Chapter 14 — the largest set, by some way. 24 multiple choice and 25 worked questions, all with solutions.
These are the Checkpoint 2 practice questions for this topic. Work them with the answers covered.
1. A transport authority asks whether mean journey time has changed from last year’s 45 minutes. The appropriate test is
(A) two-tailed (B) one-tailed, left (C) one-tailed, right (D) chosen after seeing the sample mean
2. A supplier guarantees mean battery life of at least 600 hours. As the buyer testing that claim, your alternative hypothesis is
(A) $\mu > 600$ (B) $\mu \neq 600$ (C) $\mu < 600$ (D) $\mu = 600$
3. A sample of 25 invoices gives $\bar{x} = 54.8$, $s = 7.5$, against a planning figure of $50. The $t$ statistic is
(A) 0.6400 (B) 6.4000 (C) 3.2000 (D) 4.8000
4. With that $t = 3.2$ on 24 degrees of freedom and $t_{0.025,24} = 2.0639$, at $\alpha = 0.05$ the auditor should
(A) reject the planning figure (B) accept the planning figure (C) fail to reject, as $3.2 < 2.0639$ (D) report that the test is inconclusive
5. A trial of a new process reports $p = 0.03$. The correct interpretation is
(A) data this extreme would arise 3% of the time if the process were unchanged (B) there is a 3% chance the process is unchanged (C) the improvement is small (D) the new process works 97% of the time
6. A study of 40{,}000 customers finds a 0.2-point difference significant at $p = 0.008$. The right response is
(A) accept the finding, since $p$ is small (B) reject the finding, since the effect is small (C) repeat the study with fewer customers (D) ask for the confidence interval, since a huge sample makes trivial effects significant
7. A bakery’s filling machine is set to 800 g. A sample of 16 loaves gives $\bar{x} = 793$ with $s = 12$. The $t$ statistic is
(A) -0.5833 (B) -7.0000 (C) -2.3333 (D) -9.3333
8. For that bakery test at $\alpha = 0.05$ two-tailed, with $t_{0.025,15} = 2.1314$, the conclusion is
(A) reject — the machine is off target (B) fail to reject — the machine is fine (C) accept that the mean is 800 g (D) inconclusive without the $p$-value
9. A clinic tests whether mean waiting time exceeds 20 minutes and fails to reject at $n = 18$. A later study with $n = 400$ rejects. The likeliest explanation is
(A) the first study was wrong (B) the first study lacked the power to detect a real effect (C) the effect appeared between the two studies (D) the second study used a larger $\alpha$
10. A chain claims mean wait is 25 minutes. A sample of 36 branches gives $\bar{x} = 23.6$, $s = 4.8$. The $t$ statistic is
(A) -0.2917 (B) 1.7500 (C) -3.5000 (D) -1.7500
11. For that chain, two-tailed at $\alpha = 0.05$ with $t_{0.025,35} = 2.0301$, you
(A) reject the claim (B) accept the claim (C) fail to reject the claim (D) cannot decide without the $p$-value
12. For that same chain the two-tailed $p$ is 0.0889 and the one-tailed $p$ is 0.0444. This shows
(A) the one-tailed test is more accurate (B) the tail must be fixed by the question, not chosen afterwards (C) the two-tailed test is wrong here (D) the sample was too small
13. A maker claims mean fuel economy is 40 mpg. 25 cars give $\bar{x} = 41.2$, $s = 3.0$. The $t$ statistic is
(A) 1.2000 (B) 0.4000 (C) 2.0000 (D) 1.0000
14. For those cars, two-tailed at $\alpha = 0.05$ with $t_{0.025,24} = 2.0639$, you
(A) reject (B) accept the claim (C) fail to reject (D) reject at 1% instead
15. A supplier guarantees bulbs last at least 1200 hours. 20 bulbs give $\bar{x} = 1180$, $s = 95$. The appropriate alternative is
(A) $\mu > 1200$ (B) $\mu < 1200$ (C) $\mu \neq 1200$ (D) $\mu = 1200$
16. For those bulbs $t = -0.9415$ with $t_{0.05,19} = 1.7291$. You
