Corrected practice final exam in advanced statistics (201-337)

The sum is 356, so $\overline{x}=\frac{356}{12}\approx 29.667$ minutes. The sum of the squared deviations is about $564.667$, and dividing by $n-1=11$ gives a variance of about $51.333$, so $s\approx 7.165$ minutes.

Compute the z-score of the cutoff: $z=\frac{40-29.667}{7.165}\approx 1.44$. So $P(X>40)=P(Z>1.44)\approx 0.0749$, about 7.5%.

Only one commute out of twelve exceeds 40 minutes, that is $\frac{1}{12}\approx 0.0833$, about 8.3%. The two values are close, which is rather reassuring for the model. But the conclusion must stay cautious: with twelve observations, a single commute more or less above 40 minutes would move the empirical proportion to 0 or to 16.7%. A sample of this size can neither seriously confirm nor seriously refute a normality assumption.

$z=\frac{44-29.667}{7.165}\approx 2.00$. The usual criterion puts the boundary at $|z|>2$, sometimes at $|z|>3$: the value is therefore right at the limit, and certainly not beyond all suspicion. It must not be discarded. A value is not an outlier because it is large, but because there is a reason to believe it comes from a measurement error or from another population. Here, 44 minutes is a perfectly plausible commute time.

Chebyshev's inequality states that for any distribution, whatever its shape, and for any $k>1$, the proportion of data within $k$ standard deviations of the mean is at least $1-\frac{1}{k^{2}}$. With $k=2$: the interval is $500\pm 2\times 80$, that is $[340, 660]$ grams, and it contains at least $1-\frac{1}{4}=0.75$, so at least 75% of the bags.

With $k=3$: the interval is $500\pm 240$, that is $[260, 740]$ grams, and it contains at least $1-\frac{1}{9}\approx 0.8889$, so at least 88.9% of the bags.

For $k=2$, Chebyshev guarantees 75% whereas the normal distribution would give 95.45%. For $k=3$, Chebyshev guarantees 88.9% against 99.73%. The gap is considerable, especially in the tails: beyond three standard deviations, the normal distribution leaves 0.27% of the data, while Chebyshev allows up to 11.1%, about forty times more. Giving up normality is therefore very costly in precision. In return, Chebyshev's bound holds without any condition, and that is its whole point: it can never be proved wrong.

740 grams corresponds exactly to $z=\frac{740-500}{80}=3$. Without any assumption on the shape of the distribution, only one thing can be stated: at most 11.1% of the bags lie more than three standard deviations from the mean, in either direction, so at most 11.1% weigh more than 740 grams or less than 260. If the distribution were normal, these two tails together would hold only 0.27%, and a 740-gram bag would be a remarkable event. Here it is not necessarily one.

Two mistakes. First, Chebyshev's inequality does not give 95% but at least 75%: the inspector applied the empirical rule, which assumes normality, whereas the statement rules it out. Second, and more serious, compliance has nothing to do with this calculation: it is judged against tolerances set in advance, not against the observed spread. A very irregular production always satisfies Chebyshev; that does not make it compliant.

Isolate $n$ in $E=\frac{z\sigma}{\sqrt{n}}$, which gives $n=\left(\frac{z\sigma}{E}\right)^{2}=\left(\frac{1.96\times 15}{2}\right)^{2}=\left(14.7\right)^{2}=216.09$. A sample size is always rounded UP, otherwise the target margin is not reached: 217 people are needed.

Here $n=\left(\frac{z}{E}\right)^{2}p(1-p)$, and without information on $p$ we take the value that maximizes $p(1-p)$, namely $p=0.5$, which gives $p(1-p)=0.25$. Then $n=\left(\frac{1.96}{0.03}\right)^{2}\times 0.25\approx 4268.44\times 0.25\approx 1067.11$, so 1068 people. This is the famous “plus or minus 3%” of opinion polls, which corresponds to about a thousand respondents.

With $p=0.2$, $p(1-p)=0.16$ and $n\approx 4268.44\times 0.16\approx 682.95$, so 683 people. That is 36% fewer than in b) for the same precision. Prior information, even approximate, saves about a third of the cost of the survey, because the variance of a proportion drops sharply as soon as you move away from 0.5.

No, the institute is wrong. The formulas above nowhere contain the size $N$ of the population: precision depends on the number of people surveyed, not on the share of the population they represent. As long as the sample is less than 5% of the population, which is the case in both provinces, 1068 respondents give exactly the same margin of error. It is one of the most counter-intuitive results of the course, and it is the reason why a national poll and a municipal poll survey the same number of people.

