Practice: from sample to population

Chapters 11 to 14 — sampling distributions, the CLT, estimation and hypothesis testing.

Fifteen problems. Work them with the solutions covered — reading a worked answer and producing one are different skills, and only the second is any use with a clock running.

Keep four decimal places unless a question says otherwise.

1. A population has $\mu = 80$ and $\sigma = 24$. A random sample of 36 is drawn.

(a) State $E(\bar{X})$ and the standard error. (b) Find $\mathrm{P}(\bar{X} > 88)$. (c) How large would $n$ need to be for the standard error to be 2?

2. For the population ${2, 4, 6}$:

(a) Find $\mu$ and $\sigma$. (b) List all nine samples of size 2 drawn with replacement and their means. (c) Verify $E(\bar{X}) = \mu$ and $\mathrm{SD}(\bar{X}) = \sigma/\sqrt{2}$.

3. A population is strongly left-skewed. A sample of $n = 45$ is drawn.

(a) What can you say about the distribution of $\bar{X}$? (b) What can you say about the distribution of a single observation $X$? (c) A classmate computes $\mathrm{P}(X < 30)$ using the normal table. What is wrong?

4. Explain why quadrupling the sample size only halves the standard error.

5. A sample of 25 observations gives $\bar{x} = 54.8$ and $s = 7.5$.

(a) Compute the standard error. (b) Build a 95\% confidence interval. Use $t_{0.025,24} = 2.0639$. (c) State precisely what “95\% confident” refers to.

6. Using the sample above, a planning figure assumes the mean is 50.

(a) State $H_0$ and $H_a$, and the tail. (b) Compute the test statistic. (c) Test at $\alpha = 0.05$ and state the conclusion in words. (d) Show your interval and your test agree.

7. A study wants to estimate a mean to within $\pm 1.0$ at 95\% confidence, believing $\sigma \approx 7.5$.

(a) How many observations are needed? (b) How many to get within $\pm 0.5$? (c) What does the comparison show?

8. For each, state $H_0$, $H_a$ and the tail.

(a) A supplier guarantees a mean lifetime of at least 800 hours; you are the buyer. (b) Last year’s mean was 45; the department asks whether this year differs. (c) A process is targeted at 250 g; quality control checks for drift. (d) A trainer claims a new method raises mean scores above 70.

9. A test returns $p = 0.02$.

(a) Write what this means as a sentence. (b) A classmate says “there is a 2% chance the null is true”. Correct them. (c) Another says “$p = 0.002$ would mean a bigger effect”. Correct them too.

10. Complete the table and define each cell.

11. A researcher runs 20 independent tests on data where no effect exists.

(a) How many would you expect to be “significant” at $\alpha = 0.05$? (b) What is this phenomenon called when only the significant one is reported? (c) Which of the eight steps of a hypothesis test guards against it?

12. A 95% confidence interval for a mean comes out as $(12.4,\ 18.6)$.

(a) What is the point estimate? (b) What is the margin of error? (c) Would a two-tailed test of $H_0: \mu = 19$ at $\alpha = 0.05$ reject? Explain without computing anything.

13. Two studies estimate the same quantity at 95% confidence. Study A reports $8.2 \pm 0.4$; Study B reports $8.2 \pm 3.6$.

(a) Which is more useful, and why? (b) Give two distinct reasons B’s interval might be so much wider. (c) Do the two disagree?

14. Explain why “not significant” does not mean “no effect”, and give a concrete circumstance in which a real effect would be missed.

15. A survey of 1,500 people is drawn by selecting 375 households and interviewing all adults in each. The report quotes a margin of error based on $n = 1{,}500$.

(a) Is the quoted margin too large, too small or about right? (b) Why? (c) What would you need to report an honest margin?

Solutions

1. (a) $E(\bar{X}) = 80$ and $\sigma_{\bar{X}} = 24/\sqrt{36} = 4$. (b) $z = (88-80)/4 = 2.00$, so $\mathrm{P} = 0.0228$. (c) Set $24/\sqrt{n} = 2$, so $\sqrt{n} = 12$ and $n = 144$.

2. (a) $\mu = 4$; $\sigma^2 = [(2-4)^2 + 0 + (6-4)^2]/3 = 8/3 = 2.6667$, so $\sigma = 1.6330$. (b) Means: 2, 3, 4, 3, 4, 5, 4, 5, 6 — that is $\bar{x} = 2,3,4,5,6$ with probabilities $1/9, 2/9, 3/9, 2/9, 1/9$. (c) $E(\bar{X}) = (2 + 6 + 12 + 10 + 6)/9 = 4 = \mu$. And $V(\bar{X}) = 4/3$, so $\mathrm{SD}(\bar{X}) = 1.1547 = 1.6330/\sqrt{2}$.