(A) reject the guarantee (B) fail to reject the guarantee (C) prove the guarantee holds (D) need a two-tailed test
17. A filling line targets 250 g. QC samples 16 and gets $\bar{x} = 248.5$, $s = 3.2$. The test should be
(A) one-tailed left, since the sample is low (B) one-tailed right (C) two-tailed, because drift either way is a fault (D) one-tailed, chosen from the sample mean
18. For that filling line $t = -1.8750$ against $t_{0.025,15} = 2.1314$. The line is
(A) shown to be off target (B) proven on target (C) untestable at this $n$ (D) not shown to be off target
19. Twelve students improve by $\bar{x} = 74.5$ against a prior mean of 70, $s = 6$. With $t = 2.5981$ and $t_{0.025,11} = 2.2010$, you
(A) fail to reject (B) accept the prior mean (C) reject — scores have changed (D) need a larger sample first
20. Cereal boxes are filled with known $\sigma = 6$ g. 64 boxes average 498 g against a target of 500. The right test uses
(A) $z$, because $\sigma$ is known (B) $t$, because the sample is a sample (C) $t$ with 63 degrees of freedom (D) $z$, because $n > 30$
21. For those cereal boxes $z = -2.6667$. At $\alpha = 0.01$ two-tailed (critical 2.5758) you
(A) fail to reject the target (B) reject the target (C) accept the target (D) switch to a $t$ test
22. A plant reports mean cost per unit of $18.7 against a budget of $19.5, from 40 units with $s = 2.4$. Here $t = -2.1082$ and the two-tailed $p$ is 0.0415. At $\alpha = 0.05$ the plant is
(A) exactly on budget (B) over budget (C) not distinguishable from budget (D) below budget by more than noise explains
23. A manager runs 20 independent tests at $\alpha = 0.05$ on processes that are all working correctly, then reports the one significant result. The problem is
(A) about one false positive is expected, so the finding is unremarkable (B) the tests should have used $\alpha = 0.01$ (C) 20 tests is too few to be reliable (D) the significant process really is faulty
24. A hospital sets $\alpha = 0.10$ rather than 0.05 when screening for a dangerous fault. The reasonable justification is
(A) larger $\alpha$ gives more reliable results (B) the sample size is small (C) it makes rejection less likely (D) missing a real fault costs far more than a false alarm
1. A sample of 25 gives $\bar{x} = 54.8$, $s = 7.5$. A planning figure assumes the mean is 50. (a) State $H_0$, $H_a$ and the tail. (b) Compute the test statistic. (c) Test at $\alpha = 0.05$ and conclude in words. (d) Show the interval from the estimation section agrees.
2. List the eight steps of a hypothesis test, and say which happen before you see the data.
3. For each, state $H_0$, $H_a$ and the tail: (a) a bus company advertises 45 minutes and commuters suspect longer; (b) last year’s mean was 72 and the department asks whether this year differs; (c) a machine is targeted at 500 g and QC checks for drift.
4. Explain why choosing a one-tailed test after seeing the sample mean is illegitimate.
5. Twenty independent tests are run at $\alpha = 0.05$ on data with no real effect. (a) How many ‘significant’ results would you expect? (b) What is it called if only those are reported?
6. A study of 40{,}000 people reports $p = 0.008$ for a difference of 0.2 points. What should you ask for, and why?
7. Define Type I and Type II error, state which probability you choose, and say how the other is influenced.
8. Explain why a 95% confidence interval tells you more than a two-tailed test at $\alpha = 0.05$.
9. A commute study has $n = 36$, $\bar{x} = 28.4$ minutes, $s = 6$. (a) Build a 95% interval using $t_{0.025,35} = 2.0301$. (b) Test $H_0: \mu = 25$ at $\alpha = 0.05$. (c) Show the two agree.
