Corrected exercises: estimation and hypothesis testing (201-337)

$\sigma_{\overline{x}}=\frac{\sigma}{\sqrt{n}}=\frac{20}{\sqrt{64}}=\frac{20}{8}=2.5$. Pitfall: dividing by $n$ instead of $\sqrt{n}$, which gives $0.3125$.

Standard error $\frac{10}{\sqrt{25}}=2$, margin $1.96\times 2=3.92$, so $52\pm 3.92$: the interval is $[48.08, 55.92]$.

Standard error $\frac{24}{\sqrt{36}}=4$, margin $1.645\times 4=6.58$, so $130\pm 6.58$: the interval is $[123.42, 136.58]$. Pitfall: taking $1.96$, the 95% value, out of habit.

Same standard error $4$, margin $2.576\times 4=10.304$, so $130\pm 10.30$: the interval is $[119.70, 140.30]$. It is wider than the one in c): more confidence is paid for in precision.

$\sigma$ is unknown, so we use $s$ and Student's $t$. Standard error $\frac{4}{\sqrt{16}}=1$, margin $2.131\times 1=2.131$, so $24.5\pm 2.131$: the interval is $[22.37, 26.63]$. Pitfall: taking $z=1.96$, which gives an interval that is too narrow.

$\hat{p}=\frac{260}{400}=0.65$. Standard error $\sqrt{\frac{0.65\times 0.35}{400}}=\sqrt{0.00056875}\approx 0.02385$, margin $1.96\times 0.02385\approx 0.0467$, so $0.65\pm 0.0467$: the interval is $[0.603, 0.697]$.

We require $1.96\times\frac{15}{\sqrt{n}}\leq 2$, so $\sqrt{n}\geq\frac{1.96\times 15}{2}=14.7$ and $n\geq 14.7^{2}=216.09$. Round up: $n=217$. Pitfall: rounding to the nearest integer, $216$, which gives a margin slightly above 2.

We require $1.96\sqrt{\frac{0.30\times 0.70}{n}}\leq 0.05$, so $n\geq\frac{1.96^{2}\times 0.21}{0.05^{2}}=\frac{3.8416\times 0.21}{0.0025}=322.69$. Round up: $n=323$.

Standard error $\frac{12}{\sqrt{49}}=\frac{12}{7}\approx 1.7143$, so $z=\frac{83.2-80}{1.7143}=\frac{3.2}{1.7143}\approx 1.87$.

Standard error $\frac{9}{\sqrt{36}}=1.5$, so $t=\frac{246.4-250}{1.5}=\frac{-3.6}{1.5}=-2.4$. The sign matters: the observed mean is below $\mu_{0}$. Pitfall: writing $\mu_{0}-\overline{x}$ in the numerator, which gives $+2.4$.

$\hat{p}=\frac{156}{300}=0.52$. Standard error UNDER $H_{0}$: $\sqrt{\frac{0.40\times 0.60}{300}}=\sqrt{0.0008}\approx 0.02828$, so $z=\frac{0.52-0.40}{0.02828}\approx 4.24$. Pitfall: computing the standard error with $\hat{p}=0.52$, which gives $4.16$.

The mean of the sampling distribution is that of the population: $\mu_{\overline{x}}=70$. The standard error is $\sigma_{\overline{x}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{12}{\sqrt{36}}=\dfrac{12}{6}=2.0$. Sample means are therefore six times less spread out than individual papers.

The central limit theorem states that for $n$ large enough, the distribution of $\overline{x}$ is approximately NORMAL, with mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{n}}$, WHATEVER the shape of the population distribution. Here $n=36\geq 30$: the usual practical threshold is reached. That is precisely the strength of the theorem: no assumption at all is needed about the distribution of the individual results, which may be skewed, bimodal or truncated.

$z=\dfrac{73-70}{2.0}=1.50$. The table gives 0.9332 to the left, so $P\left(\overline{x}>73\right)=1-0.9332=0.0668$, about 6.7%. Note that a SINGLE paper above 73 would be far more common: $z=\dfrac{73-70}{12}=0.25$, that is 40%. The distinction between the standard deviation of the data and the standard error of the mean is the number one source of errors in the chapter.

$\sigma_{\overline{x}}=\dfrac{12}{\sqrt{144}}=\dfrac{12}{12}=1.0$. Multiplying $n$ by four has therefore DIVIDED the standard error BY TWO, not by four, because it is $\sqrt{n}$ and not $n$ that appears in the denominator.

