Corrected exercises: hypothesis tests (201-SN1-RE)

$n\geq 30$: replace $\sigma$ by $s$ and use the normal distribution.

The sign tells on which side of $\mu_{0}$ the sample falls; keep it.

$n<30$ and normal population: Student's t distribution with $n-1=8$ degrees of freedom.

$\hat{p}=\dfrac{230}{500}=0.46$ and, under $H_{0}$, $\sigma_{\hat{P}}=\sqrt{\dfrac{0.40\times 0.60}{500}}\approx 0.02191$.

The whole area $\alpha=0.025$ is in the left tail: the critical value leaves $2.5\%$ to its left, so it is $-z_{0.025}=-1.96$.

Pitfall: taking $-1.645$ because the test is one-tailed. $1.645$ corresponds to $5\%$ in one tail, not to $2.5\%$.

$P(Z<-2.15)=1-0.9842=0.0158$, and the test is two-tailed:

$|z|=2.05>1.96$: $z$ falls in the critical region, so we reject $H_{0}$ at the $5\%$ level.

We reject when $\dfrac{\overline{x}-200}{15/\sqrt{25}}>1.645$, that is, when

So $c\approx 204.94$.

Student's t distribution with $14$ degrees of freedom: we reject when $|t|>2.145$. Here $|t|=2.40>2.145$: we reject $H_{0}$.

Parameter $\mu$, the mean yield of all treated plots.

$H_{0}:\mu=3.20$ and $H_{1}:\mu>3.20$.

Right-tailed test, because only an INCREASE in yield interests the agronomist; a decrease would not lead him to adopt the fertilizer either, but that is not what he is trying to establish.

Parameter $\mu$, the mean lead concentration at the tap in the network.

$H_{0}:\mu=8$ and $H_{1}:\mu<8$.

Left-tailed test, because it is compliance that must be PROVED: the water is declared safe only if the data place the mean below the standard.

Parameter $\mu$, the mean total phosphorus concentration of the lake this summer.

$H_{0}:\mu=20$ and $H_{1}:\mu>20$.

Right-tailed test, because the association is only worried about an exceedance: a concentration lower than the threshold calls for no action.

Parameter $\mu$, the mean of all the measurements the setup would give.

$H_{0}:\mu=9.81$ and $H_{1}:\mu\neq9.81$.

Two-tailed test, because a bias in either direction is a defect of the setup: we do not know in advance on which side it falls.

Parameter $\mu$, the mean reaction time with the catalyst.

$H_{0}:\mu=120$ and $H_{1}:\mu<120$.

Left-tailed test, because the catalyst must REDUCE the duration; a longer duration would not make it useful.

Parameter $p$, the proportion of malformed frogs in the pond.

$H_{0}:p=0.05$ and $H_{1}:p>0.05$.

Right-tailed test, because the suspicion is of a rate higher than that of the control ponds.

Parameter $\mu$, the current mean flowering duration.

$H_{0}:\mu=18$ and $H_{1}:\mu<18$.

Left-tailed test, because the question is a decrease; a longer duration would be a different study.

The hypotheses are about an unknown PARAMETER of the population: here $\mu$, the mean yield of all the plots that would be treated with the fertilizer.

$\overline{x}$ is a sample statistic: once the harvest is weighed, its value is known, and a hypothesis about it makes no sense.

The correct statement is $H_{0}:\mu=3.20$ and $H_{1}:\mu>3.20$. The equality goes in $H_{0}$ even if we think “$\mu\leq 3.20$”: it is the value under which all the calculations are done, the boundary case between the two hypotheses.

Pitfall: writing the claim to be established in $H_{0}$. A test can only reject $H_{0}$ or fail to reject it; what we want to establish therefore goes in $H_{1}$.

A test rejects $H_{0}$ only if the data are clearly incompatible with it. With $H_{1}:\mu>8$, a small, inconclusive sample leads to not rejecting $\mu=8$: contamination is not proved, and the water stays in service.

The water would therefore be declared acceptable BY DEFAULT, simply for lack of data. With $H_{1}:\mu<8$, it is the reverse: as long as compliance is not established, the water is not declared compliant.

The choice of $H_{1}$ decides who carries the burden of proof. In public health, it is compliance that must be proved, and exercise 6 puts numbers on the two errors this choice distributes.

The parameter is $\mu$, the mean yield of ALL the station's plots that would be treated with the fertilizer.