3. (a) Approximately normal, by the Central Limit Theorem, with mean $\mu$ and standard error $\sigma/\sqrt{45}$. (b) Still left-skewed. Taking a sample does not change the population. (c) The CLT applies to $\bar{X}$, not to $X$. A probability for a single observation needs the population’s actual shape, which is skewed — so the normal table gives the wrong answer.

4. The standard error is $\sigma/\sqrt{n}$, so it depends on $\sqrt{n}$ rather than $n$. Averaging cancels independent errors, but each new observation adds less than the one before — the tenth improves a mean of nine far more than the thousandth improves a mean of 999. The exponent $\tfrac{1}{2}$ is the exact rate of those diminishing returns.

5. (a) $7.5/\sqrt{25} = 1.5$. (b) $54.8 \pm 2.0639(1.5) = (51.7042,\ 57.8958)$. (c) It describes the procedure, not this interval. Repeat the whole exercise many times and 95\% of the intervals built this way would contain $\mu$. This one either does or does not, and we cannot know which.

6. (a) $H_0!: \mu = 50$ against $H_a!: \mu \neq 50$, two-tailed — the figure is “assumed”, with no direction given. (b) $t = (54.8 - 50)/1.5 = 3.2000$ with $\nu = 24$. (c) $3.2 > 2.0639$, so reject $H_0$. The sample gives strong evidence the mean is not 50; it appears about 4.8 higher. (d) The interval $(51.7042, 57.8958)$ does not contain 50, which matches the rejection. A two-tailed test at $\alpha$ rejects exactly when the $(1-\alpha)$ interval excludes the claimed value.

7. (a) $n = (1.96 \times 7.5 / 1.0)^2 = 216.08$, so 217. (b) $n = (1.96 \times 7.5 / 0.5)^2 = 864.33$, so 865. (c) Halving the margin quadruples the sample. Precision is bought at a rate of $\sqrt{n}$, so each further improvement costs more than the last.

8. (a) $H_0: \mu = 800$, $H_a: \mu < 800$, left — a guarantee is tested from the side of whoever it protects. (b) $H_0: \mu = 45$, $H_a: \mu \neq 45$, two-tailed — “differs” names no direction. (c) $H_0: \mu = 250$, $H_a: \mu \neq 250$, two-tailed — drift either way is a fault. (d) $H_0: \mu = 70$, $H_a: \mu > 70$, right — “raises above” names a direction.

9. (a) “If $H_0$ were true, data at least this extreme would arise about 2 times in 100.” (b) Backwards. The $p$-value is computed assuming $H_0$ is true, so it cannot also be the probability that $H_0$ is true. (c) A smaller $p$ means the data sits more awkwardly with $H_0$, not that the effect is larger. With a big enough sample a trivial effect produces a tiny $p$. Read the confidence interval to judge size.

10. Reject when $H_0$ is true: a Type I error, probability $\alpha$ — declaring an effect that is not there. Reject when $H_0$ is false: a correct decision, probability $1 - \beta$, called the power. Do not reject when $H_0$ is true: a correct decision. Do not reject when $H_0$ is false: a Type II error, probability $\beta$ — missing a real effect.

11. (a) About one — $20 \times 0.05 = 1$. (b) $p$-hacking (or selective reporting). The arithmetic in the reported test is perfectly correct; the omission of the other nineteen is what makes it misleading. (c) Steps 3 and 4 — setting $\alpha$ and writing the decision rule before seeing the data, which fixes the test in advance and makes selective reporting visible.

12. (a) The midpoint, $(12.4 + 18.6)/2 = 15.5$. (b) Half the width, $(18.6 - 12.4)/2 = 3.1$. (c) Yes. 19 lies outside the interval, and a two-tailed test at $\alpha = 0.05$ rejects exactly when the 95% interval excludes the claimed value.

13. (a) A — the same estimate with far more precision, so it rules out many more values. (b) Any two of: B had a much smaller sample (the margin goes as $1/\sqrt{n}$, so a nine-fold wider margin suggests a sample on the order of 80 times smaller); B’s population is more variable; B measured less precisely, inflating $s$. (c) No. They agree exactly on the point estimate. They differ in precision, not in finding — and A’s interval sits entirely inside B’s.

14. Failing to reject means the evidence was insufficient, not that $H_0$ is true. A small sample produces a large standard error, so even a substantial real difference can fall short of the critical value. Concretely: a six-point difference on a population with $\sigma = 15$ is missed more than half the time at $n = 25$ — the test simply lacks the power to detect it, and reporting that as “no effect” misrepresents what happened.

15. (a) Too small. (b) $\sigma/\sqrt{n}$ counts independent pieces of information, and adults in the same household are not independent — they share income, location and often views. The effective sample size is well below 1,500. (c) An estimate of the design effect — how much the clustering inflates the variance — obtained by comparing the variability between households with the variability within them.