Sample Size Selection - College Statistics

1. A researcher calculates the required sample size for a study and arrives at n = 541.1. For their final report, they must justify the sample size selected. Which of the following is the most appropriate action and justification?
   Answer: Use n = 542, because the sample size must be an integer and rounding down would not meet the desired margin of error. When you encounter calculated sample sizes in statistics, you're dealing with a fundamental principle: sample size formulas are designed to meet specific statistical requirements like margin of error or power, and compromising these requirements defeats the purpose of the calculation. The correct approach is D because sample sizes must be whole numbers (you can't survey 0.1 of a person), and when you round down from a calculated sample size, you're essentially saying you're willing to accept a larger margin of error or lower statistical power than originally planned. The calculation determined that 541.1 participants were needed to meet the study's requirements—rounding down to 541 means those requirements won't quite be met. A is incorrect because while standard rounding rules minimize cost, they don't preserve the statistical properties the researcher calculated. Cost savings that compromise study validity are poor research practice. B misunderstands how sample size affects margin of error. Even small changes in sample size can meaningfully impact statistical precision, especially when you're near the minimum required sample size. C introduces an irrelevant consideration. The preference for even numbers in stratified sampling depends on the specific design, not on general principles, and this reasoning ignores the statistical basis for the original calculation. Study tip: Always round calculated sample sizes up to the nearest whole number. This ensures your study maintains the statistical power and precision you planned for. Remember: it's better to slightly exceed your requirements than to fall short of them.
2. A market analyst is considering two plans to estimate the proportion of consumers who prefer a new product. Plan A requires a 90% confidence level with a 3% margin of error. Plan B requires a 95% confidence level with a 4% margin of error. Assuming no prior information is available for the population proportion, which of the following correctly compares the required sample sizes, $n_A$ and $n_B$?
   Answer: $n_A > n_B$ We must calculate the required sample size for each plan using the conservative estimate $p^*=0.5$. The formula is $n = (Z/E)^2 p^*(1-p^*)$. For Plan A (90% confidence), $Z=1.645$ and $E=0.03$. $n_A = (\frac{1.645}{0.03})^2(0.5)(0.5) \approx (54.83)^2(0.25) \approx 3006.7(0.25) \approx 751.7$, which rounds up to 752. For Plan B (95% confidence), $Z=1.96$ and $E=0.04$. $n_B = (\frac{1.96}{0.04})^2(0.5)(0.5) = (49)^2(0.25) = 2401(0.25) = 600.25$, which rounds up to 601. Since $752 > 601$, $n_A > n_B$. Even though Plan B has a higher confidence level, its larger margin of error results in a smaller required sample size.
3. An education researcher plans to estimate the proportion of college students who work full-time. They initially plan for a 95% confidence interval with a margin of error of 3%. After preliminary calculations, they decide to tighten the margin of error to 2%. Approximately what will be the ratio of the new required sample size ($n_{new}$ for E=2%) to the original required sample size ($n_{orig}$ for E=3%)?
   Answer: 2.25 The sample size $n$ for a proportion is given by $n = (Z/E)^2 p^*(1-p^*)$. The ratio of the new sample size to the original is $\frac{n_{new}}{n_{orig}} = \frac{(Z/E_{new})^2 p^*(1-p^*)}{(Z/E_{orig})^2 p^*(1-p^*)}$. The terms for Z and $p^*$ cancel out, leaving $\frac{n_{new}}{n_{orig}} = \frac{(1/E_{new})^2}{(1/E_{orig})^2} = \left(\frac{E_{orig}}{E_{new}}\right)^2$. Plugging in the values for E: $\left(\frac{0.03}{0.02}\right)^2 = (1.5)^2 = 2.25$. The new sample size will be 2.25 times larger than the original.
4. A polling organization wants to estimate the proportion of voters favoring candidate X in two different states, State A and State B. They require a 95% confidence interval with a margin of error of $\pm 3\%$ for both polls. In State A, the race is expected to be very close (around 50/50). In State B, candidate X has strong historical support around 70%. How will the required sample size $n_A$ for State A compare to $n_B$ for State B?
