Biostatistics Quiz: Standard Error
20 questions · exam conditions
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Standard ErrorQuestion 1 of 20

A quality control engineer measures the weights of manufactured parts. The sample standard deviation is s=2.4s = 2.4 grams from a sample of 36 parts. If the engineer wants to reduce the standard error of the sample mean by half, approximately how many additional parts should be sampled?

72 additional parts
108 additional parts
144 additional parts
180 additional parts
216 additional parts
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Biostatistics Quiz

Biostatistics Quiz: Standard Error

Practice Standard Error in Biostatistics 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 Standard Error, giving you a quick way to practice the rules, question types, and explanations that matter most for Biostatistics.

How to use this quiz

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.

All questions

Question 1

A quality control engineer measures the weights of manufactured parts. The sample standard deviation is s=2.4s = 2.4 grams from a sample of 36 parts. If the engineer wants to reduce the standard error of the sample mean by half, approximately how many additional parts should be sampled?

  1. 72 additional parts
  2. 108 additional parts (correct answer)
  3. 144 additional parts
  4. 180 additional parts
  5. 216 additional parts
Explanation: This question tests your understanding of the standard error formula and how sample size affects precision of estimates. The standard error of the sample mean is SE=snSE = \frac{s}{\sqrt{n}}, where ss is the sample standard deviation and nn is the sample size. Currently, with n=36n = 36 parts and s=2.4s = 2.4 grams, the standard error is SE=2.436=2.46=0.4SE = \frac{2.4}{\sqrt{36}} = \frac{2.4}{6} = 0.4 grams. To reduce this by half, you need a new standard error of 0.20.2 grams. Setting up the equation: 0.2=2.4nnew0.2 = \frac{2.4}{\sqrt{n_{new}}}. Solving for nnewn_{new}: nnew=2.40.2=12\sqrt{n_{new}} = \frac{2.4}{0.2} = 12, so nnew=144n_{new} = 144 parts total. Since you already have 36 parts, you need 144−36=108144 - 36 = 108 additional parts, which is answer B. Answer A (72 additional parts) would give you 108 total parts, yielding SE=2.4108≈0.23SE = \frac{2.4}{\sqrt{108}} \approx 0.23 grams—not quite half the original. Answer C (144 additional parts) would give you 180 total parts, creating an unnecessarily small standard error of 0.180.18 grams. Answer D (180 additional parts) would give you 216 total parts, resulting in an even smaller standard error of 0.160.16 grams. Remember: to reduce standard error by half, you must quadruple the sample size. This is because standard error involves the square root of sample size, so the relationship isn't linear.

Question 2

In a survey about voting preferences, 180 out of 400 respondents favor Candidate X. If the survey is repeated with the same methodology but double the sample size, and the sample proportion remains the same, what happens to the margin of error (approximately 2×SE2 \times SE)?

  1. It remains the same because the proportion is unchanged
  2. It decreases by a factor of 2\sqrt{2} because the sample size doubled (correct answer)
  3. It decreases by a factor of 2 because the sample size doubled
  4. It increases by a factor of 2\sqrt{2} due to increased variability
  5. It increases by a factor of 2 due to the larger absolute number of responses
Explanation: When you encounter questions about sample size and margin of error, you're dealing with the fundamental relationship between precision and sample size in statistical estimation. The margin of error is approximately 2×SE2 \times SE, where the standard error for a proportion is SE=p(1−p)nSE = \sqrt{\frac{p(1-p)}{n}}. In the original survey, p=180/400=0.45p = 180/400 = 0.45 and n=400n = 400. When the sample size doubles to 800 while keeping the same proportion (360 out of 800), the new standard error becomes SEnew=0.45(0.55)800SE_{new} = \sqrt{\frac{0.45(0.55)}{800}}. To see the relationship clearly, compare the standard errors: SEnewSEoriginal=400800=12=12\frac{SE_{new}}{SE_{original}} = \sqrt{\frac{400}{800}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}. This means the new standard error is 12\frac{1}{\sqrt{2}} times the original, so the margin of error decreases by a factor of 2\sqrt{2}. Answer A is wrong because even though the proportion stays the same, the denominator (sample size) in the standard error formula changes, affecting precision. Answer C incorrectly assumes a linear relationship—doubling sample size doesn't halve the margin of error due to the square root in the formula. Answer D contradicts basic statistical principles; larger samples reduce variability and margin of error, they don't increase it. Remember this key pattern: margin of error is inversely proportional to the square root of sample size. To halve your margin of error, you need four times the sample size, not twice.

Question 3

An epidemiologist studies the prevalence of a rare disease. In a sample of 1600 individuals, 48 have the disease. The epidemiologist wants to estimate the standard error of the sample prevalence rate. Which calculation is correct?