10. A supplier guarantees mean battery life of at least 500 hours. A sample of 25 gives $\bar{x} = 486$, $s = 30$. Test at $\alpha = 0.05$, using $t_{0.05,24} = 1.7109$, and explain why the critical value is not 2.0639.
11. A chain claims mean wait is 25 minutes. A sample of 36 branches gives $\bar{x} = 23.6$, $s = 4.8$. (a) State $H_0$, $H_a$ and the tail. (b) Compute $t$. (c) Test at $\alpha = 0.05$ with $t_{0.025,35} = 2.0301$, concluding in words about waits.
12. For that same chain, the 95% interval is $(21.9759, 25.2241)$. Show it agrees with your test, and state the general rule.
13. For that chain the two-tailed $p$ is 0.0889 and the one-tailed $p$ is 0.0444. A colleague proposes reporting the one-tailed result because it is significant. Respond.
14. A maker claims 40 mpg. 25 cars give $\bar{x} = 41.2$, $s = 3.0$. (a) Compute $t$. (b) Test two-tailed at $\alpha = 0.05$ ($t_{0.025,24} = 2.0639$). (c) Would the verdict change at $\alpha = 0.10$ ($t_{0.05,24} = 1.7109$)?
15. A supplier guarantees at least 1200 hours. 20 bulbs give $\bar{x} = 1180$, $s = 95$, giving $t = -0.9415$ against $t_{0.05,19} = 1.7291$. State the conclusion carefully, and say what it does NOT establish.
16. A filling line targets 250 g. 16 items give $\bar{x} = 248.5$, $s = 3.2$. (a) Why is this two-tailed? (b) Compute $t$ and test at $\alpha = 0.05$ ($t_{0.025,15} = 2.1314$).
17. Cereal boxes have known $\sigma = 6$ g. 64 boxes average 498 g against a 500 g target. (a) Should you use $z$ or $t$, and why? (b) Compute the statistic. (c) Test at $\alpha = 0.01$, critical 2.5758.
18. Write the $p$-value 0.0415 from the cost-per-unit test as a sentence a plant manager would understand.
19. State the eight steps of a hypothesis test and explain why steps 3 and 4 must come before step 5.
20. Explain the difference between ‘fail to reject $H_0$’ and ‘accept $H_0$’, and why the second is never written.
21. A researcher tests at $\alpha = 0.05$, obtains $p = 0.06$, and writes that the result is ‘marginally significant, so the effect is probably real’. Critique this in two or three sentences.
22. Define Type I and Type II error for a drug trial where $H_0$ is ‘the drug has no effect’, and say which error a regulator should fear more.
23. A study of 50{,}000 people reports that a 0.1-point difference in a score is significant at $p = 0.002$. Explain what is and is not established.
24. Two analysts test the same claim on the same data. One uses $\alpha = 0.05$ and rejects; the other uses $\alpha = 0.01$ and does not. Is either wrong?
25. A quality manager wants to detect a shift of half a standard deviation with reasonable confidence. A pilot at $n = 10$ fails to reject. Advise.
