Corrected exercises: confidence intervals (201-SN1-RE)

Standard error: $\dfrac{s}{\sqrt{n}}=\dfrac{9}{6}=1.5$.

Margin: $E=1.96\times 1.5=2.94$.

With $n\ge 30$, the unknown $\sigma$ is replaced by $s$ and we keep $z$.

$E=1.645\times 1.5=2.4675$, so $[61.73, 66.67]$.

Less confidence, narrower interval.

$E=2.576\times 1.5=3.864$, so $[60.34, 68.06]$.

More confidence, wider interval.

A 98% level leaves $2\%$ outside the interval, that is $1\%$ in each tail.

So we look for $z$ such that $P(Z\le z)=0.99$. The table gives $0.9901$ for $z=2.33$: $z_{0.01}\approx 2.33$ (finer value: $2.326$).

Pitfall: looking for $P(Z\le z)=0.98$, which gives $2.05$, the critical value of a 96% interval.

Standard error: $\dfrac{1.8}{\sqrt{9}}=0.6$.

Margin: $E=2.306\times 0.6=1.3836$.

Small sample and $\sigma$ unknown: Student with $n-1=8$ degrees of freedom, under the assumption of a normal population.

$\hat{p}=\dfrac{126}{300}=0.42$; conditions: $n\hat{p}=126\ge 5$ and $n(1-\hat{p})=174\ge 5$.

Standard error: $\sqrt{\dfrac{0.42\times 0.58}{300}}\approx 0.02850$.

Margin: $E=1.96\times 0.02850\approx 0.0559$, so $[0.364, 0.476]$.

The interval is symmetric about the estimate: $\overline{x}=\dfrac{12.6+15.0}{2}=13.8$ and $E=\dfrac{15.0-12.6}{2}=1.2$.

$1.96\dfrac{6}{\sqrt{n}}\le 1 \iff n\ge(1.96\times 6)^{2}=138.2976$, so $n=139$.

Pitfall: rounding to $138$ gives a margin of $1.0011$, above the target.

$n\ge\dfrac{1.96^{2}\times 0.5\times 0.5}{0.03^{2}}\approx 1067.1$, so $n=1068$.

$p=0.5$ makes $p(1-p)$ as large as possible: the size obtained works whatever the true proportion.

$\mu=\dfrac{92+98+104+110}{4}=\dfrac{404}{4}=101$ mm.

Deviations from the mean: $-9$, $-3$, $3$, $9$.

$\sigma^{2}=\dfrac{81+9+9+81}{4}=\dfrac{180}{4}=45$ mm², so $\sigma=\sqrt{45}\approx 6.71$ mm.

Divide by $N$ and not by $N-1$: these are ALL the values of the population, not a sample.

Each draw has $4$ equally likely outcomes, so there are $4\times 4=16$ ordered samples, each with probability $\frac{1}{16}$.

The sum $X_{1}+X_{2}$ goes from $184$ to $220$ in steps of $6$, so $\overline{X}$ takes $7$ values: $92$, $95$, $98$, $101$, $104$, $107$, $110$.

Count the samples: $1$, $2$, $3$, $4$, $3$, $2$, $1$ out of $16$. For example $\overline{X}=101$ comes from $(92,110)$, $(110,92)$, $(98,104)$ and $(104,98)$: $P(\overline{X}=101)=\frac{4}{16}=0.25$.

The population had four equally likely values; the mean has seven, and it concentrates in the centre.

By symmetry about $101$: $E(\overline{X})=101=\mu$.

Deviations from $101$: $\pm 9$ (probability $\frac{1}{16}$ each), $\pm 6$ ($\frac{2}{16}$), $\pm 3$ ($\frac{3}{16}$), $0$ ($\frac{4}{16}$).

And $\dfrac{\sigma^{2}}{2}=\dfrac{45}{2}=22.5$: the variance of the mean of two draws is exactly half that of the population.

This is the thread of the set: the standard deviation of one scallop is $6.71$ mm, that of the mean of two scallops $\sqrt{22.5}\approx 4.74$ mm.

With replacement, $X_{1},\dots,X_{n}$ all have the distribution of the population, so $E(X_{i})=\mu$ and $V(X_{i})=\sigma^{2}$, and they are independent.

Expected value, by linearity (true with no assumption at all): $E(\overline{X})=\dfrac{1}{n}\big(E(X_{1})+\dots+E(X_{n})\big)=\dfrac{n\mu}{n}=\mu$.

Variance: $V(aY)=a^{2}V(Y)$ gives $V(\overline{X})=\dfrac{1}{n^{2}}V(X_{1}+\dots+X_{n})$.

Since the $X_{i}$ are independent, the variance of their sum is the sum of the variances: $V(X_{1}+\dots+X_{n})=n\sigma^{2}$.