  1. 48×1552/1600\sqrt{48 \times 1552 / 1600}
  2. 0.03×0.97/1600\sqrt{0.03 \times 0.97 / 1600} (correct answer)
  3. 48/1600\sqrt{48 / 1600}
  4. 0.03/16000.03 / \sqrt{1600}
  5. 48/1600\sqrt{48} / 1600
Explanation: When you encounter questions about standard error for prevalence or proportion data, you're working with the standard error formula for a binomial proportion: SE=p(1−p)nSE = \sqrt{\frac{p(1-p)}{n}}, where p is the sample proportion and n is the sample size. First, calculate the sample proportion: p=481600=0.03p = \frac{48}{1600} = 0.03. Then apply the standard error formula: SE=0.03×(1−0.03)1600=0.03×0.971600SE = \sqrt{\frac{0.03 \times (1-0.03)}{1600}} = \sqrt{\frac{0.03 \times 0.97}{1600}}. This matches option B exactly. Let's examine why the other options are incorrect. Option A uses 48×1552/1600\sqrt{48 \times 1552 / 1600}, which appears to use raw counts rather than proportions. While 48 × 1552 = 48 × (1600-48), this isn't the correct formula structure and yields a much larger value than the true standard error. Option C uses 48/1600\sqrt{48/1600}, which is simply p\sqrt{p} – this ignores the variance component (1-p) entirely and isn't a valid standard error calculation. Option D shows 0.03/16000.03/\sqrt{1600}, which resembles the standard error of a mean (σ/n\sigma/\sqrt{n}) rather than a proportion. This treats the proportion like a continuous variable instead of recognizing its binomial nature. Remember this key distinction: proportions and means have different standard error formulas. For proportions, you always need both p and (1-p) in the numerator under the square root, while the denominator contains n. Watch for this pattern whenever you're calculating uncertainty around prevalence, success rates, or any binary outcome data.

Question 4

A medical researcher measures serum cholesterol levels in patients. From a pilot study of 16 patients, the sample mean is 220 mg/dL with sample standard deviation 40 mg/dL. For the main study, the researcher wants the standard error of the sample mean to be no more than 5 mg/dL. What is the minimum sample size needed?

  1. 48 patients
  2. 64 patients (correct answer)
  3. 80 patients
  4. 96 patients
  5. 128 patients
Explanation: When you encounter questions about sample size and standard error, you're dealing with the fundamental relationship between sample size and precision of estimates. The standard error of the sample mean decreases as sample size increases, following the formula SE=snSE = \frac{s}{\sqrt{n}}, where s is the sample standard deviation and n is the sample size. To find the minimum sample size needed for a standard error of no more than 5 mg/dL, you set up the inequality: 40n≤5\frac{40}{\sqrt{n}} \leq 5. Solving for n: n≥405=8\sqrt{n} \geq \frac{40}{5} = 8, so n≥64n \geq 64. Therefore, the minimum sample size is 64 patients. Let's examine why the other options are incorrect. Option A (48 patients) is too small—plugging into the formula gives SE=4048=5.77SE = \frac{40}{\sqrt{48}} = 5.77 mg/dL, which exceeds the required 5 mg/dL maximum. Options C (80 patients) and D (96 patients) would indeed achieve the desired precision (giving standard errors of 4.47 and 4.08 mg/dL respectively), but the question asks for the minimum sample size needed. These represent unnecessarily large samples that would increase costs and time without additional benefit. The correct answer is B (64 patients), which gives exactly SE=4064=5SE = \frac{40}{\sqrt{64}} = 5 mg/dL. Remember this pattern: sample size calculations often involve setting up an equation or inequality with the standard error formula, then solving for n. Always check whether the question asks for minimum, maximum, or exact values to avoid selecting a technically correct but non-optimal answer.

Question 5

In a genetics study, the frequency of a particular allele is estimated from a sample. The sample contains 80 individuals (160 alleles total), and 24 copies of the target allele are observed. If this study is replicated with 320 individuals, and the true allele frequency remains the same, what is the expected standard error of the new sample allele frequency?