1. A — ‘Changed’ names no direction, and the tail is never chosen from the data.
2. C — A guarantee is tested from the side of whoever it protects.
3. C — The first distractor divides by $s$ rather than the standard error.
4. A — The statistic is beyond the critical value. Note you never accept a null hypothesis.
5. A — The $p$-value is computed assuming the null, so it cannot be a probability about it.
6. D — Significance is a statement about $n$ as much as about the effect.
7. C — SE $= 12/4 = 3$, so $t = (793-800)/3$. -0.5833 divides by $s$ rather than the standard error.
| 8. A — $ | -2.3333 | > 2.1314$. Note that ‘accept’ is never the verdict. |
9. B — Failing to reject is not evidence of absence; a small sample misses real effects routinely.
10. D — SE $= 0.8$, so $t = (23.6-25)/0.8$. The first distractor divides by $s$.
| 11. C — $ | -1.75 | < 2.0301$. ‘Accept’ is never the verdict. |
12. B — Same data, opposite verdicts at 5%. Choosing the tail after seeing the data halves the $p$-value for nothing.
13. C — SE $= 0.6$, so $t = 1.2/0.6$.
14. C — $2.0000 < 2.0639$ — narrowly. The one-tailed $p$ of 0.0285 would reject, which is why the tail matters.
15. B — A guarantee is tested from the side of whoever it protects — the buyer needs evidence it fails.
16. B — $-0.9415$ is nowhere near $-1.7291$. Failing to reject is not proof the guarantee holds — 20 bulbs simply cannot settle it.
17. C — A target is missed in both directions. The sample mean never chooses the tail.
| 18. D — $ | -1.875 | < 2.1314$. Note the careful wording: not shown to be off target is not the same as shown to be on target. |
19. C — $2.5981 > 2.2010$, so the change is larger than sampling noise explains.
20. A — The question is whether the denominator is known or estimated — not the sample size.
| 21. B — $ | -2.6667 | > 2.5758$, so the fill is off target even at 1%. |
22. D — $p < 0.05$ and the sample mean is below 19.5, so costs are genuinely lower.
23. A — $20 \times 0.05 = 1$. The arithmetic in the reported test is correct; omitting the other nineteen is what misleads.
24. D — Choosing $\alpha$ is a judgement about the relative cost of the two errors.
1. (a) $H_0: \mu = 50$ against $H_a: \mu \neq 50$, two-tailed. (b) $t = 3.2000$, $\nu = 24$. (c) $3.2 > 2.0639$, so reject — the mean is not 50, and appears about 4.8 higher. (d) The interval $(51.7042, 57.8958)$ excludes 50, matching the rejection.
2. State the hypotheses; choose the test; set $\alpha$; write the decision rule; get the data; compute the statistic and $p$-value; reach the statistical conclusion; say what it means. Steps 1 to 5 happen before you look at a single number — that ordering is the protection against fooling yourself.
3. (a) $\mu = 45$ against $\mu > 45$, right. (b) $\mu = 72$ against $\mu \neq 72$, two-tailed. (c) $\mu = 500$ against $\mu \neq 500$, two-tailed — drift either way is a fault.
4. It halves the $p$-value without any new evidence. The tail must follow from the question, not the data; choosing afterwards means the stated error rate no longer describes the procedure actually used.
5. (a) About one: $20 \times 0.05$. (b) $p$-hacking, or selective reporting. The arithmetic in the reported test is correct; the omission of the other nineteen is what misleads.
6. The confidence interval. With a very large $n$ the standard error is tiny, so a trivial difference clears any significance threshold. The $p$-value cannot distinguish an important finding from a meaningless one; the interval shows the plausible size.
7. Type I is rejecting a true $H_0$, with probability $\alpha$, which you choose. Type II is failing to reject a false $H_0$, with probability $\beta$, which you influence mainly through the sample size. For fixed $n$ they trade off.
8. The test answers one claimed value; the interval tests every possible value at once — anything inside survives, anything outside is rejected — and reports the plausible range for $\mu$. The test does add the $p$-value, which says how far past the line you landed.
9. (a) SE $= 1$, so $28.4 \pm 2.0301 = (26.3699, 30.4301)$. (b) $t = (28.4-25)/1 = 3.4000$ on 35 df, beyond $2.0301$, so reject. (c) The interval excludes 25, which matches the rejection — a two-tailed test at $\alpha$ rejects exactly when the $(1-\alpha)$ interval excludes the claim.
