Quiz: Confidence Intervals: Estimating with Uncertainty
Test your understanding of confidence intervals, the t-distribution, and the interpretation of "95% confidence." Try to answer each question before revealing the answer.
1. A confidence interval is best described as:
(a) A range of values that contains 95% of the data (b) A range of plausible values for a population parameter (c) A range that has a 95% probability of containing the population mean (d) The distance between the sample mean and the population mean
Answer
**(b) A range of plausible values for a population parameter.** A confidence interval estimates a population parameter (like $\mu$ or $p$), not individual data values. It combines the point estimate with a margin of error to produce a range. Option (a) confuses the CI with a range of data values. Option (c) makes the common mistake of assigning probability to the parameter. Option (d) describes the error, not the interval.2. Which of the following is the correct interpretation of a 95% confidence interval?
(a) There is a 95% probability that the population parameter is in the interval (b) 95% of sample values fall within the interval (c) If we repeated the sampling process many times, about 95% of the resulting intervals would contain the true parameter (d) The interval is 95% accurate
Answer
**(c) If we repeated the sampling process many times, about 95% of the resulting intervals would contain the true parameter.** This is the repeated sampling interpretation — the defining feature of frequentist confidence intervals. The "95%" describes the long-run performance of the *method*, not the probability that any single interval is correct. The population parameter is fixed; the intervals are random.3. A researcher reports: "The 95% confidence interval for the mean commute time is (32, 42) minutes." Which statement is correct?
(a) 95% of commuters take between 32 and 42 minutes (b) The true mean commute time is definitely between 32 and 42 minutes (c) We are 95% confident that the true mean commute time is between 32 and 42 minutes (d) The sample mean is between 32 and 42 minutes
Answer
**(c) We are 95% confident that the true mean commute time is between 32 and 42 minutes.** Option (a) confuses the CI for the mean with a range of individual values. Option (b) expresses certainty — but the whole point of a CI is that there's a 5% chance the interval misses. Option (d) is technically true (the sample mean is the center of the interval, at 37), but it's not what the CI tells you — you already know the sample mean exactly.4. The sample mean for a study is $\bar{x} = 37$. Which is the correct sample mean for the CI (32, 42)?
(a) 35 (b) 37 (c) 38 (d) Cannot be determined from the interval alone
Answer
**(b) 37.** The sample mean is always the center of the confidence interval: $(32 + 42)/2 = 37$. The margin of error is 5 (the distance from the center to either endpoint).5. The t-distribution is used instead of the normal distribution when:
(a) The sample size is very large (b) The population standard deviation $\sigma$ is unknown and estimated by $s$ (c) The population is not normally distributed (d) The confidence level is 99% instead of 95%
Answer
**(b) The population standard deviation $\sigma$ is unknown and estimated by $s$.** When we replace $\sigma$ with $s$, we introduce extra uncertainty. The t-distribution's heavier tails account for this. As $n$ grows, $s$ becomes a better estimate of $\sigma$, and the t-distribution approaches the normal. The t-distribution isn't about sample size, population shape, or confidence level specifically — it's about estimating $\sigma$.6. As degrees of freedom increase, the t-distribution:
(a) Becomes more spread out (b) Approaches the standard normal distribution (c) Becomes more skewed (d) Produces narrower intervals regardless of sample size
Answer
**(b) Approaches the standard normal distribution.** More degrees of freedom means a larger sample, which means $s$ is a better estimate of $\sigma$, which means there's less extra uncertainty to account for. The t-distribution's heavy tails thin out and converge to the standard normal. At $df = \infty$, $t^* = z^*$ exactly.7. A 95% confidence interval for a population mean is (18.5, 23.5). The margin of error is:
(a) 2.0 (b) 2.5 (c) 5.0 (d) 21.0
Answer
**(b) 2.5.** The margin of error is half the width of the interval: $(23.5 - 18.5)/2 = 2.5$. Equivalently, it's the distance from the center ($\bar{x} = 21.0$) to either endpoint: $23.5 - 21.0 = 2.5$.8. A sample of 64 observations has $\bar{x} = 40$ and $s = 8$. The 95% confidence interval for $\mu$ is approximately:
(a) (38.0, 42.0) (b) (39.0, 41.0) (c) (32.0, 48.0) (d) (36.0, 44.0)
Answer
**(a) (38.0, 42.0).** $\text{SE} = s/\sqrt{n} = 8/\sqrt{64} = 8/8 = 1$. With $df = 63$, $t^* \approx 2.00$ (close to $z^* = 1.96$ for large $df$). Margin of error $\approx 2.00 \times 1 = 2.0$. So the CI is $40 \pm 2 = (38, 42)$.9. Which of the following would make a confidence interval narrower?