$n$ must be multiplied by $10^{2}=100$. Going from 36 to 3600 papers only reduces the standard error by a factor of 10. The practical consequence is decisive: the precision of a poll improves very slowly with cost. Doubling the precision requires FOUR TIMES the sample, so roughly four times the budget. That is why almost all opinion polls stop at around 1000 people: beyond that, the gain in precision no longer justifies the expense.

$\overline{x}\pm z\dfrac{\sigma}{\sqrt{n}}=70\pm 1.96\times 2.0=70\pm 3.92$. The interval is therefore $\left[66.08, 73.92\right]$, of width 7.84.

At 99%, $z=2.576$, so the margin is $2.576\times 2.0=5.15$ and the interval is $\left[64.85, 75.15\right]$, of width 10.3. The 99% interval is therefore WIDER. This makes sense: demanding more certainty of capturing the true parameter forces us to cast a wider net. There is a permanent trade-off between CONFIDENCE and PRECISION, and both can be improved only by increasing $n$.

The statement is incorrect because it treats $\mu$ as a random variable. But $\mu$ is a FIXED number, though unknown: it is either in the interval or it is not, and there is no probability involved. What is random is the INTERVAL, which changes with every sample. The correct statement is: “if the sampling were repeated a large number of times, 95% of the intervals built this way would contain $\mu$”. In practice this is shortened to “I am 95% confident that $\mu$ lies between 66.08 and 73.92”, which is acceptable when speaking but must be justified correctly in a written answer.

We require $E=z\dfrac{\sigma}{\sqrt{n}}\leq 2$, hence $\sqrt{n}\geq\dfrac{1.96\times 12}{2}=11.76$ and $n\geq 138.3$. Since $n$ is an integer and the condition must hold, we ROUND UP: $n=139$ papers. Rounding to 138 would give a margin slightly above 2, which would break the requirement; it is the only case in the course where rounding is never to the nearest value.

Because the margin $E=z\dfrac{\sigma}{\sqrt{n}}$ vanishes only as $n$ tends to infinity. As long as the sample is a proper subset of the population, an irreducible sampling uncertainty remains. The only way to reach a zero margin would be to survey the ENTIRE population, which is no longer sampling. On top of that, every real survey carries a share of non-sampling error, due to the wording of the questions, to non-response and to selection bias, which the formula does not measure at all and which a very large sample does nothing to improve.

$\hat{p}=\dfrac{540}{1000}=0.540$. The standard error is $\sqrt{\dfrac{0.540\times 0.460}{1000}}=\sqrt{2.484\times 10^{-4}}=0.01576$. The 95% margin is $1.96\times 0.01576=0.0309$. The interval is therefore $\left[0.509, 0.571\right]$.

The margin is 0.0309, that is about $\pm 3.1$ percentage points. This is exactly the “plus or minus 3 points” stated by every polling firm, and we have just recomputed it from scratch.

Yes, but only just. The LOWER bound of the interval, 0.509, is above 0.500: even in the least favourable case compatible with the data at 95%, the measure keeps its majority. The conclusion is therefore supported, but with a margin of only 0.9 points above the decisive threshold. Had the poll given 52% instead of 54%, the interval would have contained 50% and we would have had to conclude that the poll does NOT settle the question, a statement the media almost always get wrong.

In the worst case $\hat{p}=0.5$, we require $1.96\sqrt{\dfrac{0.25}{n}}\leq 0.02$, hence $\sqrt{n}\geq\dfrac{1.96\times 0.5}{0.02}=49$ and $n\geq 2401$. The sample must therefore more than DOUBLE, from 1000 to 2401 people, to gain barely more than one point of precision.

For two reasons that combine. First, the sample size is almost always chosen around 1000, a figure that comes from the cost-precision trade-off seen in d. Second, the factor $\sqrt{\hat{p}\left(1-\hat{p}\right)}$ varies very little: it is 0.500 at its maximum at $\hat{p}=0.5$, and still 0.458 for $\hat{p}=0.30$, barely 8% less. The margin is therefore very insensitive to the observed result, and for $n=1000$ it always stays close to $\dfrac{1.96\times 0.5}{\sqrt{1000}}=0.031$. Polling firms in fact publish the maximum margin, computed at $\hat{p}=0.5$, which makes it strictly constant for a given sample size.

$H_{0}: \mu=500$ g, the manufacturer's statement is accurate. $H_{1}: \mu\neq 500$ g. The test is TWO-SIDED because the inspector is checking that the labelling is accurate: a gap in either direction is a defect, a systematic overfilling also being a process anomaly. Had the mandate been to detect only fraud at the consumer's expense, we would have set $H_{1}: \mu<500$ and run a one-sided test.