10. $H_0: \mu = 500$ against $H_a: \mu < 500$, one-tailed, because a guarantee is tested from the side it protects. SE $= 6$, so $t = -2.3333$. Since $-2.3333 < -1.7109$, reject. The value 2.0639 puts 2.5% in each tail; here all 5% sits in one tail, so the cutoff is closer in.
| 11. (a) $H_0: \mu = 25$ against $H_a: \mu \neq 25$, two-tailed — the claim gives no direction. (b) SE $= 0.8$, so $t = -1.7500$, $\nu = 35$. (c) $ | -1.75 | < 2.0301$, so fail to reject: the sample does not give enough evidence that mean wait differs from 25 minutes. |
12. The interval contains 25, matching the failure to reject. A two-tailed test at level $\alpha$ rejects exactly when the $(1-\alpha)$ interval excludes the claimed value, because both are built from the same inequality.
13. The tail must follow from the question asked, fixed before the data is seen. Nothing in ‘is the mean 25?’ names a direction, so two-tailed is correct. Switching afterwards halves the $p$-value without any new evidence — that is $p$-hacking, and the stated 5% error rate would no longer describe the procedure actually used.
14. (a) SE $= 0.6$, $t = 2.0000$. (b) $2.0000 < 2.0639$, so fail to reject — narrowly. (c) Yes: $2.0000 > 1.7109$, so at 10% we would reject. The verdict sits between the two conventions, which is exactly why $\alpha$ must be fixed in advance.
15. Fail to reject: there is insufficient evidence that mean life falls below 1200 hours. It does not establish that the guarantee holds — with only 20 bulbs and $s = 95$ the test has little power, and a real shortfall of 20 hours would often go undetected.
| 16. (a) A target can be missed in either direction, and both are faults. (b) SE $= 0.8$, $t = -1.8750$; $ | -1.875 | < 2.1314$, so fail to reject — no evidence of drift. |
| 17. (a) $z$: $\sigma$ is known rather than estimated from these boxes. (b) SE $= 0.75$, so $z = -2.6667$. (c) $ | -2.6667 | > 2.5758$, so reject — the fill is below target even at the 1% level. |
18. “If cost per unit really were $19.5 on average, a sample this far below it would turn up about 4 times in 100. So the saving is larger than ordinary sampling variation explains.”
19. State the hypotheses; choose the test; set $\alpha$; write the decision rule; get the data; compute the statistic and $p$-value; reach the statistical conclusion; say what it means. Setting $\alpha$ and the decision rule before seeing the data is what stops the threshold being chosen to suit the result; a marker can see from the order of your working whether you did.
20. Failing to reject means the evidence was insufficient to rule $H_0$ out. Accepting would claim positive evidence that $H_0$ is true, which the test never provides — a small sample fails to reject almost regardless of the truth. The asymmetry mirrors ‘not guilty’ against ‘innocent’.
21. At the stated $\alpha$ the result is simply not significant, and ‘marginally’ is a way of rejecting without having earned it. If 0.06 is persuasive, $\alpha$ should have been set at 0.10 beforehand. The honest report is that the evidence is insufficient at the 5% level, accompanied by a confidence interval showing which effect sizes remain plausible.
22. Type I: concluding the drug works when it does not — an ineffective treatment reaches patients. Type II: concluding it does not work when it does — a useful treatment is abandoned. A regulator usually fears Type I more, which is why $\alpha$ is set low; but for a terminal illness with no alternative the balance can reasonably shift.
23. Established: the difference is larger than sampling variation would produce by chance. Not established: that it matters. With $n = 50{,}000$ the standard error is tiny, so trivial differences clear any threshold. The confidence interval would show the plausible size, which is the quantity anyone making a decision actually needs.
24. Neither. The $p$-value lies between 0.01 and 0.05, so the verdicts differ purely by the threshold each chose. What would be wrong is choosing the threshold after seeing $p$ — the level must be fixed in advance and reported alongside the conclusion.
25. The failure most likely reflects low power rather than the absence of a shift: at $n = 10$ an effect of $0.5\sigma$ is missed most of the time. Increasing the sample is the remedy — around 30 or more gives a realistic chance of detecting a shift that size. Lowering $\alpha$ would make matters worse, since for fixed $n$ it raises $\beta$.