  1. 0.0122
  2. 0.0173 (correct answer)
  3. 0.0244
  4. 0.0346
  5. 0.0488
Explanation: When you encounter allele frequency problems, you're working with population genetics principles where sample size directly affects the precision of your estimates through the standard error. First, calculate the original allele frequency: p^=24160=0.15\hat{p} = \frac{24}{160} = 0.15. The standard error for allele frequency follows the formula: SE=p(1−p)2nSE = \sqrt{\frac{p(1-p)}{2n}}, where pp is the allele frequency and nn is the number of individuals (the factor of 2 accounts for diploid organisms having two alleles per individual). For the new study with 320 individuals: SE=0.15×0.852×320=0.1275640=0.000199=0.0173SE = \sqrt{\frac{0.15 \times 0.85}{2 \times 320}} = \sqrt{\frac{0.1275}{640}} = \sqrt{0.000199} = 0.0173 Option A (0.0122) represents a calculation error where someone likely used the total number of alleles (640) instead of twice the number of individuals in the denominator. Option C (0.0244) is exactly twice the correct answer, suggesting someone forgot the factor of 2 in the denominator entirely. Option D (0.0346) appears to use an incorrect formula altogether, possibly confusing this with a different type of standard error calculation. Remember that standard error decreases with the square root of sample size, so quadrupling the sample size (from 80 to 320 individuals) halves the standard error. Always verify your formula includes the factor of 2 for diploid genetics problems—this distinguishes allele frequency calculations from simple proportion estimates.

Question 6

An environmental scientist measures mercury levels in fish samples. The current sampling protocol yields a standard error of 0.8 ppm for the sample mean mercury level. Budget constraints will reduce the sample size by 75%. What will be the new standard error, assuming the population standard deviation remains constant?

  1. 0.2 ppm
  2. 1.0 ppm
  3. 1.6 ppm (correct answer)
  4. 2.4 ppm
  5. 3.2 ppm
Explanation: This question tests your understanding of how sample size affects the standard error of the mean, a fundamental concept in sampling theory. The standard error measures the precision of your sample mean as an estimate of the population mean. The key relationship is: SE=σnSE = \frac{\sigma}{\sqrt{n}}, where σ is the population standard deviation and n is the sample size. Notice that standard error is inversely proportional to the square root of sample size. If the sample size is reduced by 75%, you're left with 25% of the original sample size, or nnew=0.25nn_{new} = 0.25n. To find the new standard error, you can use the ratio: SEnewSEold=noldnnew=n0.25n=4=2\frac{SE_{new}}{SE_{old}} = \sqrt{\frac{n_{old}}{n_{new}}} = \sqrt{\frac{n}{0.25n}} = \sqrt{4} = 2 Therefore: SEnew=2×0.8=1.6SE_{new} = 2 × 0.8 = 1.6 ppm. Answer A (0.2 ppm) incorrectly assumes the standard error decreases proportionally with sample size reduction (0.8 × 0.25). Answer B (1.0 ppm) might result from incorrectly using the square root of the proportion directly (0.8 × √0.25 × 2). Answer D (2.4 ppm) comes from multiplying by 3 instead of 2, perhaps confusing the 75% reduction with the multiplier. Remember this pattern: when sample size decreases by a factor of k, standard error increases by a factor of √(1/k). Cutting your sample size to one-fourth quadruples the variance of your estimate and doubles the standard error.

Question 7

A market researcher surveys consumer preferences. In a pilot study of 25 consumers, 8 prefer Brand A. The researcher wants to design a main study where the standard error of the sample proportion preferring Brand A is at most 0.03. What minimum sample size is needed for the main study?

  1. 256 consumers
  2. 342 consumers (correct answer)
  3. 400 consumers
  4. 512 consumers
  5. 625 consumers
Explanation: When you encounter sample size questions involving proportions, you're dealing with the standard error formula for sample proportions. The key relationship is that standard error decreases as sample size increases, following the formula SE=p(1−p)nSE = \sqrt{\frac{p(1-p)}{n}}, where p is the proportion and n is the sample size. From the pilot study, you have p^=825=0.32\hat{p} = \frac{8}{25} = 0.32. To find the minimum sample size where the standard error is at most 0.03, you set up: 0.03≥0.32(1−0.32)n0.03 \geq \sqrt{\frac{0.32(1-0.32)}{n}} Squaring both sides: 0.0009≥0.32×0.68n0.0009 \geq \frac{0.32 \times 0.68}{n} Solving for n: n≥0.21760.0009=241.8n \geq \frac{0.2176}{0.0009} = 241.8 Since you need at least 242 consumers, answer B (342 consumers) is the smallest option that satisfies this requirement. Answer A (256 consumers) falls short of providing adequate precision—while close to the calculated minimum, it doesn't account for the conservative approach typically used in study design. Answer C (400 consumers) and D (512 consumers) would certainly work but represent unnecessarily large (and expensive) samples that exceed the requirements. Study tip: For proportion sample size problems, always use the pilot study proportion if available rather than assuming the worst-case scenario of p = 0.5. This gives you a more realistic and often smaller required sample size. Remember that the answer should be the smallest option that meets or exceeds your calculated minimum.

Question 8

A psychologist studies reaction times in cognitive tests. The sample consists of 144 participants with sample variance s2=64s^2 = 64 ms². Due to budget cuts, future studies can only include 36 participants. Assuming the population variance remains the same, how will the standard error of the sample mean change?