The standard error is $\dfrac{\sigma}{\sqrt{n}}=\dfrac{15}{\sqrt{40}}=2.372$ g. The statistic is $z=\dfrac{494-500}{2.372}=-2.530$.

Since the test is two-sided, the p-value is $2\times P\left(Z<-2.530\right)=2\times 0.0057=0.0114$. Exact meaning: IF the bags really weighed 500 g on average, then a sample of 40 bags differing from that value by at least 6 g, in either direction, would occur in only 1.14% of cases. The p-value is NOT the probability that $H_{0}$ is true, a confusion that is the most widespread misinterpretation in statistics.

At the level $\alpha=0.05$: $0.0114<0.05$, we REJECT $H_{0}$ and conclude that the mean weight differs from 500 g. At the level $\alpha=0.01$: $0.0114>0.01$, we do NOT reject $H_{0}$, for lack of sufficient evidence. The same data therefore lead to two opposite conclusions. There is no contradiction here: the significance level expresses the risk of a type I error we are willing to accept, and it is a decision made in advance, according to the consequences. This case shows why the p-value itself must ALWAYS be reported rather than the bare word “significant”, and why setting the level after seeing the data is a serious methodological fault.

TYPE I error: concluding that the bags do not weigh 500 g when they actually do, so wrongly accusing an honest manufacturer. TYPE II error: concluding that nothing is wrong when the bags are actually underfilled, so letting a deception go through. A consumer protection agency will first try to minimize the TYPE II error, the one that lets fraud go unpunished, even if it means accepting a higher $\alpha$. A court, bound by the presumption of innocence, would make the opposite trade-off. For a fixed $n$, reducing one automatically increases the other: the only way to reduce both at once is to INCREASE the sample.

Because the z-score assumes $\sigma$ is KNOWN. Replacing it by $s$ brings in a second source of variability: from one sample to the next, not only does $\overline{x}$ change, but $s$ changes too, and an $s$ that is too small would produce a misleadingly narrow interval. The ratio $\frac{\overline{x}-\mu}{s/\sqrt{n}}$ therefore does not follow a normal distribution but a STUDENT t distribution with $n-1=11$ degrees of freedom, more spread out than the normal, with heavier tails. One degree of freedom is lost because $s$ was computed from $\overline{x}$, itself drawn from the same data.

Standard error: $\frac{s}{\sqrt{n}}=\frac{3.2}{\sqrt{12}}\approx 0.924$ mg. Margin: $(2.201)(0.924)\approx 2.03$ mg. Interval: $48.6\pm 2.03$, that is about $[46.6, 50.6]$ mg.

With $z=1.96$: margin $(1.96)(0.924)\approx 1.81$ mg, so $[46.8, 50.4]$. The correct interval is wider by the ratio $\frac{2.201}{1.96}\approx 1.12$, that is about 12% more. The gap must go in this direction: Student's t distribution is MORE SPREAD OUT than the normal, so its critical values are always larger. Using z when $\sigma$ is unknown amounts to claiming a precision you do not have, and therefore to being wrong more often than the stated confidence level suggests.

With 29 degrees of freedom, $t_{0.025}=2.045$ against $z=1.96$: the ratio drops to $\frac{2.045}{1.96}\approx 1.043$, a gap of only 4%, against 12% with 12 measurements. Student's t distribution CONVERGES to the normal as the number of degrees of freedom grows, because $s$ then estimates $\sigma$ better and better. That is where the threshold of 30, found in every textbook, comes from: beyond it, the gap between the two distributions falls below a few percent, negligible compared with the other approximations of a real study. It is not a mathematical threshold but a convention of convenience; today, since software uses Student's t distribution whatever the size, the question only arises on exams.

On such a small sample, validity rests on the NORMALITY of the original population, because the central limit theorem has not had time to act with only 12 data values. In practice, the Student procedure is robust: it copes well with a moderate departure from normality from about fifteen observations on, but it fails with strong skewness or with outliers. The reflex is therefore to PLOT the data before computing, with a box plot or a histogram. If the distribution is clearly skewed, two routes are open: transform the variable, for example by taking its logarithm, or turn to a nonparametric method, which assumes no shape for the distribution.

Step 1: state $H_{0}$ and $H_{1}$. Step 2: set the significance level $\alpha$ and check the conditions. Step 3: compute the test statistic. Step 4: compute the p-value, or compare with the critical value, and decide. Step 5: conclude in plain language, in the words of the problem, without jargon.

