Chapter 13. 6 multiple choice and 8 worked questions, all with solutions.
These are the Checkpoint 2 practice questions for this topic. Work them with the answers covered.
1. A sample of 25 invoices has $s = 7.5$. The standard error of the mean invoice value is
(A) 7.5000 (B) 0.3000 (C) 1.5309 (D) 1.5000
2. With $\bar{x} = 54.8$ and $t_{0.025,24} = 2.0639$, the 95% interval for the mean invoice value runs
(A) (39.3208, 70.2792) (B) (53.3000, 56.3000) (C) (51.8601, 57.7399) (D) (51.7042, 57.8958)
3. A market report widens its interval from 90% to 99% confidence. The interval becomes
(A) wider (B) narrower (C) unchanged (D) narrower then wider
4. An auditor wants a narrower interval for mean order value. Which actually achieves it?
(A) raising the confidence level (B) sampling more orders (C) reporting the variance instead (D) sampling fewer but larger orders
5. A consultant reports a 95% interval for mean wait time of $(4.2, 6.8)$ minutes. The correct reading is
(A) 95% of intervals built this way contain the true mean (B) the true mean is in this interval with probability 0.95 (C) 95% of waits fall between 4.2 and 6.8 (D) the estimate is accurate to within 95%
6. A firm wants mean invoice value estimated within $\pm$1$ at 95\% confidence, believing $\sigma \approx 7.5$. It must sample
(A) 153 (B) 55 (C) 216 (D) 217
1. A sample of 25 invoices gives $\bar{x} = 54.8$ and $s = 7.5$. (a) Compute the standard error. (b) Build a 95% confidence interval using $t_{0.025,24} = 2.0639$. (c) State precisely what ‘95% confident’ refers to.
2. Using $\sigma \approx 7.5$, find the sample size needed to estimate a mean within $\pm 1.0$, then within $\pm 0.5$, both at 95% confidence. Comment.
3. Explain why a point estimate reported without a margin of error is of little use.
4. A sample of 16 gives $\bar{x} = 50$, $s = 8$. Build a 95% interval using $t_{0.025,15} = 2.1314$, and say how it would change at 99%.
5. Name the three ways to reduce the margin of error and say which is the only honest one.
6. Explain why the sample size formula always rounds up.
7. A firm wants its mean order value estimated to within $3 at 99\% confidence, and believes $\sigma \approx 25$. (a) Find the required sample size. (b) The budget allows 300 orders. What margin does that buy, at the same confidence?
8. Explain why a 95% confidence interval built from a skewed population at $n = 36$ may contain $\mu$ less than 95% of the time, and what fixes it.
1. D — $s/\sqrt{n}$. The first distractor reports $s$ itself.
2. D — The first distractor forgets to divide by $\sqrt{n}$; the last uses $z$ where $\sigma$ was estimated.
3. A — More confidence is bought with more vagueness.
4. B — Only $n$, the confidence level and $\sigma$ move the width, and only $n$ improves precision without giving something up.
5. A — The true mean is fixed; the interval is what was random.
6. D — $(1.96(7.5)/1)^2 = 216.08$, rounded up. The first uses the 90% value; the last forgets to round up.
1. (a) $7.5/5 = 1.5$. (b) $54.8 \pm 2.0639(1.5) = (51.7042, 57.8958)$. (c) The procedure, not this interval: repeat the exercise many times and 95% of the intervals built this way contain $\mu$.
2. $(1.96(7.5)/1.0)^2 = 216.08 \Rightarrow 217$; $(1.96(7.5)/0.5)^2 = 864.33 \Rightarrow 865$. Halving the margin quadruples the sample — precision costs at the rate $\sqrt{n}$.
3. $\mathrm{P}(\bar{X} = \mu) = 0$, so the point estimate is certainly not exactly right. More practically, it hides how much the figure would move with a different sample: $n = 10$ and $n = 10{,}000$ can give the same $\bar{x}$ and do not deserve the same confidence.
4. SE $= 2$, so $50 \pm 2.1314(2) = (45.7371, 54.2629)$. At 99% the critical value rises to 2.9467, giving $\pm 5.89$ — a wider interval for the same data.
5. Raise $n$; accept lower confidence; reduce $\sigma$. Raising $n$ is the only one that buys genuine precision — lowering confidence just means being wrong more often, and reducing $\sigma$ means changing what you measure or how.
6. $n$ must be a whole number, and rounding down would leave the margin larger than the target, failing the requirement the calculation was made to meet.
7. (a) $n = (2.5758 \times 25/3)^2 = 460.75$, so 461. (b) $E = 2.5758(25/\sqrt{300}) = 3.72$ — so 300 orders buys about $3.72, short of the $3 target.
8. The interval’s coverage relies on $\bar{X}$ being normal. At $n = 36$ a strongly skewed population has not finished converging, so the $t$ interval is slightly too narrow and under-covers — in simulation, about 92% rather than 95%. Raising $n$ fixes it, because the CLT approximation improves with $\sqrt{n}$.