  1. It will decrease by a factor of 2
  2. It will increase by a factor of 2 (correct answer)
  3. It will decrease by a factor of 4
  4. It will increase by a factor of 4
  5. It will remain unchanged because the population variance is constant
Explanation: When you encounter questions about how sample size affects standard error, you're dealing with one of the most fundamental relationships in statistics. The standard error of the sample mean measures how much variability you expect in sample means across different samples. The standard error formula is SE=σnSE = \frac{\sigma}{\sqrt{n}}, where σ is the population standard deviation and n is the sample size. Since the sample variance is 64 ms², the standard deviation is s=64=8s = \sqrt{64} = 8 ms. For the original study: SE1=8144=812=0.67SE_1 = \frac{8}{\sqrt{144}} = \frac{8}{12} = 0.67 ms For the smaller study: SE2=836=86=1.33SE_2 = \frac{8}{\sqrt{36}} = \frac{8}{6} = 1.33 ms The ratio is SE2SE1=1.330.67=2\frac{SE_2}{SE_1} = \frac{1.33}{0.67} = 2, so the standard error increases by a factor of 2. Option A incorrectly suggests the standard error decreases by a factor of 2 — this reverses the direction of change. Option C claims it decreases by a factor of 4, which both gets the direction wrong and confuses the sample size ratio (144÷36 = 4) with the standard error change. Option D gets the factor of 4 wrong because it fails to account for the square root in the denominator — while the sample size changes by a factor of 4, the standard error only changes by 4=2\sqrt{4} = 2. Remember: standard error is inversely proportional to the square root of sample size. Smaller samples always mean larger standard errors, making your estimates less precise.

Question 9

In a manufacturing process, the standard error of the sample mean diameter is 0.05 mm when 100 parts are measured. Due to equipment constraints, only 25 parts can be measured in future quality checks. By what factor will the standard error increase?

  1. 1.5
  2. 2.0 (correct answer)
  3. 3.0
  4. 4.0
  5. 0.5
Explanation: This question tests your understanding of how sample size affects the standard error of the mean. The standard error measures the precision of your sample mean as an estimate of the population mean, and it's directly related to sample size through a specific mathematical relationship. The standard error of the mean follows the formula: SE=σnSE = \frac{\sigma}{\sqrt{n}}, where σ is the population standard deviation and n is the sample size. The key insight is that standard error is inversely proportional to the square root of sample size. When sample size decreases from 100 to 25 parts, you need to find the ratio of the new standard error to the old standard error. Since the population standard deviation remains constant, this ratio equals 10025=105=2\frac{\sqrt{100}}{\sqrt{25}} = \frac{10}{5} = 2. The standard error will be twice as large, so the factor of increase is 2.0, making B correct. Looking at the wrong answers: A (1.5) has no mathematical basis in the standard error formula. C (3.0) might tempt you if you incorrectly think the relationship is linear rather than involving square roots. D (4.0) is what you'd get if you mistakenly used the ratio of original to new sample sizes (100/25 = 4) without applying the square root. Remember this pattern: when sample size changes by a factor of k, standard error changes by a factor of 1k\frac{1}{\sqrt{k}}. Smaller samples always mean larger standard errors, reflecting decreased precision in your estimates.

Question 10

An agricultural scientist measures crop yields from experimental plots. Plot Type 1 yields have sample mean 85 kg with standard error 3.2 kg based on 16 plots. Plot Type 2 yields have sample mean 92 kg with standard error 2.8 kg based on 25 plots. If the scientist wants to compare these means, what is the standard error of the difference in sample means, assuming the plot types are independent?

  1. 0.4 kg
  2. 4.2 kg (correct answer)
  3. 6.0 kg
  4. 8.96 kg
  5. 36.0 kg
Explanation: When comparing two independent sample means, you need to find the standard error of their difference. This is a fundamental concept in hypothesis testing and confidence intervals for two-sample problems. The standard error of the difference between two independent sample means follows the formula: SEdiff=SE12+SE22SE_{diff} = \sqrt{SE_1^2 + SE_2^2}, where SE1SE_1 and SE2SE_2 are the standard errors of the individual samples. For this problem, you have:
  • Plot Type 1: SE1=3.2SE_1 = 3.2 kg
  • Plot Type 2: SE2=2.8SE_2 = 2.8 kg
Calculating: SEdiff=(3.2)2+(2.8)2=10.24+7.84=18.08=4.2SE_{diff} = \sqrt{(3.2)^2 + (2.8)^2} = \sqrt{10.24 + 7.84} = \sqrt{18.08} = 4.2 kg Looking at the wrong answers: Choice A (0.4 kg) represents subtracting the standard errors rather than using the proper formula—this ignores the variability principle. Choice C (6.0 kg) comes from simply adding the two standard errors (3.2 + 2.8), which incorrectly assumes perfect positive correlation between the samples. Choice D (8.96 kg) is the sum of the squared standard errors without taking the square root, showing incomplete application of the formula. The key insight is that when combining independent random variables, their variances add (not their standard deviations). Since standard error is essentially a standard deviation of the sampling distribution, you square both values, add them, then take the square root. Remember this "square, add, square root" pattern for any problems involving the standard error of sums or differences of independent samples.