$H_{0}: \mu\leq 4.0$ minutes, the company's claim is true. $H_{1}: \mu>4.0$ minutes. The test is RIGHT-TAILED because the association is trying to establish only one thing: that the wait is LONGER than claimed. A waiting time shorter than promised is of no interest to it and would be no grievance. The alternative hypothesis must always reflect what we are trying to show.

Conditions: $n=50\geq 30$, the central limit theorem applies and $\sigma$ is known. The standard error is $\dfrac{1.5}{\sqrt{50}}=0.2121$ minutes. The statistic is $z=\dfrac{4.4-4.0}{0.2121}=1.886$. Since the test is right-tailed, the p-value is $P\left(Z>1.886\right)=0.0296$. As $0.0296<0.05$, we REJECT $H_{0}$. Written conclusion: at the 5% level, the data collected allow us to conclude that the mean waiting time really exceeds the 4.0 minutes claimed by the company; the observed gap of 0.4 minutes is too large to be put down to sampling chance alone.

In a two-sided test, the p-value would double: $2\times 0.02967=0.0593$. As $0.0593>0.05$, we would NOT reject $H_{0}$. The conclusion is therefore reversed. This shows that a one-sided test is more POWERFUL when there is a legitimate reason to orient the hypothesis, since it puts all of the risk $\alpha$ on one side. But it also shows the danger: choosing a one-sided test AFTER seeing the direction of the gap amounts to quietly doubling your significance level, which is a methodological fault. The direction of the test must be set before the data are collected.

The power, that is $1-\beta$, the probability of detecting a gap that is really there, would INCREASE. Going from 50 to 200 calls divides the standard error by $\sqrt{4}=2$, down to 0.106 minutes. The same observed gap of 0.4 minutes would then produce a statistic twice as large, $z=3.77$, and therefore a much smaller p-value. In other words, with 200 calls the association would detect SMALLER gaps than those it can establish with 50, and its conclusion would hold up whatever the significance level and whether the test is one-sided or two-sided. Increasing $n$ is the only way to reduce both types of error at once.

$\hat{p}=\frac{172}{200}=0.860$. Hypotheses: $H_{0}:p=0.90$, the manufacturer's claim, which is the default hypothesis, rejected only on evidence; and $H_{1}:p<0.90$. The test is LEFT-TAILED because the customer is only worried about a quality BELOW the claim: a proportion above 90% would be no ground for a dispute. The direction of $H_{1}$ must be set before looking at the data, otherwise the significance level loses all meaning.

Conditions: $np_{0}=(200)(0.90)=180\geq 5$ and $n(1-p_{0})=(200)(0.10)=20\geq 5$, so the normal approximation is legitimate. Standard error under $H_{0}$: $\sigma_{\hat{p}}=\sqrt{\frac{p_{0}(1-p_{0})}{n}}=\sqrt{\frac{(0.90)(0.10)}{200}}=\sqrt{0.00045}\approx 0.0212$. Test statistic: $z=\frac{\hat{p}-p_{0}}{\sigma_{\hat{p}}}=\frac{0.860-0.900}{0.0212}\approx -1.89$.

Left-tailed test: p-value $=P(Z<-1.886)\approx 0.0297$, about 3.0%, keeping $z$ unrounded; the table, read at $z=-1.89$, gives $0.0294$, which changes nothing in what follows. Exact meaning: IF the manufacturer were telling the truth, that is, if the true proportion really were 0.90, then a sample of 200 parts would give a result this unfavourable or worse in 3% of cases. It is NOT the probability that the manufacturer is wrong: the p-value is computed assuming $H_{0}$ is true, so it can say nothing about the probability of $H_{0}$.

At the 5% level: $0.0297<0.05$, we REJECT $H_{0}$, the data are significantly below the claim. At the 1% level: $0.0297>0.01$, we do NOT reject. Such a difference is not a contradiction: it simply signals that the evidence is moderate, and that the conclusion depends on the standard of proof chosen. Good practice is to report the p-value itself rather than a yes-or-no verdict, so that readers can judge with their own level. It is also a reminder that the level must be set BEFORE the calculation: choosing it afterwards amounts to choosing the conclusion.

In a TEST, we reason assuming $H_{0}$ is true from start to finish: it is this hypothesis that supplies the value of $p$ to use, hence 0.90. Using 0.86 would partly assume the very thing we are putting to the test. In a confidence INTERVAL, on the other hand, there is no starting hypothesis: we have only the sample, and the only value available to estimate the standard error is $\hat{p}$. The two procedures do not start from the same place, hence two formulas that look alike without being identical, and a very common confusion on exams.