   Answer: $n_A > n_B$ The sample size formula for a proportion is $n = (Z/E)^2 p^*(1-p^*)$. In this problem, Z (for 95% confidence) and E (0.03) are the same for both states. The only difference is the planning value $p^*$. The term $p^*(1-p^*)$ represents the variance and is maximized when $p^*=0.5$. For State A, $p^*=0.5$, so $p^*(1-p^*) = 0.5 \times 0.5 = 0.25$. For State B, $p^*=0.7$, so $p^*(1-p^*) = 0.7 \times 0.3 = 0.21$. Since $0.25 > 0.21$, the required sample size for State A will be larger than for State B.
5. A medical researcher is planning a study to compare the mean recovery time for two drugs, A and B. They want to estimate the difference in means with a margin of error of 2 days and 95% confidence. The standard deviation of recovery time is assumed to be 5 days for both drug groups. If the sample sizes for the two groups are to be equal ($n_A = n_B = n$), what is the minimum number of patients required for each group?
   Answer: 49 For a two-sample confidence interval for the difference in means with equal sample sizes $n$, the margin of error is $E = Z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n} + \frac{\sigma_2^2}{n}}$. We are given $E=2$, $Z=1.96$ (for 95% confidence), and $\sigma_1=\sigma_2=5$. The formula becomes $2 = 1.96 \sqrt{\frac{5^2}{n} + \frac{5^2}{n}} = 1.96 \sqrt{\frac{50}{n}}$. To solve for $n$, we rearrange: $\frac{2}{1.96} = \sqrt{\frac{50}{n}}$. Squaring both sides gives $(\frac{2}{1.96})^2 = \frac{50}{n}$, which simplifies to $1.0412 \approx \frac{50}{n}$. Solving for $n$ gives $n \approx \frac{50}{1.0412} \approx 48.019$. We must round up to the next integer, so $n=49$ patients are required for each group.
6. A research team is planning a study to estimate the average household income in a city. Their goal is a 95% confidence interval with a margin of error of $\pm \$1,500$. Which of the following pieces of information is LEAST likely to be required to calculate the necessary sample size?
   Answer: The total number of households in the city. When you encounter sample size calculations for confidence intervals, focus on the core formula: $n = \left(\frac{z \cdot \sigma}{E}\right)^2$, where $n$ is sample size, $z$ is the critical value, $\sigma$ is the population standard deviation, and $E$ is the margin of error. Looking at this formula reveals what's actually needed. You need the margin of error ($1,500), the critical value (1.96 for 95% confidence), and an estimate of variability (standard deviation). Notice that the total population size doesn't appear in this formula at all. Let's examine each choice: Choice A ($1,500 margin of error) is essential—it's the $E$ in our formula. Choice B (variability estimate) is critical because without knowing how spread out household incomes are, you can't determine how many observations you need for reliable estimation. Choice C (the z-score of 1.96 for 95% confidence) directly affects the required sample size through the numerator of our formula. Choice D (total households in the city) is the least necessary. Sample size calculations for confidence intervals typically assume sampling from an infinite population. The population size only matters when you're sampling a large fraction of a finite population (usually over 5%), which would require a finite population correction factor. For a city's household income study, you'd likely need only a small fraction of all households, making the total population size irrelevant. Study tip: Remember that confidence interval sample sizes depend on precision goals (margin of error), confidence level, and variability—not population size, unless you're sampling most of the population.
7. A company wants to estimate the proportion of employees who are satisfied with their benefits package. A preliminary study of 50 employees found 35 were satisfied. Using this information as a planning value, what is the minimum sample size needed to create a 90% confidence interval with a width of at most 0.10?
   Answer: 228 First, determine the planning value $p^*$ from the preliminary study: $p^* = 35/50 = 0.70$. Second, determine the margin of error $E$ from the desired width. Width = 2E, so $E = 0.10 / 2 = 0.05$. For a 90% confidence level, the critical value $Z$ is 1.645. Now, use the sample size formula: $n = \left(\frac{Z}{E}\right)^2 p^*(1-p^*) = \left(\frac{1.645}{0.05}\right)^2 (0.70)(0.30) = (32.9)^2 (0.21) = 1082.41 \times 0.21 \approx 227.3$. We must round up to the next whole number, so the required sample size is 228.