Question 11

An economist analyzes household income data. Sample A consists of 100 households with sample standard deviation $12,000. Sample B consists of 400 households with sample standard deviation $24,000. Both samples are drawn from populations with the same income distribution. Which statement about their standard errors is correct?

  1. Sample A has a smaller standard error because its sample standard deviation is half that of Sample B
  2. Sample B has a smaller standard error because its sample size is four times larger than Sample A
  3. Sample A and Sample B have equal standard errors due to the offsetting effects of sample size and variability (correct answer)
  4. Sample A has a standard error twice as large as Sample B
  5. Sample B has a standard error twice as large as Sample A
Explanation: When you encounter questions about standard errors, remember that the standard error of the mean depends on both the sample standard deviation and the sample size according to the formula: SE=snSE = \frac{s}{\sqrt{n}}, where ss is the sample standard deviation and nn is the sample size. Let's calculate both standard errors. For Sample A: SEA=12,000100=12,00010=1,200SE_A = \frac{12,000}{\sqrt{100}} = \frac{12,000}{10} = 1,200. For Sample B: SEB=24,000400=24,00020=1,200SE_B = \frac{24,000}{\sqrt{400}} = \frac{24,000}{20} = 1,200. Both samples have identical standard errors of $1,200, making answer choice C correct—the effects of sample size and variability perfectly offset each other. Answer choice A incorrectly focuses only on the sample standard deviations while ignoring sample size. While Sample A's standard deviation is indeed half of Sample B's, this doesn't determine the standard error by itself. Answer choice B makes the opposite mistake, considering only sample size. Yes, Sample B has four times more observations, but its sample standard deviation is also twice as large, which increases the standard error and cancels out the benefit of the larger sample size. Answer choice D gets the relationship backwards—if there were a difference, it would be in the opposite direction given the relative sample sizes. The key insight is that standard error involves a trade-off between variability and sample size. Always calculate both components rather than focusing on just one factor, as exam questions often test whether you understand this dual relationship.

Question 12

A pharmaceutical company tests a new drug on patients with high blood pressure. The sample mean reduction in systolic blood pressure is 15.2 mmHg with a sample standard deviation of 8.6 mmHg from 49 patients. What is the standard error of the sample mean reduction?

  1. 0.175 mmHg
  2. 1.229 mmHg (correct answer)
  3. 1.756 mmHg
  4. 8.600 mmHg
  5. 15.200 mmHg
Explanation: When you encounter a question asking for the standard error of a sample mean, you're being tested on your understanding of sampling distributions and how sample statistics relate to population parameters. The standard error of the sample mean measures how much the sample mean would vary if you repeated the sampling process many times. It's calculated using the formula: SE=snSE = \frac{s}{\sqrt{n}}, where s is the sample standard deviation and n is the sample size. In this problem, you have a sample standard deviation of 8.6 mmHg and 49 patients. Plugging into the formula: SE=8.649=8.67=1.229SE = \frac{8.6}{\sqrt{49}} = \frac{8.6}{7} = 1.229 mmHg. This confirms answer B is correct. Let's examine why the other options are wrong. Option A (0.175 mmHg) appears to result from incorrectly dividing the sample standard deviation by the sample size itself (8.6 ÷ 49), rather than by the square root of the sample size. Option C (1.756 mmHg) might come from computational errors or using an incorrect denominator. Option D (8.600 mmHg) is simply the sample standard deviation itself – a common trap for students who confuse standard deviation with standard error. Remember this key distinction: the standard deviation describes the variability in your sample data, while the standard error describes the precision of your sample mean as an estimate of the population mean. Always divide by the square root of n, not n itself, when calculating standard error.

Question 13

In a clinical trial, the sample proportion of patients experiencing side effects is p^=0.12\hat{p} = 0.12 based on n=300n = 300 patients. What is the standard error of this sample proportion?

