Inference from Sample Statistics and Margin of Error Flashcards

A polling organization surveys 400 registered voters and finds that 52% support a ballot measure, with a margin of error of ±4.9% at a 95% confidence level. A rival organization surveys 900 voters and finds 49% support, with a margin of error of ±3.3%. Which of the following conclusions is best supported by these two surveys taken together?

Answer: The true level of support cannot be determined to be above or below 50% with confidence, since the confidence intervals from both surveys overlap and include 50%.

The first survey gives a 95% CI of approximately 47.1%–56.9%, and the second gives approximately 45.7%–52.3%. Both intervals contain 50%, meaning neither survey provides sufficient evidence that support is definitively above or below 50%. When confidence intervals overlap and include a critical threshold, no firm conclusion about which side of that threshold the true value falls on is warranted.

A researcher increases her sample size from 100 to 900 while keeping all other conditions the same. By what factor does the margin of error change?

Answer: It decreases by a factor of 3.

The margin of error is proportional to 1/√n. When n increases from 100 to 900, √n increases from 10 to 30 — a factor of 3. Therefore the margin of error decreases by a factor of 3, regardless of the population proportion. Answer C is a tempting distractor but the factor-of-3 reduction applies for any proportion, not just 0.5.

A 95% confidence interval for the mean number of hours students study per week is calculated as (12.4, 15.6). A student argues: 'There is a 95% probability that the true population mean is between 12.4 and 15.6.' Which of the following identifies the flaw in this reasoning?

Answer: The flaw is that once the interval is computed, the true mean is either inside it or not — the 95% refers to the long-run proportion of such intervals that would contain the true mean, not the probability for this specific interval.

The true population mean is a fixed (though unknown) constant, not a random variable. After computing a specific interval, it either contains the true mean or it doesn't — there is no probabilistic 'chance' about it. The correct interpretation is that if this procedure were repeated many times, 95% of the resulting intervals would contain the true mean. The student's statement confuses confidence level with posterior probability.

Two candidates are running for office. Candidate A receives 54% support and Candidate B receives 46% support in a poll of 600 likely voters. The margin of error is ±4% at a 95% confidence level. A news outlet reports: 'Candidate A has a statistically significant lead.' Is this claim supported?

Answer: No, because the margin of error applies to each candidate's estimate separately, and the margin of error on the difference between two proportions is approximately ±8%, making the lead not statistically significant.

When comparing two proportions from the same sample, the margin of error on the difference is roughly twice the individual margin of error (since both estimates carry uncertainty). Here, ±4% × 2 = ±8%. The observed difference is 54% − 46% = 8%, which exactly equals the combined margin of error — meaning the lead is not statistically significant at the 95% level. The news outlet's claim is not well-supported.

A study reports that a 90% confidence interval for the proportion of adults who exercise regularly is (0.38, 0.52). If the researchers had instead used a 99% confidence interval with the same sample data, which of the following would be true?

Answer: The interval would be wider, such as (0.33, 0.57), because a higher confidence level requires capturing a larger range of plausible values.

A higher confidence level requires a larger critical value (z* increases from 1.645 for 90% to 2.576 for 99%), which widens the interval. There is an inherent trade-off: greater confidence comes at the cost of precision. The sample size and point estimate (0.45) stay the same, so the interval doesn't shift — it expands symmetrically around the same center.

A school district wants to estimate the proportion of students who bring lunch from home with a margin of error no greater than 2% at a 95% confidence level. They have no prior estimate of the proportion. What is the minimum sample size required?

Answer: 2,401 students, using the formula n = (1.96)²(0.5)(0.5)/(0.02)²

With no prior estimate, the most conservative (largest) sample size uses p = 0.5, which maximizes p(1−p) = 0.25. The formula is n = (z*)²·p(1−p)/E², where z* = 1.96 for 95% confidence and E = 0.02. This gives n = (1.96)²(0.25)/(0.0004) = (3.8416)(0.25)/0.0004 = 0.9604/0.0004 = 2,401. Answer A uses the wrong z* (1.645 is for 90%), and Answer C uses the wrong margin of error (0.03 instead of 0.02).