8. A sample size of $n$ is calculated to estimate a population proportion with a margin of error $E$ at a 95% confidence level, using the conservative planning value $p^*=0.5$. If the researcher decides to aim for half the original margin of error ($E/2$) but also lowers the confidence level to 90%, what will be the approximate new required sample size, $n_{new}$?
   Answer: $n_{new} \approx 2.8n$ When you encounter sample size problems involving changes to both margin of error and confidence level, you need to understand how each component affects the required sample size formula: $n = \frac{(z_{\alpha/2})^2 \cdot p^*(1-p^*)}{E^2}$ Let's analyze how each change affects the sample size. First, halving the margin of error (from $E$ to $E/2$) means the denominator becomes $(E/2)^2 = E^2/4$. Since $E^2$ is now divided by 4, this multiplies the sample size by 4. Second, lowering the confidence level from 95% to 90% changes the critical value from $z_{0.025} = 1.96$ to $z_{0.05} = 1.645$. The ratio of squared critical values is $\frac{(1.645)^2}{(1.96)^2} = \frac{2.706}{3.842} \approx 0.704$. Combining both effects: $n_{new} = n \times 4 \times 0.704 \approx 2.8n$, confirming answer D. Here's why the other options fail: Option A ($0.7n$) only accounts for the confidence level change while ignoring the margin of error reduction. Option B ($1.4n$) incorrectly assumes the margin of error change doubles rather than quadruples the sample size. Option C ($4.0n$) considers only the margin of error change while ignoring the confidence level reduction. Study tip: Remember that margin of error has a squared relationship with sample size (halving $E$ requires 4× the sample), while confidence level changes affect the critical value. Always calculate both effects separately, then multiply them together.
9. A quality control engineer wants to estimate the mean weight of a product to within $\pm 0.5$ grams with 99% confidence. A pilot study involving 30 items yielded a sample standard deviation of 2.5 grams. The engineer will use these initial 30 items as part of the final sample. How many additional items must be sampled to meet the desired precision?
   Answer: 136 First, calculate the total sample size needed. The formula for the sample size to estimate a mean is $n = \left(\frac{Z_{\alpha/2} \sigma}{E}\right)^2$. For 99% confidence, $Z_{\alpha/2} = 2.576$. The margin of error $E$ is 0.5, and we use the pilot study's standard deviation as an estimate for $\sigma$, so $\sigma \approx 2.5$. The required total sample size is $n = \left(\frac{2.576 \times 2.5}{0.5}\right)^2 = (12.88)^2 = 165.8944$. We must round up to 166. The question asks for the additional items needed. Since 30 items have already been sampled, the additional number is $166 - 30 = 136$.
10. A market researcher wants to estimate the mean household income in a certain area. They want the estimate to be within 2% of the true mean, with 95% confidence. A pilot study suggests the mean income is about $60,000 with a standard deviation of $15,000. What is the minimum required sample size?
    Answer: 601 First, calculate the absolute margin of error $E$. The desired margin of error is 2% of the mean income, so $E = 0.02 \times \$60,000 = \$1,200$. For 95% confidence, $Z=1.96$. The estimated standard deviation $\sigma$ is $15,000. Use the sample size formula for a mean: $n = \left(\frac{Z_{\alpha/2} \sigma}{E}\right)^2 = \left(\frac{1.96 \times 15000}{1200}\right)^2 = (1.96 \times 12.5)^2 = (24.5)^2 = 600.25$. We must round up to the next integer, so the minimum sample size is 601.
11. A researcher calculated a required sample size of $n=500$ to estimate a population mean. Which of the following independent changes to the study's parameters would result in the largest new required sample size?