  1. 0.0188 (correct answer)
  2. 0.0324
  3. 0.0544
  4. 0.1056
  5. 0.2000
Explanation: When you encounter questions about sample proportions in clinical trials, you're dealing with the sampling distribution of proportions and need to calculate the standard error to understand the variability of your estimate. The standard error of a sample proportion measures how much the sample proportion p^\hat{p} would vary from sample to sample. The formula is: SE(p^)=p^(1−p^)nSE(\hat{p}) = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} With p^=0.12\hat{p} = 0.12 and n=300n = 300: SE(p^)=0.12×0.88300=0.1056300=0.000352=0.0188SE(\hat{p}) = \sqrt{\frac{0.12 \times 0.88}{300}} = \sqrt{\frac{0.1056}{300}} = \sqrt{0.000352} = 0.0188 This confirms answer A is correct. Looking at the wrong answers: Answer B (0.0324) appears to use an incorrect denominator, possibly calculating 0.12×0.88100\sqrt{\frac{0.12 \times 0.88}{100}} instead of 300. Answer C (0.0544) might result from forgetting to take the square root entirely, giving just 0.12×0.88300\frac{0.12 \times 0.88}{300} without the square root. Answer D (0.1056) is simply the numerator p^(1−p^)=0.12×0.88\hat{p}(1-\hat{p}) = 0.12 \times 0.88 without dividing by nn or taking the square root. Remember this formula pattern: standard error always involves the square root of variance divided by sample size. For proportions, the variance component is p^(1−p^)\hat{p}(1-\hat{p}). Double-check that you've included both the division by nn and the square root—these are the most common calculation errors on biostatistics exams.

Question 14

A sociologist surveys residents about community satisfaction. In the first district, 156 out of 400 residents report high satisfaction. In the second district, 117 out of 300 residents report high satisfaction. Which district's sample proportion has the larger standard error?

  1. First district, because it has more satisfied residents in absolute numbers
  2. Second district, because it has a higher proportion of satisfied residents
  3. Second district, because it has a smaller sample size and similar proportion variance (correct answer)
  4. First district, because larger samples always have larger standard errors
  5. They have approximately equal standard errors due to the offsetting effects of sample size and proportion
Explanation: When comparing standard errors of sample proportions, you need to understand that the standard error measures how much variability you'd expect if you repeated the sampling process many times. The standard error for a sample proportion is calculated as SE=p(1−p)nSE = \sqrt{\frac{p(1-p)}{n}}, where p is the sample proportion and n is the sample size. Let's calculate both districts' values: First district: p₁ = 156/400 = 0.39, so SE1=0.39(0.61)400=0.0244SE_1 = \sqrt{\frac{0.39(0.61)}{400}} = 0.0244 Second district: p₂ = 117/300 = 0.39, so SE2=0.39(0.61)300=0.0282SE_2 = \sqrt{\frac{0.39(0.61)}{300}} = 0.0282 The second district has the larger standard error (0.0282 vs 0.0244). This occurs because both districts have nearly identical proportions (both 0.39), making the p(1-p) terms essentially equal, but the second district has a smaller sample size (300 vs 400). Since sample size appears in the denominator, smaller samples produce larger standard errors. Answer A is wrong because absolute numbers don't determine standard error—proportions and sample sizes do. Answer B incorrectly focuses on the proportion difference, but both proportions are virtually identical (0.39). Answer D reflects a fundamental misunderstanding—larger samples actually have smaller standard errors, not larger ones. Remember this key principle: when comparing standard errors of proportions, if the proportions are similar, the smaller sample will always have the larger standard error. Standard error decreases as sample size increases, reflecting greater precision with more data.

Question 15

A researcher collects a simple random sample of 64 observations from a population with variance σ2=144\sigma^2 = 144. If the sample size is increased to 256 while the population variance remains unchanged, by what factor does the standard error of the sample mean decrease?

  1. 2 (correct answer)
  2. 4
  3. 8
  4. 16
  5. The standard error increases by a factor of 2
Explanation: When you encounter questions about standard error and sample size changes, you're dealing with the fundamental relationship between sampling variability and sample size. The standard error of the sample mean is given by SE=σnSE = \frac{\sigma}{\sqrt{n}}, where σ\sigma is the population standard deviation and nn is the sample size. Let's calculate the standard errors for both scenarios. Initially, with n1=64n_1 = 64 and σ2=144\sigma^2 = 144 (so σ=12\sigma = 12), the standard error is SE1=1264=128=1.5SE_1 = \frac{12}{\sqrt{64}} = \frac{12}{8} = 1.5. After increasing the sample size to n2=256n_2 = 256, the new standard error becomes SE2=12256=1216=0.75SE_2 = \frac{12}{\sqrt{256}} = \frac{12}{16} = 0.75. The factor by which the standard error decreases is SE1SE2=1.50.75=2\frac{SE_1}{SE_2} = \frac{1.5}{0.75} = 2. Looking at the wrong answers: B (4) represents the ratio of the sample sizes (256/64), which is tempting but incorrect since standard error involves the square root of sample size. C (8) might come from incorrectly comparing the square roots directly (256−64=16−8=8\sqrt{256} - \sqrt{64} = 16 - 8 = 8), but this arithmetic difference isn't the multiplicative factor we need. D (16) represents the difference in the denominators (16 - 8), another arithmetic trap. The correct answer is A (2). Remember this key pattern: when sample size increases by a factor of k2k^2, the standard error decreases by a factor of kk. Since 256=42×64256 = 4^2 \times 64, the standard error decreases by a factor of 4/2 = 2.