$V(\overline{X})=\dfrac{n\sigma^{2}}{n^{2}}=\dfrac{\sigma^{2}}{n}$, so $\sigma_{\overline{X}}=\dfrac{\sigma}{\sqrt{n}}$.

Pitfall: invoking independence for the expected value. It is needed only for the variance; without it, covariances are added to the sum of the variances.

Without replacement, the pairs $\{92,98\}$, $\{92,104\}$, $\{92,110\}$, $\{98,104\}$, $\{98,110\}$, $\{104,110\}$ give $\overline{X}=95$, $98$, $101$, $101$, $104$, $107$.

$E(\overline{X})=101$ again; deviations $-6$, $-3$, $0$, $0$, $3$, $6$.

$V(\overline{X})=\dfrac{36+9+0+0+9+36}{6}=\dfrac{90}{6}=15$ mm²

We recover $22.5\times\dfrac{N-n}{N-1}=22.5\times\dfrac{2}{3}=15$: without replacement, the second draw cannot repeat the first, the draws are dependent and the mean varies less. This correction factor goes beyond the minimum of the official course description.

On a large population (usual rule: $n$ below $5\%$ of $N$), removing one individual changes almost nothing, $\frac{N-n}{N-1}$ is close to $1$ and the formula with replacement remains a good approximation. That is what allows it to be used on a lake, a forest or a bird population.

$\sigma_{\overline{X}}=\dfrac{\sigma}{\sqrt{8}}=\dfrac{\sqrt{45}}{\sqrt{8}}=\sqrt{5.625}\approx 2.37$ mm

To divide $\dfrac{\sigma}{\sqrt{n}}$ by $2$, $\sqrt{n}$ must be multiplied by $2$, so $n$ by $4$: $32$ scallops instead of $8$.

The precision of a mean is expensive: it improves like the square root of the effort.

In the normal model, $z=\dfrac{0-0.42}{0.22}\approx -1.91$.

Nearly $3\%$ of the fish would have a negative content, which is impossible. The histogram, very skewed with a long tail towards high contents, confirms it: the normal distribution does not describe the content of a single walleye.

Pitfall: believing that the central limit theorem makes the data “normal”. It is about the MEAN, never about the individuals.

The sampling is treated as $40$ independent draws from the population (exercise 1, question d).

$E(\overline{X})=\mu=0.42$ mg/kg

$\sigma_{\overline{X}}=\dfrac{\sigma}{\sqrt{n}}=\dfrac{0.22}{\sqrt{40}}\approx 0.0348$ mg/kg

These two results require no assumption about the shape of the distribution.

Central limit theorem: if $X_{1},\dots,X_{n}$ are independent, identically distributed, with finite expected value $\mu$ and standard deviation $\sigma$, then, for $n$ large enough, $\overline{X}$ approximately follows the normal distribution with mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{n}}$.

“Large enough” translates in practice into $n\ge 30$; the more skewed the population, the more observations are needed for the approximation to be good.

Here the draws are treated as independent and $n=40\ge 30$: $\overline{X}$ approximately follows $N\big(0.42, 0.0348^{2}\big)$, even though $X$ is not normal.

Standardize with the standard error, not with $\sigma$:

A random sample of $40$ walleyes has only about a $1\%$ chance of having a mean above $0.5$ mg/kg, whereas the figure shows that a good share of the walleyes taken one by one exceed this value.

Pitfall: dividing by $\sigma=0.22$, which gives $z\approx 0.36$ and a probability of $0.36$: that is the probability that a single walleye exceeds the reference value in a normal model that does not even suit it.

The condition threatened is independence. Walleyes caught together in the same bay ate the same prey in the same waters: their contents resemble each other, and the sample poorly represents the other parts of the reservoir.

The real variability of the mean of such a sample, around the mean of the reservoir, is then larger than $\dfrac{\sigma}{\sqrt{40}}$, and the bay may even be more or less contaminated than the rest.

The calculation in d), which assumes $40$ independent pieces of information, is therefore probably too optimistic: it only holds under its conditions.

At least $121$ walleyes are needed, three times more than $40$, for a standard deviation not even half as large.

The parameter is the mean $\mu$ of the ozone maxima over the $920$ summer days, a fixed but unknown number.

Its point estimate is the sample mean: $\overline{x}=45.9$ ppb.

The mean of the $36$ days drawn is not the parameter: it would change with another draw.

$\dfrac{s}{\sqrt{n}}=\dfrac{10.9}{\sqrt{36}}=\dfrac{10.9}{6}\approx 1.82$ ppb

A summer day typically deviates by $10.9$ ppb from the mean; the mean of $36$ days, six times less.