(a) Increasing the confidence level from 95% to 99% (b) Decreasing the sample size (c) Increasing the sample size (d) Increasing the sample standard deviation
Answer
**(c) Increasing the sample size.** Larger $n$ decreases the standard error ($s/\sqrt{n}$), which decreases the margin of error and narrows the interval. Options (a), (b), and (d) all make the interval wider: higher confidence requires a larger critical value, smaller $n$ increases SE, and larger $s$ increases SE.10. A researcher wants to estimate a proportion with a margin of error of ±0.03 at 95% confidence. Using the conservative estimate $\hat{p} = 0.5$, the minimum sample size is approximately:
(a) 267 (b) 752 (c) 1,068 (d) 1,111
Answer
**(c) 1,068.** $n = (z^*/E)^2 \cdot \hat{p}(1-\hat{p}) = (1.96/0.03)^2 \times 0.25 = (65.33)^2 \times 0.25 = 4268.44 \times 0.25 = 1067.1$. Rounding up: $n = 1{,}068$.11. In a random sample of 400 registered voters, 220 favor a ballot measure. The 95% confidence interval for the true proportion is approximately:
(a) (0.50, 0.60) (b) (0.48, 0.57) (c) (0.51, 0.59) (d) (0.55, 0.55)
Answer
**(a) (0.50, 0.60).** $\hat{p} = 220/400 = 0.55$. $\text{SE} = \sqrt{0.55 \times 0.45 / 400} = \sqrt{0.000619} = 0.0249$. Margin of error $= 1.96 \times 0.0249 = 0.0488 \approx 0.05$. CI: $0.55 \pm 0.05 = (0.50, 0.60)$.12. If a 95% CI for a proportion includes 0.50, this means:
(a) The sample proportion is 0.50 (b) We cannot be 95% confident that the true proportion differs from 0.50 (c) The population is evenly split (d) The sample size is too small
Answer
**(b) We cannot be 95% confident that the true proportion differs from 0.50.** If 0.50 is within the CI, it's a plausible value for the true proportion — we can't rule it out at the 95% level. This doesn't mean the proportion *is* 0.50 (option c), nor that the sample proportion is 0.50 (option a), nor that the sample is necessarily too small (option d — it might be an appropriate sample that simply reflects a proportion close to 0.50).13. To halve the margin of error while keeping the confidence level the same, you must:
(a) Double the sample size (b) Triple the sample size (c) Quadruple the sample size (d) Halve the standard deviation
Answer
**(c) Quadruple the sample size.** MOE $= t^* \times s / \sqrt{n}$. To halve MOE, you need to double $\sqrt{n}$, which means quadrupling $n$. This is the "diminishing returns" property from Chapter 11. Option (d) would also work mathematically, but you can't control $s$ by changing the sample size — it's a property of the population.14. A researcher computes a 95% CI of (12.3, 15.7) and a 90% CI of (12.6, 15.4) from the same data. A colleague asks: "Didn't you say 95% is more confident? Then why is the 95% interval bigger — doesn't that mean you know less?" How should the researcher respond?