$H_{0}:\mu=3.20$ t/ha (the fertilizer changes nothing) and $H_{1}:\mu>3.20$ t/ha (it increases the mean yield). Right-tailed test.

The central limit theorem is about the MEAN: for $n\geq 30$ independent draws, $\overline{X}$ is approximately normal with mean $\mu$ and standard deviation $\dfrac{\sigma}{\sqrt{n}}$, whatever the shape of the distribution of yields, provided $\sigma$ is finite.

The skewness of the histogram describes the DATA; it does not rule anything out for the distribution of $\overline{X}$. With $n=40$, the normal approximation is reasonable, all the more so since the skewness remains moderate.

For the same reason, with $n\geq 30$, $s$ estimates $\sigma$ precisely enough for us to keep the normal distribution: this is the course's convention for a mean with $n\geq 30$.

Pitfall: concluding that a test is impossible because the data are not normal. That condition is needed only for $n<30$ (exercise 3).

$\sigma_{\overline{X}}\approx\dfrac{s}{\sqrt{n}}=\dfrac{0.52}{\sqrt{40}}\approx 0.0822$ t/ha

The calculation is done UNDER $H_{0}$: the distribution of $\overline{X}$ is centred on $\mu_{0}=3.20$, and it is the gap from this value that we measure in standard errors.

Right-tailed test at the $5\%$ level: we reject $H_{0}$ when $z>1.645$.

In terms of yield: $\dfrac{\overline{x}-3.20}{0.0822}>1.645 \iff \overline{x}>3.20+1.645\times 0.0822\approx 3.335$ t/ha.

The critical region is set BEFORE looking at $\overline{x}$: it depends only on $\mu_{0}$, the standard error and the level.

$z\approx 2.19>1.645$, or equivalently $\overline{x}=3.38>3.335$: we reject $H_{0}$.

Conclusion: at the $5\%$ level, the data from the $40$ plots allow us to conclude that the fertilizer increases the mean yield of this variety at this station.

The conclusion is about the mean yield and the station studied; it says nothing about another variety or another soil.

If the fertilizer had no effect, a mean yield of at least $3.38$ t/ha over $40$ plots would appear in only about $1.4\%$ of samples.

$p=0.0143<0.05$: we reject $H_{0}$, as in e). The two rules, critical region and $p$-value, always give the same decision.

At the $1\%$ level, we reject when $z>2.326$, that is, $\overline{x}>3.20+2.326\times 0.0822\approx 3.391$ t/ha.

$\overline{x}=3.38<3.391$ (or $p=0.0143>0.01$): we do not reject $H_{0}$.

Conclusion: at the $1\%$ level, the data do not allow us to conclude that the mean yield has increased.

Pitfall: writing “at the $1\%$ level, the fertilizer has no effect”. Not rejecting $H_{0}$ does not prove it: the same data show a likely effect, just not clear enough for such a demanding level.

$\overline{x}=\dfrac{211.0}{10}=21.10$ µg/L.

$s=\sqrt{\dfrac{\sum (x_{i}-\overline{x})^{2}}{n-1}}=\sqrt{\dfrac{42.02}{9}}\approx 2.16$ µg/L

On the calculator: statistics mode, list of the ten values, and read $s_{x}$ (not $\sigma_{x}$, which divides by $n$).

With $n=10<30$, the central limit theorem no longer guarantees that $\overline{X}$ is normal, and $s$, computed on ten values, is an uncertain estimate of $\sigma$.

If the population of concentrations is NORMAL, the statistic $T=\dfrac{\overline{X}-\mu_{0}}{S/\sqrt{n}}$ follows exactly a Student's t distribution with $n-1=9$ degrees of freedom: that is the condition that replaces $n\geq 30$.

The plot does not prove normality, but it makes it plausible: a single hump, no strong skewness, no value isolated far from the others.

Pitfall: using $z_{0.05}=1.645$. The tails of Student's t distribution are heavier; its critical value, $1.833$, is larger.

Right-tailed test, $9$ degrees of freedom: we reject $H_{0}$ if $t>1.833$.

$t\approx 1.61<1.833$: we do not reject $H_{0}$.

Conclusion: at the $5\%$ level, these ten measurements do not allow us to conclude that the mean phosphorus concentration of the summer exceeds $20$ µg/L.

Pitfall: writing “the lake meets the threshold”. The observed mean, $21.10$, is above $20$; the test only says that this gap may come from the chance of ten samples.