$H_{0}:\mu_{1}=\mu_{2}$, that is $\mu_{1}-\mu_{2}=0$: the two methods give the same mean result. $H_{1}:\mu_{1}\neq\mu_{2}$. The test is TWO-SIDED because the question asked is whether the methods DIFFER, without having decided in advance which one would be better. Choosing a one-sided test after noticing that group 1 has the higher mean would be a fault: it would amount to halving the p-value in light of the data.

$\sigma_{\overline{x_{1}}-\overline{x_{2}}}=\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{\frac{72.25}{40}+\frac{84.64}{35}}=\sqrt{1.806+2.418}=\sqrt{4.225}\approx 2.055$. We add because it is the VARIANCES that add up for independent variables, whether these variables are added or subtracted. The intuition is simple: subtracting two uncertain quantities does not remove the uncertainty, it accumulates it, each of the two means contributing to the variation of the difference. The standard deviation of the difference is larger than each of the two taken separately, never smaller.

$z=\frac{(74.2-70.1)-0}{2.055}=\frac{4.1}{2.055}\approx 1.99$. Two-sided test: p-value $=2\,P(Z>1.99)\approx 2(0.0230)=0.046$.

At the 5% level: $0.046<0.05$, we reject $H_{0}$, but only just. At the 1% level: $0.046>0.01$, we do not reject. The result is therefore borderline: it is a clue, not a proof, and it calls for a replication on a larger sample rather than a firm conclusion. Announcing “method 1 is better” on this basis would go far beyond what the data allow.

Interval: $(74.2-70.1)\pm(1.96)(2.055)=4.1\pm 4.03$, that is about $[0.07, 8.13]$. It excludes zero, only just, which agrees exactly with the narrow rejection at the 5% level: the two procedures are two readings of the same calculation, and a 95% interval that excludes zero is always equivalent to a p-value below 0.05 in a two-sided test. The interval nevertheless adds a great deal: it does not only say that a difference exists, it gives the set of plausible values of that difference, here between almost nothing and eight points. In other words, the data are compatible with a negligible effect as well as with a large one, which the bare statement “significant at 5%” would completely hide.

The table crosses attendance with passing: regular attenders, 90 passes and 30 failures, total 120; irregular attenders, 40 and 40, total 80. Column totals: 130 passes and 70 failures, for 200 students. Pass rates: $\frac{90}{120}=75$% among regular attenders against $\frac{40}{80}=50$% among the others, a gap of 25 points. $H_{0}$: the two variables are INDEPENDENT, attendance has no link with passing. $H_{1}$: they are related.

Under $H_{0}$, the pass rate must be the same everywhere, namely the overall rate $\frac{130}{200}=65$%. Among regular attenders we therefore expect $(120)(0.65)=78$ passes, which can be written $\frac{(120)(130)}{200}$: EXPECTED count $=\frac{\text{row total}\times\text{column total}}{\text{grand total}}$. The four expected values are thus 78 and 42 for regular attenders, 52 and 28 for irregular attenders. Check: each row and each column adds back up to its original total, which must always be the case.

$\chi^{2}=\sum\frac{(O-E)^{2}}{E}=\frac{(90-78)^{2}}{78}+\frac{(30-42)^{2}}{42}+\frac{(40-52)^{2}}{52}+\frac{(40-28)^{2}}{28}=1.846+3.429+2.769+5.143\approx 13.19$. The gaps all equal 12 in absolute value, as is always the case in a 2 by 2 table, but they weigh more when the expected count is small, hence the division by $E$. Degrees of freedom: $(2-1)(2-1)=1$, because once one cell is fixed, the totals determine the other three.

$13.19>3.841$: we reject $H_{0}$ at the 5% level. $13.19>6.635$: we also reject it at the 1% level, and by a wide margin, the p-value being of the order of 0.0003. The conclusion is therefore solid, unlike the borderline case of a narrow rejection. Conditions of validity: all EXPECTED counts must be at least 5; here the smallest is 28, so the condition is easily met. Note that it applies to the expected counts, never to the observed counts.

No. The chi-square test establishes a statistical ASSOCIATION between two variables; it says nothing about the direction of causation, or even whether there is any. Three readings remain possible. Attendance may indeed help students pass, which is plausible. Passing may also feed attendance: a student who understands and succeeds takes a liking to the course and keeps coming, while one who falls behind stops showing up, so that the arrow would go from passing to attendance. Finally, a CONFOUNDING variable may produce both: motivation, personal organization, or simply not working thirty hours a week alongside one's studies, would explain both attendance and results, without either one causing the other. Only an experiment in which attendance was imposed at random could settle the matter, and it would hardly be practicable. It is exactly the same caution as the one a correlation coefficient demands.