation, what sample size would be required to estimate the mean lifespan with a margin of error of just 20 hours at 99% confidence?

  1. Approximately 432
  2. Approximately 1562
  3. Approximately 2699 (correct answer)
  4. Approximately 403
Explanation: This is a two-step problem. First, determine the sample standard deviation (s) from the information provided. The margin of error (ME) for the initial 95% confidence interval is (8150−8050)/2=50(8150 - 8050) / 2 = 50 hours. Using the formula ME=z∗snME = z^* \frac{s}{\sqrt{n}}, we have 50=1.96×s25050 = 1.96 \times \frac{s}{\sqrt{250}}. Solving for s gives s=50×2501.96≈403.35s = \frac{50 \times \sqrt{250}}{1.96} \approx 403.35 hours. Second, use this value of s to calculate the required sample size for the new conditions. The desired ME is 20 hours and the confidence level is 99% (z∗≈2.576z^* \approx 2.576). The sample size formula is n=(z∗sME)2n = \left(\frac{z^* s}{ME}\right)^2. Plugging in the values: n=(2.576×403.3520)2≈(51.94)2≈2698.3n = \left(\frac{2.576 \times 403.35}{20}\right)^2 \approx (51.94)^2 \approx 2698.3. Since sample size must be an integer, we round up to 2699. Distractor A uses the old ME of 50 in the new calculation. Distractor B uses the old z-score of 1.96 in the new calculation. Distractor D incorrectly identifies the calculated standard deviation as the sample size.

Question 18

An e-commerce company wants to determine the sample size needed to estimate its average order value to within $5.00 at 95% confidence. The population standard deviation of order values is unknown. The company takes a small pilot sample and finds its standard deviation is $28.00. However, a business analyst points out that this pilot sample included a few unusually large orders. If these were removed, the standard deviation would be $21.00. How does the required sample size change if the analyst uses the smaller, outlier-removed standard deviation for planning?

  1. It decreases by approximately 25%.
  2. It decreases by approximately 44%. (correct answer)
  3. It is now 56% larger than the original estimate.
  4. It decreases by approximately 75%.
Explanation: The required sample size nn is calculated by n=(z∗σME)2n = \left(\frac{z^* \sigma}{ME}\right)^2. Since z∗z^* and MEME are constant, the sample size is directly proportional to the square of the standard deviation (n∝σ2n \propto \sigma^2). Let n1n_1 be the size calculated with σ1=28\sigma_1 = 28 and n2n_2 be the size with σ2=21\sigma_2 = 21. The ratio of the new sample size to the original is n2n1=σ22σ12=212282=441784=0.5625\frac{n_2}{n_1} = \frac{\sigma_2^2}{\sigma_1^2} = \frac{21^2}{28^2} = \frac{441}{784} = 0.5625. This means the new required sample size is 56.25% of the original. The percentage decrease is 100100% - 56.25% = 43.75%, which is approximately 44%. Distractor A incorrectly uses the ratio of the standard deviations, not their squares (21/28=0.7521/28 = 0.75, a 25% decrease). Distractor C misinterprets the ratio of 0.5625. Distractor D incorrectly squares the ratio of the standard deviations ((21/28)2=0.752=0.5625(21/28)^2 = 0.75^2 = 0.5625, then subtracts from 1, but uses 0.75 in the text).

Question 19

A research team is planning a survey to estimate a population mean. They calculate a required sample size of n=400n=400 to achieve their desired margin of error and confidence level. If the team discovers their budget is cut and they can only survey 300 people, but they are unwilling to reduce their confidence level, what will be the new margin of error relative to their original target?

  1. The new margin of error will be 75% of the original target.
  2. The new margin of error will be 133% of the original target.
  3. The new margin of error will be 115% of the original target. (correct answer)
  4. The new margin of error will be 150% of the original target.
Explanation: The formula for margin of error is ME=z∗σnME = z^* \frac{\sigma}{\sqrt{n}}. Since z∗z^* and σ\sigma are constant, MEME is proportional to 1/n1/\sqrt{n}. Let ME1ME_1 be the original margin of error with n1=400n_1 = 400, and ME2ME_2 be the new margin of error with n2=300n_2 = 300. The ratio of the new ME to the old ME is ME2ME1=1/n21/n1=n1n2\frac{ME_2}{ME_1} = \frac{1/\sqrt{n_2}}{1/\sqrt{n_1}} = \sqrt{\frac{n_1}{n_2}}. Plugging in the values: ME2ME1=400300=1.333...≈1.1547\frac{ME_2}{ME_1} = \sqrt{\frac{400}{300}} = \sqrt{1.333...} \approx 1.1547. Therefore, the new margin of error will be approximately 115% of the original target. Distractor B incorrectly uses the direct ratio 400/300=1.33400/300 = 1.33. Distractor A uses the inverse ratio 300/400=0.75300/400 = 0.75. Distractor D is a simple miscalculation.

Question 20

A pharmaceutical company is planning a clinical trial to estimate the proportion of patients who experience side effects from a new medication. They want to construct a 95% confidence interval with a margin of error of 0.05. A similar drug showed a 28% side effect rate. If the company decides to increase the confidence level to 99% while keeping the same margin of error, how will this affect the required sample size?

  1. The sample size will increase by approximately 44%, from 310 to 447 participants (correct answer)
  2. The sample size will increase by approximately 78%, from 310 to 552 participants
  3. The sample size will increase by approximately 62%, from 310 to 502 participants
  4. The sample size will increase by approximately 35%, from 310 to 419 participants
Explanation: For 95% CI: n₁ = (1.96² × 0.28 × 0.72)/(0.05²) = 309.5 ≈ 310. For 99% CI: n₂ = (2.576² × 0.28 × 0.72)/(0.05²) = 447.1 ≈ 447. The increase is (447-310)/310 = 0.442 or about 44%. Choice B overestimates the increase. Choice C provides an intermediate value that doesn't match the calculation. Choice D underestimates the impact of increasing the confidence level.