Let $D$ be the event “the person has the disease” and $T$ the event “the test is positive.” The statement gives $P(D)=0.008$, the sensitivity $P(T\mid D)=0.95$, and the false positive rate $P(T\mid\overline{D})=0.04$, which is equivalent to a specificity $P(\overline{T}\mid\overline{D})=0.96$.

By the law of total probability: $P(T)=0.008\times 0.95+0.992\times 0.04=0.0076+0.03968=0.04728$, about 4.73%.

$P(D\mid T)=\frac{P(D\cap T)}{P(T)}=\frac{0.0076}{0.04728}\approx 0.1607$. A person who tests positive therefore has only about a 16% chance of having the disease: more than four positives out of five are false alarms. The reason is arithmetic, not medical: healthy people are so numerous that their 4% of false positives, about 39.7 people per thousand, swamp the 7.6 true positives per thousand.

$P(\overline{T})=0.008\times 0.05+0.992\times 0.96=0.0004+0.95232=0.95272$, so $P(\overline{D}\mid\overline{T})=\frac{0.95232}{0.95272}\approx 0.9996$. A negative result, on the other hand, is almost certain. The test is excellent at reassuring and poor at raising the alarm.

“95% reliable” refers to the sensitivity, that is, a conditional probability GIVEN that the person has the disease. But the patient who receives the result is asking the reverse question: what is the probability of having the disease given that the test is positive? The two numbers, 95% and 16%, are both correct and do not describe the same thing; confusing them is the error of inverting conditional probabilities. In a high-risk population where $P(D)=0.3$, we would have $P(T)=0.3\times 0.95+0.7\times 0.04=0.285+0.028=0.313$ and $P(D\mid T)=\frac{0.285}{0.313}\approx 0.9105$. The same test, without any of its characteristics changing, then becomes highly conclusive. This is why a rare disease is not screened for in the general population: the value of a test depends as much on who is tested as on the test itself.

The same two-outcome trial, defective or not, is repeated 200 times with a constant probability and independent draws: $X$ follows the binomial distribution with parameters $n=200$ and $p=0.02$. Its expected value is $np=4$ parts and its standard deviation $\sqrt{np(1-p)}=\sqrt{3.92}\approx 1.9799$.

$P(X=3)=\binom{200}{3}(0.02)^{3}(0.98)^{197}$. With $\binom{200}{3}=1\,313\,400$, we get $P(X=3)\approx 0.1963$.

The Poisson approximation uses $\lambda=np=4$: $P(X=3)\approx\frac{\mathrm{e}^{-4}\times 4^{3}}{3!}=\frac{64\,\mathrm{e}^{-4}}{6}\approx 0.1954$. The gap with the exact value is less than one thousandth, a relative error of about 0.5%. The conditions are indeed met since $n=200\geq 50$ and $p=0.02\leq 0.1$.

No. The condition $np\geq 5$ is not satisfied: $np=4$. This is consistent with the shape of the distribution, which is very skewed here since it stops at zero on one side and spreads out on the other, whereas the normal distribution is symmetric. Approximating such a distribution by a normal one would even give nonzero probabilities to negative values of $X$, which makes no sense for a count.

With $p=0.4$: $np=80$ and $n(1-p)=120$, both far above 5, so the normal approximation becomes perfectly legitimate. On the other hand, the condition $p\leq 0.1$ no longer holds, and the Poisson approximation stops being valid. The two approximations are therefore complementary rather than competing: Poisson is for rare events over a large number of trials, the normal for situations where the expected value and its complement are both comfortably large. In between, the exact binomial is always available.

The player stakes 2 units in every case. The player therefore gains $3-2=1$ with probability 0.3, $12-2=10$ with probability 0.1, and $0-2=-2$ with the remaining probability $1-0.3-0.1=0.6$. The distribution of $X$ thus takes the values $-2$, $1$ and $10$ with respective probabilities $0.6$, $0.3$ and $0.1$.

$E(X)=-2\times 0.6+1\times 0.3+10\times 0.1=-1.2+0.3+1=0.1$ unit. Then $E(X^{2})=4\times 0.6+1\times 0.3+100\times 0.1=2.4+0.3+10=12.7$, so $\mathrm{Var}(X)=12.7-0.1^{2}=12.69$ and $\sigma\approx 3.5623$. Since the expected value is strictly positive, the game is even slightly favourable to the player, which is unusual. But the standard deviation is thirty-five times the expected value: the advantage is drowned in randomness.