Margin: $E=1.96\times 1.817\approx 3.56$ ppb.

$[45.9-3.56, 45.9+3.56]\approx[42.3, 49.5]$ ppb

Write-up: “At the 95% confidence level, the mean of the ozone maxima of summer days lies between $42.3$ and $49.5$ ppb.”

At 90%: $E=1.645\times 1.817\approx 2.99$, so $[42.9, 48.9]$, width $6.0$ ppb.

At 99%: $E=2.576\times 1.817\approx 4.68$, so $[41.2, 50.6]$, width $9.4$ ppb.

Widths: $6.0<7.1<9.4$ ppb. With the same sample, greater confidence is obtained only by widening the interval: what is gained in safety is lost in precision.

To be both safer and more precise, more data are needed (exercise 6).

With $n=36\ge 30$, two things come into play. The central limit theorem makes $\overline{X}$ approximately normal, whatever the shape of the histogram. And $s$, computed on $36$ values, is a stable enough estimate of $\sigma$ to be substituted for $\sigma$ without changing the distribution.

Student's t distribution with $35$ degrees of freedom would in fact give $t_{0.025}\approx 2.03$ instead of $1.96$: the difference is small, and the $n\ge 30$ convention neglects it.

Pitfall: justifying $z$ by “the concentrations are normal”. Nothing says so, and the method does not need it here.

Consecutive days are not independent: a heat and smog episode lasts several days, and the whole series shares the same weather.

A single summer does not represent all ten either: if it was hotter, the estimate is biased, and no margin formula corrects a bias.

Drawing at random from the $920$ days is precisely what gives the right to write “95%”.

$\overline{x}=\dfrac{27.6}{12}=2.30$ °Brix.

Deviations from $2.30$: $-0.2$, $0.3$, $-0.5$, $0.1$, $0.6$, $-0.1$, $-0.4$, $0.2$, $0$, $-0.3$, $0.2$, $0.1$.

Sum of squares: $1.10$, so $s^{2}=\dfrac{1.10}{11}=0.1$ and $s=\sqrt{0.1}\approx 0.316$ °Brix.

Pitfall: dividing by $12$, which gives $0.303$ and underestimates the spread of the population.

The standard deviation $\sigma$ of the population is unknown and $n=12<30$: $s$ varies a lot from one sample to another, and the ratio $T=\dfrac{\overline{X}-\mu}{S/\sqrt{n}}$ does not follow a normal distribution.

If the population is NORMAL, $T$ follows Student's t distribution with $n-1=11$ degrees of freedom, more spread out than the normal.

On the figure, the contents spread from $1.8$ to $2.9$ with no strong skewness and no isolated value: the normality assumption is plausible. Twelve points do not prove it; they only allow us not to reject it.

Standard error: $\dfrac{s}{\sqrt{n}}=\dfrac{0.316}{\sqrt{12}}\approx 0.0913$ °Brix.

Margin: $E=2.201\times 0.0913\approx 0.201$ °Brix.

$[2.30-0.201, 2.30+0.201]\approx[2.10, 2.50]$ °Brix

At the 95% level, the mean sugar content of the sap in the sugar bush, that morning, lies between $2.10$ and $2.50$ °Brix.

With $1.96$: $E=1.96\times 0.0913\approx 0.179$, so $[2.12, 2.48]$.

This interval contains $\mu$ exactly when $|T|<1.96$, which happens with probability $0.924$ with $11$ degrees of freedom.

His method is therefore reliable only $92.4\%$ of the time while it claims $95\%$: the interval is too narrow because it ignores that $s$ itself is uncertain.

Pitfall: taking $z$ “because it is simpler”. On a small sample, that claims a precision one does not have.

The function $x\mapsto\dfrac{86}{x}$ is strictly decreasing on $(0, +\infty)$. So $2.10\le\mu\le 2.50$ is equivalent to $\dfrac{86}{2.50}\le\dfrac{86}{\mu}\le\dfrac{86}{2.10}$: the two events are THE SAME, and have the same probability of covering the true value, $95\%$.

$\dfrac{86}{2.501}\approx 34.4$ L and $\dfrac{86}{2.099}\approx 41.0$ L

The interval $[34.4, 41.0]$ is not centred on the point estimate $\dfrac{86}{2.30}\approx 37.4$ L, because the transformation is not affine.

It concerns $\dfrac{86}{\mu}$, the requirement computed from the mean content, and not the mean of the $\dfrac{86}{x}$ of the tapholes, which is a different quantity.

Each ash tree drawn is infested with probability $p$, independently of the others (with replacement): $X$ counts the successes of $250$ Bernoulli trials, so $X$ follows the binomial distribution $B(250, p)$, with $E(X)=np$ and $V(X)=np(1-p)$.