Chapters 11 and 12. 9 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. Weekly study hours have $\sigma = 24$. For a sample of 36 students the standard error of the mean is
(A) 24.0000 (B) 4.0000 (C) 0.6667 (D) 0.8165
2. If the true mean is 80 hours, the probability that a sample of 36 averages above 88 is closest to
(A) 0.3694 (B) 0.0228 (C) 0.9772 (D) 0.5000
3. How many students must be sampled for that standard error to fall to 2?
(A) 12 (B) 48 (C) 24 (D) 144
4. A market researcher quadruples her sample. The margin of error
(A) quarters (B) stays the same (C) halves (D) falls by three quarters
5. Household incomes in a city are strongly right-skewed. For a sample of 50 households, which is approximately normal?
(A) a single household’s income only (B) the sample mean income only (C) both (D) neither
6. A factory’s output per shift has $\sigma = 40$. A manager wants the standard error of mean output below 2.5. The smallest number of shifts is
(A) 16 (B) 256 (C) 64 (D) 1024
7. Claim sizes at an insurer have mean 60 and $\sigma = 15$. For 9 claims, the range containing the middle 95% of sample means is closest to
(A) (30.6, 89.4) (B) (55.1, 64.9) (C) (50.2, 69.8) (D) (58.4, 61.6)
8. A pollster draws 300 people by selecting 60 households and interviewing every adult. The true standard error is
(A) equal to $\sigma/\sqrt{300}$ (B) larger than $\sigma/\sqrt{300}$ (C) smaller than $\sigma/\sqrt{300}$ (D) not defined
9. An agronomist averages yields from 4 plots drawn from a strongly skewed population. The distribution of that average is
(A) exactly normal (B) approximately normal (C) still noticeably skewed (D) uniform
1. A population has $\mu = 80$, $\sigma = 24$, and a sample of 36 is drawn. (a) State $E(\bar{X})$ and the standard error. (b) Find $\mathrm{P}(\bar{X} > 88)$. (c) Find $\mathrm{P}(76 < \bar{X} < 84)$.
2. For the population ${3, 6, 9}$, list all nine samples of size 2 drawn with replacement, and verify $E(\bar{X}) = \mu$ and $\mathrm{SD}(\bar{X}) = \sigma/\sqrt{2}$.
3. Explain why quadrupling the sample size only halves the standard error.
4. A skewed population is sampled at $n = 4$. Of 1{,}000 simulated means, 95.1% fall inside the nominal 95% interval, but the tails are 0.049 and 0.000. Explain why the headline figure is misleading.
5. Distinguish sampling error from non-sampling error, and say which a larger sample fixes.
6. State the three things the CLT tells you about $\bar{X}$, and which of them needs the theorem.
7. In Lab 6 we drew 1{,}000 samples of 36 from a population and found the standard deviation of the sample means close to $\sigma/6$. Explain what that simulation demonstrates and what it cannot demonstrate.
8. A population has $\mu = 450$, $\sigma = 60$. (a) For $n = 100$, find $\mathrm{P}(\bar{X} < 440)$. (b) Find $\mathrm{P}(X < 440)$ for one observation, or explain why you cannot.
1. B — $\sigma/\sqrt{n}$. 0.6667 divides by $n$ rather than $\sqrt{n}$.
2. B — The first distractor uses $\sigma$ instead of the standard error — the commonest error in this material.
3. D — $\sigma/\sqrt{n}=2$ gives $\sqrt{n}=12$. 12 stops at $\sqrt{n}$.
4. C — It falls with $\sqrt{n}$.
5. B — The CLT is about $\bar{X}$. Sampling does not reshape the population.
6. B — $40/\sqrt{n}\le 2.5$ gives $\sqrt{n}\ge 16$. 16 stops at $\sqrt{n}$.
7. C — SE $= 5$, so $60 \pm 1.96(5)$. (30.6, 89.4) uses $\sigma = 15$ instead of the standard error.
8. B — People in a household are not independent draws, so 300 clustered people carry less information than 300 unrelated ones.
9. C — At $n = 4$ the CLT has barely begun. Convergence depends on how far the population starts from normal.
1. (a) $E(\bar{X}) = 80$, $\sigma_{\bar{X}} = 4$. (b) $z = 2.00$, so $0.0228$. (c) $z = \mp 1$, so $0.6827$.
2. $\mu = 6$, $\sigma^2 = 6$, $\sigma = 2.4495$. The nine means are 3, 4.5, 6, 4.5, 6, 7.5, 6, 7.5, 9 — so $\bar{x} = 3, 4.5, 6, 7.5, 9$ with probabilities $1/9, 2/9, 3/9, 2/9, 1/9$. Then $E(\bar{X}) = 6 = \mu$ and $V(\bar{X}) = 3$, so $\mathrm{SD}(\bar{X}) = 1.7321 = 2.4495/\sqrt{2}$.
3. The standard error depends on $\sqrt{n}$. Averaging cancels independent errors, but each new observation adds less than the one before, so precision accumulates at the rate $\sqrt{n}$ rather than $n$.
4. The two tails are wrong in opposite directions and the errors cancel in the total. A two-sided statement survives by luck; any one-tailed probability would be badly wrong — the lower tail is about double what it should be and the upper tail is empty.
5. Sampling error is $\bar{x} - \mu$, arising because you measured a sample; it is random and shrinks as $n$ grows. Non-sampling error is mistyped data, leading questions, non-response; it is often systematic and does not shrink with $n$. A bigger sample fixes only the first.
6. Centre $E(\bar{X}) = \mu$; spread $\sigma/\sqrt{n}$; shape approximately normal. The first two follow from the rules for expectation and variance and hold for any population. Only the third needs the CLT.
7. It demonstrates the spread result, $\mathrm{SD}(\bar X) = \sigma/\sqrt{n}$, which holds for any population. It cannot by itself demonstrate the CLT’s shape claim — for that you must look at the histogram of the means and check it is bell-shaped, not merely narrow.
8. (a) SE $= 6$, so $z = -1.6667$ and $\mathrm{P} = 0.0478$. (b) You cannot, unless the population is normal. The CLT licenses a statement about $\bar{X}$ only; a probability for a single observation needs the population’s own shape, which was not given.