By the central limit theorem, the mean of 100 independent, identically distributed variables is approximately normal, whatever the original distribution, which here is clearly not normal since it takes only three values. This mean has expected value $0.1$ and standard error $\frac{3.5623}{\sqrt{100}}\approx 0.3562$.

Losing overall means that the mean gain is negative. Compute $z=\frac{0-0.1}{0.3562}\approx -0.28$, so the probability is about $0.3897$. In other words, despite a favourable game, the player has nearly a 39% chance of walking away a loser after a hundred plays.

The expected value per play has the same size, but the opposite sign for the operator, who LOSES 0.1 unit on average per play made against them: as designed, this game is a losing one for the house. Above all, the standard error decreases like $\frac{1}{\sqrt{n}}$: for the player and their 100 plays it is 0.356, three and a half times the expected value, whereas for the operator and their 100,000 plays it is $\frac{3.5623}{\sqrt{100\,000}}\approx 0.0113$, about one ninth of the expected value. Randomness dominates for one and vanishes for the other. This is exactly the mechanism casinos and insurance companies rely on: the same expected value becomes a near certainty once there are enough repetitions. Here it works against the operator, who is almost sure to lose about 0.1 unit per play over 100,000 plays; a real casino sets its prices so that this near certainty is a gain.

The standard error is $\frac{\sigma}{\sqrt{n}}=\frac{8}{\sqrt{25}}=1.6$ grams. The decision rule rejects $H_{0}$ when $\overline{x}<500-1.645\times 1.6=500-2.632=497.368$ grams.

The type I error consists in concluding that the machine underfills when it is in fact correctly set to 500 grams. Its probability is $\alpha=0.05$ by construction: the inspector chose it by setting the significance level. In concrete terms, a perfectly compliant machine will be wrongly accused once in twenty.

The type II error consists in not rejecting $H_{0}$, and so letting the machine pass, when it really does underfill. If the true mean is 496, $\beta=P\left(\overline{X}>497.368\ |\ \mu=496\right)=P\left(Z>\frac{497.368-496}{1.6}\right)=P(Z>0.855)\approx 0.1963$. The power is therefore $1-\beta\approx 0.8037$: the test detects this misadjustment about four times out of five.

With $\mu=495$: $z=\frac{497.368-495}{1.6}=1.48$, so $\beta\approx 0.0694$ and the power reaches about $0.9306$. The power increases as the misadjustment gets worse, which is reassuring and perfectly logical: the larger the gap to detect, the easier it is to detect. A test therefore never has ONE power, it has one for each value of the alternative hypothesis, and stating a power without saying against what makes no sense.

Three options. Lowering $\alpha$ reduces the first risk but moves the cutoff down, and so increases $\beta$: for a fixed sample size the two risks move in opposite directions, you only slide the cursor, and the trade-off is then settled by the cost of each error, a compliant machine wrongly accused against an underserved customer who goes unnoticed. Accepting a larger misadjustment as the detection threshold reduces $\beta$ but amounts to giving up on detecting small deviations. The only option that truly improves both at once is to INCREASE $n$: the standard error decreases like $\frac{1}{\sqrt{n}}$, the two distributions tighten and overlap less, so that $\alpha=0.05$ can be kept while $\beta$ goes down. It is the only lever in statistics that trades no risk for another, and it is paid for in data collection.

The population standard deviation is unknown and replaced by the sample's: the statistic then follows a Student t distribution with $n-1=15$ degrees of freedom, not a normal distribution. We compute $t=\frac{\overline{x}-\mu_{0}}{s/\sqrt{n}}=\frac{104.5-100}{8/4}=\frac{4.5}{2}=2.25$.

At the 5% level, $|t|=2.25>2.131$: reject $H_{0}$, the mean lifetime differs significantly from 100 hours. At the 1% level, $|t|=2.25<2.947$: do not reject $H_{0}$. The conclusion therefore depends on the level, which is normal and must be stated, never hidden.

The interval is $\overline{x}\pm t\times\frac{s}{\sqrt{n}}$, with $\frac{s}{\sqrt{n}}=2$. At 95%: $104.5\pm 2.131\times 2=104.5\pm 4.262$, that is $[100.238, 108.762]$. At 99%: $104.5\pm 2.947\times 2=104.5\pm 5.894$, that is $[98.606, 110.394]$.

The 95% interval does NOT contain the value 100, and the test at the 5% level rejects. The 99% interval contains 100, and the test at the 1% level does not reject. The two approaches agree perfectly, and this is no coincidence: a two-sided test at level $\alpha$ rejects $H_{0}:\mu=\mu_{0}$ if and only if the confidence interval at level $1-\alpha$ does not contain $\mu_{0}$. Both calculations handle the same quantity, one starting from $\mu_{0}$ and going toward the data, the other starting from the data to delimit the compatible values of $\mu_{0}$. The interval is in fact more informative: it does not only say that 100 is rejected, it gives all the values that would not be.