    Answer: Increasing the confidence level from 95% to 99%. Let's analyze the multiplicative factor on the original sample size for each change, based on $n \propto (Z\sigma/E)^2$. A) Change from 95% (Z=1.96) to 99% (Z=2.576). Factor = $(2.576/1.96)^2 \approx (1.314)^2 \approx 1.73$. The new $n \approx 500 \times 1.73 = 865$. B) New E is $0.85E$. Factor = $(1/0.85)^2 \approx (1.176)^2 \approx 1.38$. The new $n \approx 500 \times 1.38 = 690$. C) New $\sigma$ is $1.15\sigma$. Factor = $(1.15)^2 = 1.3225$. The new $n \approx 500 \times 1.3225 = 661$. D) New E is $0.9E$ and new $\sigma$ is $1.1\sigma$. Factor = $(1.1/0.9)^2 \approx (1.222)^2 \approx 1.49$. The new $n \approx 500 \times 1.49 = 745$. Comparing the factors (1.73, 1.38, 1.32, 1.49), the change in confidence level has the largest impact.
12. A sociologist wants to estimate the mean number of hours of television watched per week by high school students, with a margin of error of 1 hour and 95% confidence. Previous studies indicate the weekly hours watched can range from 0 to 28 hours. Using the range rule of thumb ($\sigma \approx \text{Range}/4$) to estimate the standard deviation, what is the minimum required sample size?
    Answer: 189 First, estimate the standard deviation $\sigma$ using the range rule of thumb. The range is $28 - 0 = 28$ hours. The estimate is $\sigma \approx \frac{\text{Range}}{4} = \frac{28}{4} = 7$. Next, use the sample size formula for a mean: $n = \left(\frac{Z_{\alpha/2} \sigma}{E}\right)^2$. For 95% confidence, $Z=1.96$. The desired margin of error $E$ is 1. Plugging in the values: $n = \left(\frac{1.96 \times 7}{1}\right)^2 = (13.72)^2 = 188.2384$. The sample size must be an integer, so we round up to 189.
13. A manufacturer wants to estimate the mean lifespan of a new type of battery. They want to be 95% confident that their estimate is within 10 hours of the true mean. An initial test on a small batch of 20 batteries shows a sample standard deviation of 45 hours. However, due to budget constraints, the final sample size cannot exceed 75. What is the most direct consequence of this sample size limitation?
    Answer: The achieved margin of error will be greater than the desired 10 hours. First, let's calculate the required sample size for the desired parameters: $n = (Z\sigma/E)^2 = (1.96 \times 45 / 10)^2 = (8.82)^2 \approx 77.8$, which rounds up to 78. The study requires a sample of at least 78 batteries to achieve a 10-hour margin of error at 95% confidence. Since the budget limits the sample size to 75, which is less than 78, the study cannot achieve its goal. With a smaller sample size, the margin of error will be larger than the desired 10 hours. Lowering the confidence level is one way to compensate, but it's not a direct consequence of the sample size limit itself. The sample size does not introduce bias. The initial standard deviation estimate is the best information available.
14. An auditor wants to determine the percentage of invoices containing errors. They need to be 99% confident that their estimate is within $\pm 5$ percentage points of the true value. Last year's audit found errors in 15% of invoices. What is the minimum number of invoices they must sample?
    Answer: 339 The formula for sample size for a proportion is $n = \left(\frac{Z_{\alpha/2}}{E}\right)^2 p^*(1-p^*)$. The confidence level is 99%, so $Z=2.576$. The margin of error $E$ is 5 percentage points, or 0.05. The prior estimate of the proportion $p^*$ is 0.15. Calculation: $n = \left(\frac{2.576}{0.05}\right)^2 (0.15)(1-0.15) = (51.52)^2 (0.15)(0.85) \approx 2654.31 \times 0.1275 \approx 338.42$. Since the sample size must be a whole number, we round up to 339.
15. An environmental scientist plans to estimate the mean concentration of a pollutant. Based on a standard deviation estimate of 6 ppm, they calculate a required sample size of 100. After collecting 100 samples, they find the sample standard deviation is actually 9 ppm. What is the direct consequence for the margin of error of their confidence interval, compared to what was originally planned?
    Answer: It will be 50% larger than planned. The margin of error for a mean is calculated as $E = Z \frac{\sigma}{\sqrt{n}}$. In this formula, the margin of error $E$ is directly proportional to the standard deviation $\sigma$. The planned standard deviation was 6 ppm, but the actual standard deviation was 9 ppm. The ratio of the actual to planned standard deviation is $9 / 6 = 1.5$. Therefore, the actual margin of error will be 1.5 times the planned margin of error, which is an increase of 50%.