Question 16

A clinical researcher measures biomarker levels in blood samples. The standard error of the sample mean is currently 1.2 units based on 36 samples. If the researcher increases the sample size to 81 samples while the population variability remains constant, what will be the new standard error?

  1. 0.53 units
  2. 0.80 units (correct answer)
  3. 1.08 units
  4. 1.35 units
  5. 2.70 units
Explanation: When you encounter questions about standard error and sample size changes, remember that standard error measures how much sample means vary around the true population mean, and it has an inverse relationship with sample size. The standard error of the mean follows the formula: SE=σnSE = \frac{\sigma}{\sqrt{n}}, where σ is the population standard deviation and n is the sample size. The key insight is that standard error is inversely proportional to the square root of sample size. Since the population variability remains constant, you can use the relationship: SE1SE2=n2n1\frac{SE_1}{SE_2} = \sqrt{\frac{n_2}{n_1}}. With the original standard error of 1.2 units from 36 samples, and the new sample size of 81 samples: 1.2SE2=8136=2.25=1.5\frac{1.2}{SE_2} = \sqrt{\frac{81}{36}} = \sqrt{2.25} = 1.5 Therefore: SE2=1.21.5=0.80SE_2 = \frac{1.2}{1.5} = 0.80 units. Looking at the wrong answers: A) 0.53 units results from incorrectly using the ratio of sample sizes (81/36 = 2.25) instead of the square root ratio. C) 1.08 units comes from using an incorrect square root calculation or applying the formula backwards. D) 1.35 units represents multiplying by the square root factor instead of dividing. Study tip: Remember that doubling sample size doesn't halve standard error—it reduces it by a factor of √2 ≈ 0.71. Always take the square root when dealing with sample size effects on standard error, and larger samples always mean smaller standard errors.

Question 17

Two independent samples are drawn: Sample A has nA=50n_A = 50 observations with sA=8s_A = 8, and Sample B has nB=200n_B = 200 observations with sB=12s_B = 12. Which sample has the smaller standard error of the sample mean?

  1. Sample A, because it has a smaller sample standard deviation
  2. Sample B, because it has a larger sample size despite the larger standard deviation (correct answer)
  3. Sample A, because the ratio sA/nAs_A/\sqrt{n_A} is smaller than sB/nBs_B/\sqrt{n_B}
  4. Sample B, because the larger sample size always results in smaller standard error
  5. They have equal standard errors because the samples are independent
Explanation: When you encounter questions about standard error of the sample mean, remember that this measures how much sample means vary around the true population mean. The standard error depends on both the population standard deviation (estimated by sample standard deviation) and sample size. The standard error of the sample mean is calculated as SE=snSE = \frac{s}{\sqrt{n}}, where ss is the sample standard deviation and nn is the sample size. Let's calculate both: Sample A: SEA=850=87.07=1.13SE_A = \frac{8}{\sqrt{50}} = \frac{8}{7.07} = 1.13 Sample B: SEB=12200=1214.14=0.85SE_B = \frac{12}{\sqrt{200}} = \frac{12}{14.14} = 0.85 Sample B has the smaller standard error (0.85 vs 1.13), making answer B correct. Despite having a larger standard deviation, Sample B's much larger sample size (200 vs 50) more than compensates for this difference. Answer A incorrectly focuses only on standard deviation while ignoring sample size. Answer C contains a calculation error—when you compute sA/nAs_A/\sqrt{n_A} versus sB/nBs_B/\sqrt{n_B}, Sample A's ratio is actually larger, not smaller. Answer D makes an overly broad claim that larger sample size "always" results in smaller standard error, which isn't true if the standard deviation differences are extreme enough. Study tip: Always calculate both standard errors when comparing samples. The square root relationship means sample size has diminishing returns—quadrupling sample size only halves the standard error—but large sample size differences often outweigh moderate standard deviation differences.

Question 18

A researcher compares two methods for estimating population variance. Method 1 uses the sample variance s12s_1^2 from n1=25n_1 = 25 observations. Method 2 pools data from two independent samples of sizes n2=16n_2 = 16 and n3=9n_3 = 9 to calculate a pooled variance sp2s_p^2. If all samples are drawn from the same population, which variance estimator has the smaller standard error?