With $n=16$, we must assume that the lifetime itself follows an approximately normal distribution in the population: Student's t distribution is exact only under this condition, and the central limit theorem has not had time to act on such a small sample. With $n=200$, this assumption would become practically unnecessary, because the central limit theorem makes the mean approximately normal whatever the original distribution. This is the real issue with small samples: it is not only that the intervals are wide, it is that their very validity rests on an assumption that cannot be checked with so little data.

$H_{0}$: the die is fair, each face has probability $\frac{1}{6}$. $H_{1}$: at least one face departs from this probability. Under $H_{0}$, the expected frequency is $120\times\frac{1}{6}=20$ for each of the six faces. The validity condition is satisfied since all the expected frequencies exceed 5.

The squared deviations are $25$, $4$, $16$, $49$, $36$ and $4$, with a sum of 134. So $\chi^{2}=\frac{134}{20}=6.7$. Since $6.7<11.070$, we do not reject $H_{0}$: nothing allows us to claim that the die is loaded. Be careful with the wording: we do not conclude that the die is fair.

The expected frequencies become $\frac{600}{6}=100$. The squared deviations are $625$, $100$, $400$, $1225$, $900$ and $100$, with a sum of 3350, so $\chi^{2}=\frac{3350}{100}=33.5$. This time $33.5>11.070$ and $H_{0}$ is rejected by a wide margin: the die is loaded.

Because the chi-square statistic is proportional to the sample size when the proportions stay constant. Multiplying the frequencies by 5 multiplies each deviation $O-E$ by 5 as well, and its square by 25, while the denominator $E$ is only multiplied by 5: the ratio is therefore multiplied by 5, and we do get $5\times 6.7=33.5$. The critical value, for its part, depends only on the number of degrees of freedom and does not move. In other words, the same relative imbalance becomes detectable as soon as it is observed long enough: a test does not measure the size of a deviation, it measures our ability to rule out chance, and that ability grows with the data.

The difference lies in how the expected frequencies are computed and in the number of degrees of freedom. In a goodness-of-fit test, the theoretical distribution is given in advance, here equal probabilities, and $k$ categories are compared, hence $k-1=5$ degrees of freedom. In a test of independence, no distribution is supplied: the expected frequencies are built from the marginal totals of the table itself, by $E=\frac{\text{row total}\times\text{column total}}{\text{grand total}}$, and the number of degrees of freedom becomes $(r-1)(c-1)$, for $r$ rows and $c$ columns. The statistic is then computed the same way in both cases.

Program A: men obtain $\frac{240}{400}=60.0\%$ and women $\frac{65}{100}=65.0\%$. Program B: men obtain $\frac{20}{100}=20.0\%$ and women $\frac{100}{400}=25.0\%$. In each of the two programs, without exception, women are admitted more often than men.

Overall, men total $240+20=260$ admissions for $400+100=500$ applications, that is $52.0\%$. Women total $65+100=165$ admissions for $100+400=500$ applications, that is $33.0\%$. Overall, then, it is men who are admitted far more often, and the gap is huge.

Program A admits $\frac{305}{500}=61.0\%$ of its applicants, program B only $\frac{120}{500}=24.0\%$: A is not very selective, B is highly selective. Now men submitted 400 of their 500 applications to A, that is 80%, while women submitted 400 of their 500 to B, also 80%, but in the difficult program.

The overall rate of a group is an average of the rates per program, weighted by how its applications are distributed. Men applied massively where 61% of applicants are admitted, women where only 24% are. This difference in weighting, which is considerable, outweighs the five-point advantage women enjoy in each program. This is Simpson's paradox: an association observed in every subgroup can reverse in the aggregate, as soon as a third variable, here the choice of program, is linked both to the group and to the outcome.

The accusation, as worded, is not justified: in admissions, that is, once the program is chosen, women are favoured everywhere. What the data show is an overall gap of 19 points produced entirely by how the applications are distributed. What the data cannot tell us is WHY this distribution is what it is, nor whether some upstream mechanism, academic guidance, information, stereotypes, steers women toward the more selective program. The question therefore remains open, but it moves from admission to application. The methodological lesson is general: never aggregate subgroups without checking whether membership in a subgroup itself depends on the variable under study.