16. A research firm has a budget of $20,000 for a survey. There is a fixed setup cost of $2,000, and the cost per respondent is $30. The firm wants to estimate a population proportion with 95% confidence and needs to determine the tightest margin of error achievable within this budget. Assuming they use the most conservative value for the population proportion, what is the approximate margin of error?
    Answer: 4.0% First, determine the maximum sample size $n$ allowed by the budget. The available budget for respondents is $\$20,000 - \$2,000 = \$18,000$. At $30 per respondent, the sample size is $n = \frac{18000}{30} = 600$. The formula for margin of error for a proportion is $E = Z_{\alpha/2} \sqrt{\frac{p^*(1-p^*)}{n}}$. For 95% confidence, $Z=1.96$. The most conservative estimate for the proportion is $p^*=0.5$. Plugging in the values: $E = 1.96 \sqrt{\frac{0.5(1-0.5)}{600}} = 1.96 \sqrt{\frac{0.25}{600}} \approx 1.96 \times 0.02041 \approx 0.0400$. This corresponds to a margin of error of 4.0%.
17. A team is planning a survey to estimate the proportion of likely voters who support a ballot measure. They have a fixed budget, which limits their maximum sample size. Which of the following would allow them to achieve a smaller margin of error without increasing their sample size?
    Answer: Decreasing the confidence level for the interval from 99% to 95%. The formula for the margin of error for a proportion is $E = Z_{\alpha/2} \sqrt{\frac{p^*(1-p^*)}{n}}$. To decrease $E$ with a fixed $n$, one must decrease the other terms. A) Using $p^*=0.5$ yields the largest possible value for $p^*(1-p^*)$, which would increase, not decrease, the margin of error. B) Decreasing the confidence level from 99% ($Z=2.576$) to 95% ($Z=1.96$) reduces the Z-score, which directly reduces the margin of error. C) A more heterogeneous population with respect to the measured trait would likely have a proportion closer to 0.5, increasing variability and thus increasing the margin of error. D) Estimating for a sub-group means the effective sample size is smaller than the total sample, which would increase the margin of error for that specific estimate.
18. A political pollster conducted a preliminary survey of 100 voters and found that 60% intended to vote for a certain candidate. For a follow-up poll, the pollster wants to be 95% confident that the estimated proportion is within 2 percentage points of the true proportion. Based on the preliminary results, what is the minimum sample size required for the follow-up poll?
    Answer: 2,305 The formula for the sample size to estimate a proportion is $n = \left(\frac{Z_{\alpha/2}}{E}\right)^2 p^*(1-p^*)$. For a 95% confidence level, the critical value $Z_{\alpha/2}$ is 1.96. The desired margin of error $E$ is 0.02. The planning value $p^*$ from the preliminary survey is 0.60. Plugging these values in: $n = \left(\frac{1.96}{0.02}\right)^2 (0.60)(1-0.60) = (98)^2 (0.60)(0.40) = 9604 \times 0.24 = 2304.96$. Since the sample size must be a whole number, we must round up to the next integer, which is 2305.
19. A university administrator wants to estimate the mean student loan debt for graduates. They require the total width of a 95% confidence interval to be no more than $2,000. If the standard deviation of loan debt is estimated to be $8,000, what is the minimum number of graduates that must be surveyed?
    Answer: 246 First, determine the margin of error $E$. The total width of a confidence interval is twice the margin of error (Width = 2E). Given the width is $2,000, the margin of error is $E = \$2,000 / 2 = \$1,000$. For a 95% confidence interval, the z-score is 1.96. The estimated standard deviation $\sigma$ is $8,000. Using the sample size formula for a mean: $n = \left(\frac{Z_{\alpha/2} \sigma}{E}\right)^2 = \left(\frac{1.96 \times 8000}{1000}\right)^2 = (15.68)^2 = 245.86$. Since the sample size must be an integer, it must be rounded up to 246.

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