  1. Method 1, because it uses a single sample which reduces bias
  2. Method 2, because pooling always improves precision regardless of sample sizes
  3. Method 2, because the effective sample size (25) equals Method 1's sample size
  4. Method 1, because single samples have lower standard errors than pooled estimates
  5. They have equal standard errors because the total sample information is the same (correct answer)
Explanation: When comparing variance estimators, you need to consider how the standard error of the variance estimate decreases as sample size increases. The standard error of a sample variance is approximately proportional to σ2/2n\sigma^2/\sqrt{2n}, where nn is the degrees of freedom. Method 1 uses n1=25n_1 = 25 observations, giving df1=24df_1 = 24 degrees of freedom. Method 2 pools two independent samples with n2=16n_2 = 16 and n3=9n_3 = 9, providing df2=(16−1)+(9−1)=23df_2 = (16-1) + (9-1) = 23 degrees of freedom. Since the standard error decreases with the square root of degrees of freedom, Method 1 has the smaller standard error because 24>23\sqrt{24} > \sqrt{23}. Choice A incorrectly focuses on bias rather than precision—both methods are unbiased estimators of population variance. Choice B makes the false claim that pooling always improves precision regardless of sample sizes; precision depends on total degrees of freedom, not the pooling process itself. Choice C incorrectly states that Method 2's "effective sample size" equals Method 1's—the total observations (25) don't determine precision, the degrees of freedom do. Choice D correctly identifies Method 1 as better but gives an incorrect general rule about single versus pooled samples. Study tip: For variance estimator precision, always count degrees of freedom, not just sample sizes. The estimator with more degrees of freedom will have the smaller standard error, making it more precise.

Question 19

In a longitudinal study, researchers measure biomarker levels at baseline and follow-up for n=36n = 36 subjects. The change scores di=xi,follow-up−xi,baselined_i = x_{i,\text{follow-up}} - x_{i,\text{baseline}} have mean dˉ=2.4\bar{d} = 2.4 and standard deviation sd=7.2s_d = 7.2. However, the baseline measurements have a known systematic bias of +0.8+0.8 units (baseline values are consistently 0.8 units too high). After correcting for this bias, what is the standard error of the corrected mean change?

  1. 7.22+0.8236=1.25\sqrt{\frac{7.2^2 + 0.8^2}{36}} = 1.25
  2. (7.2+0.8)236=1.33\sqrt{\frac{(7.2 + 0.8)^2}{36}} = 1.33
  3. 7.236=1.20\frac{7.2}{\sqrt{36}} = 1.20 (correct answer)
  4. (7.2−0.8)236=1.07\sqrt{\frac{(7.2 - 0.8)^2}{36}} = 1.07
Explanation: Correcting for systematic bias means adding a constant (+0.8) to each change score, since baseline values were 0.8 units too high. Adding a constant to all observations shifts the mean but does not change the variance or standard deviation. Therefore, sds_d remains 7.2, and SE = sdn=7.236=1.20\frac{s_d}{\sqrt{n}} = \frac{7.2}{\sqrt{36}} = 1.20. Choice A incorrectly adds the bias to the variance. Choice B adds the bias to the standard deviation. Choice D subtracts the bias from the standard deviation.

Question 20

A researcher estimates the prevalence of a disease using a two-stage sampling design. In stage 1, m=20m = 20 clusters are randomly selected from a population of clusters. In stage 2, n=15n = 15 individuals are randomly sampled from each selected cluster. The overall sample proportion is p^=0.12\hat{p} = 0.12. The between-cluster variance of proportions is estimated as sb2=0.0084s_b^2 = 0.0084, and the average within-cluster variance is estimated as sˉw2=0.0960\bar{s}_w^2 = 0.0960. What is the design-adjusted standard error of p^\hat{p}?

  1. sb2m+sˉw2mn=0.024\sqrt{\frac{s_b^2}{m} + \frac{\bar{s}_w^2}{mn}} = 0.024 (correct answer)
  2. p^(1−p^)mn=0.019\sqrt{\frac{\hat{p}(1-\hat{p})}{mn}} = 0.019
  3. sb2+sˉw2mn=0.018\sqrt{\frac{s_b^2 + \bar{s}_w^2}{mn}} = 0.018
  4. sb2m+sˉw2m=0.035\sqrt{\frac{s_b^2}{m} + \frac{\bar{s}_w^2}{m}} = 0.035
Explanation: For two-stage cluster sampling, the variance has two components: between-cluster variance (sb2m\frac{s_b^2}{m}) and within-cluster variance (sˉw2mn\frac{\bar{s}_w^2}{mn}). The standard error is 0.008420+0.096020×15=0.00042+0.00032=0.00074=0.024\sqrt{\frac{0.0084}{20} + \frac{0.0960}{20 \times 15}} = \sqrt{0.00042 + 0.00032} = \sqrt{0.00074} = 0.024. Choice B uses simple random sampling formula, ignoring clustering. Choice C incorrectly pools variances before dividing by total sample size. Choice D incorrectly divides within-cluster variance by mm instead of mnmn.