We have $\overline{x}=\frac{36}{6}=6$ and $\overline{y}=\frac{398}{6}\approx 66.333$. Then $S_{xx}=\sum(x-\overline{x})^{2}=16+4+1+1+4+16=42$, $S_{yy}\approx 573.333$ and $S_{xy}=155$. The slope is $b=\frac{155}{42}\approx 3.6905$ and the y-intercept $a=66.333-3.6905\times 6\approx 44.1905$. The line is therefore $\hat{y}\approx 3.6905x+44.1905$: each additional hour of study is worth about 3.7 points.

$r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\frac{155}{\sqrt{42\times 573.333}}\approx 0.99886$, so $r^{2}\approx 0.9977$. About 99.8% of the variation in marks is explained by the variation in the number of hours. That is extraordinarily high for human data, and it should raise suspicion: the marks of a real group of students never line up this well, and a fit computed on only six points says nothing about whether the relationship stays linear outside the observed hours, nor whether it would hold up on another group.

$S_{yy}-bS_{xy}=573.333-3.6905\times 155\approx 1.3095$, the residual sum of squares. So $s_{e}=\sqrt{\frac{1.3095}{6-2}}=\sqrt{0.3274}\approx 0.5722$ point.

The point prediction is $\hat{y}=3.6905\times 6+44.1905\approx 66.333$, and since $x_{0}=\overline{x}$ the term $(x_{0}-\overline{x})^{2}$ vanishes. The confidence interval for the mean is $\hat{y}\pm t\,s_{e}\sqrt{\frac{1}{n}}=66.333\pm 2.776\times 0.5722\times 0.4082\approx 66.333\pm 0.648$, that is $[65.685, 66.982]$. The prediction interval is $\hat{y}\pm t\,s_{e}\sqrt{1+\frac{1}{n}}=66.333\pm 2.776\times 0.5722\times 1.0801\approx 66.333\pm 1.716$, that is $[64.618, 68.049]$. The second is almost three times wider because it combines two uncertainties: the one about the position of the line, the only one present in the first, AND the individual spread of students around this line, measured by $s_{e}$. That is the 1 under the square root. A mean can be known very precisely while an individual remains impossible to predict.

The model gives $\hat{y}=3.6905\times 20+44.1905\approx 118$, which exceeds the maximum of 100 and therefore makes no sense. Two mistakes add up. First, extrapolation: the data cover the range from 2 to 10 hours, and nothing guarantees that the relationship stays linear beyond it, whereas we would be doubling the observed range. Second, common sense: a mark is bounded, so the relationship must bend and then level off, which a straight line will never do. A regression model is valid only on the domain over which it was fitted, and a coefficient of determination of 99.8% says strictly nothing about what happens outside it.

False: it is the coefficient of DETERMINATION that is interpreted as a percentage of variation explained, and it equals $r^{2}=0.9^{2}=0.81$, that is 81% and not 90%. The confusion is all the more stubborn as the two numbers are close when $r$ is high, but it becomes glaring elsewhere: for $r=0.5$, only 25% of the variation is explained, not half.

False as stated. In the classical approach, the true mean is a fixed, unknown number: it is either in the interval or it is not, and there is no probability to assign to it. What is worth 95% is the reliability of the PROCEDURE: if the sampling were repeated a large number of times and such an interval built each time, 95% of them would contain the true mean. The correct wording is about the method, not about the particular interval in front of you.

False, and it is the same inversion of conditional probabilities as in exercise 4. The p-value is the probability of observing data at least as extreme as ours IF $H_{0}$ were true, that is $P(\text{data}\ |\ H_{0})$. The claim reads it as $P(H_{0}\ |\ \text{data})$, which is an entirely different quantity and which, in the classical framework, does not even make sense since $H_{0}$ is not a random variable.

False. The margin of error depends on the size of the SAMPLE, not on that of the population, as long as the sample represents less than 5% of the population, which is very much the case here. With $n=1000$ in both countries, the margin is about $\frac{1.96\times 0.5}{\sqrt{1000}}\approx 0.031$, that is about 3.1 points, the same in both. This is exactly the point established in exercise 3.

False, and it is the error with the heaviest consequences in the whole course. Not rejecting $H_{0}$ means that the data do not provide sufficient evidence of a difference, which is not evidence that there is no difference. A test can fail to reject simply because it lacks power: in exercise 7, a real misadjustment of 4 grams went unnoticed nearly two times out of ten. To claim equivalence, you need either a specific equivalence test, or at the very least to show that the confidence interval for the difference lies entirely within a zone judged negligible. The correct wording is always: “these data do not allow us to conclude that there is a difference.”