Answer
The researcher should explain that the wider interval reflects *more* caution, not less knowledge. A 95% CI is wider because it needs to capture the true parameter more often (95% vs. 90% of the time). Think of it as casting a wider net to be more sure of catching the fish. The data are the same — what changes is how much certainty you demand. Both intervals use the same information. The 90% interval is narrower but accepts a higher risk (10% vs. 5%) of missing the parameter. The tradeoff between width and confidence is fundamental: you can't increase confidence without widening the interval (unless you get more data).15. Why do we use $z^*$ (from the normal distribution) for confidence intervals for proportions but $t^*$ (from the t-distribution) for confidence intervals for means?
Answer
For means, we must estimate the population standard deviation $\sigma$ using the sample standard deviation $s$, which introduces extra uncertainty. The t-distribution accounts for this extra variability with heavier tails. For proportions, the standard error formula $\sqrt{\hat{p}(1-\hat{p})/n}$ is a self-contained expression — it doesn't require a separate estimate of a population spread parameter. The variability of $\hat{p}$ is fully determined by $\hat{p}$ and $n$, so the standard normal distribution is appropriate. (Note: Some statisticians do advocate using $t^*$ for proportions as well, or using improved intervals like the Wilson interval. But the standard approach uses $z^*$.)16. A quality control engineer takes a sample of 25 widgets from a production line and finds the average weight is 12.3 grams with $s = 0.8$ grams. She constructs a 95% CI using $t^*$ with $df = 24$. Her colleague argues she should use $z^* = 1.96$ instead. Who is right?
(a) The engineer — she should use $t^*$ because $\sigma$ is unknown (b) The colleague — $z^*$ is always correct for 95% intervals (c) It doesn't matter — the results will be the same (d) Neither — she needs a larger sample before constructing a CI
Answer
**(a) The engineer — she should use $t^*$ because $\sigma$ is unknown.** With $n = 25$, $df = 24$, and $t^* = 2.064$ (compared to $z^* = 1.96$), there's a meaningful difference. Using $z^*$ would produce a narrower interval that doesn't adequately account for the uncertainty in estimating $\sigma$ with a small sample. The t-distribution is the correct choice here. For $n > 100$, the difference is negligible, but for $n = 25$, it matters.17. A study reports a 95% CI for the difference in average test scores between two teaching methods as (-2.4, +8.6). Which conclusion is best supported?
(a) The new method is definitely better (b) There is no difference between the methods (c) We cannot conclude at 95% confidence that there is a meaningful difference (d) The study was poorly designed
Answer
**(c) We cannot conclude at 95% confidence that there is a meaningful difference.** The interval includes 0 (which represents "no difference"), so we cannot rule out the possibility that the methods perform equally well. However, the interval also includes positive values up to 8.6, so the new method *might* be substantially better. The data are simply not conclusive enough to distinguish a real effect from random variation. This is not the same as concluding "no difference" (option b) — absence of evidence is not evidence of absence.18. Dr. Maya Chen constructs a 95% CI for mean blood pressure: (124.9, 131.7). Later, she realizes she forgot to include 3 additional observations. Adding them changes the interval to (125.2, 131.0). What happened?
(a) The interval got wider because she added data (b) The interval got narrower because larger $n$ reduces the margin of error (c) The interval shifted because the sample mean changed (d) Both (b) and (c)
Answer
**(d) Both (b) and (c).** Adding observations does two things: it increases $n$ (which decreases the standard error and narrows the interval), and it potentially changes $\bar{x}$ and $s$ (which can shift the center and affect the width). In this case, the interval both narrowed (from width 6.8 to width 5.8) and shifted slightly to the right, suggesting the new observations had values near or slightly above the original mean.19. A researcher wants to estimate a population mean with 95% confidence and a margin of error of ±5. She estimates $\sigma \approx 30$. How many observations does she need?
(a) 36 (b) 69 (c) 139 (d) 554
Answer
**(c) 139.** $n = (z^* \cdot \sigma / E)^2 = (1.96 \times 30 / 5)^2 = (11.76)^2 = 138.3$. Rounding up: $n = 139$.20. Suppose you compute 20 independent 95% confidence intervals. Approximately how many would you expect to not contain the true parameter?
(a) 0 (b) 1 (c) 5 (d) 19