The $p$-value is the area to the right of $t=1.61$ under Student's t distribution with $9$ degrees of freedom.

$1.383<1.61<1.833$, so $0.05<p<0.10$.

A Student table only gives a bracket; software would give $p\approx 0.071$, consistent with the bracket.

Without $17.2$: $\overline{x}\approx 21.53$, $s\approx 1.77$, $n=9$, so $8$ degrees of freedom and

This second test would reject $H_{0}$, but it is worthless: the measurement was removed AFTER seeing that it stood in the way of the desired conclusion. The data were chosen to get the result.

A measurement may be discarded only for a documented experimental reason (contaminated bottle, protocol error), decided without looking at the effect on the test. Otherwise, the decision remains that of d).

Parameter: $p$, the proportion of malformed frogs in the whole pond.

$H_{0}:p=0.05$ and $H_{1}:p>0.05$: right-tailed.

$np_{0}=220\times 0.05=11\geq 5$ and $n(1-p_{0})=220\times 0.95=209\geq 5$: under $H_{0}$, $\hat{P}$ is approximately normal.

The conditions are checked with $p_{0}$, the value under which we calculate. The captures must also be random and independent: if the team had caught only the slow frogs, more often malformed, the proportion would be biased, and no calculation would fix it.

We reject $H_{0}$ if $z>2.326$. Here $z\approx 2.47>2.326$: we reject $H_{0}$.

Conclusion: at the $1\%$ level, the data allow us to conclude that the proportion of malformed frogs in this pond exceeds the $5\%$ rate of the control ponds.

If the pond's rate were $5\%$, we would find $19$ or more malformed frogs out of $220$ in only about $0.7\%$ of samples. $p<0.01$: same decision as in d).

With $\hat{p}$: $\sqrt{\dfrac{0.0864\times 0.9136}{220}}\approx 0.0189$ and $z\approx\dfrac{0.0364}{0.0189}\approx 1.92<2.326$: he would not reject $H_{0}$.

Yet the whole test reasons UNDER $H_{0}$: we compute the distribution of $\hat{P}$ assuming $p=0.05$, and in that world the standard error is $\sqrt{\frac{0.05\times 0.95}{220}}$. Putting $\hat{p}$ in mixes two hypotheses in the same calculation.

Pitfall: the confidence interval formula, which uses $\hat{p}$ because there is no assumed value. Here the mistake costs the whole conclusion, as the figure shows.

The further $\hat{p}$ is from $p_{0}$, the more the two standard errors differ, and the more this choice weighs on the decision.

The test establishes only one thing: in this pond, the proportion of malformed frogs exceeds $5\%$, with a Type I error risk of $1\%$.

It says nothing about the CAUSE: the study is observational. The pond differs from the control ponds by the neighbouring fields, but perhaps also by its temperature, its parasites (some trematodes cause limb malformations) or its fauna.

To attribute the increase to the pesticides, one would need to measure their concentration in several ponds, or run a controlled laboratory experiment. The report should say “is associated with the proximity of the fields, without the cause being established”.

Parameter: $\mu$, the mean of all the values of $g$ this setup would give. $H_{0}:\mu=9.81$ m/s² (no bias) and $H_{1}:\mu\neq 9.81$ m/s².

Two-tailed: we are looking for a defect of the setup, and we do not know in advance whether it overestimates or underestimates $g$. The choice is made BEFORE seeing that $\overline{x}<9.81$.

$n=50\geq 30$: $\sigma_{\overline{X}}\approx\dfrac{0.092}{\sqrt{50}}\approx 0.0130$ m/s².

Two-tailed at the $5\%$ level: we reject if $|z|>1.96$. Here $|z|\approx 2.77>1.96$: we reject $H_{0}$.

If the setup had no bias, a mean of $50$ measurements at least as far from $9.81$, on one side or the other, would appear in only about $0.56\%$ of cases.

The interval is $[9.748,\,9.800]$ m/s²: it does not contain $9.81$, which gives the same decision as in b).

Let $e=\sigma_{\overline{X}}$ be the standard error, the same in the test and in the interval. The two-tailed test rejects $H_{0}$ if and only if

That is exactly the confidence interval of level $1-\alpha$. Each step is an equivalence, so the two rules always give the same decision.

Two conditions hold this equivalence: a TWO-TAILED test, and the same standard error on both sides. For a proportion, the test uses $p_{0}$ and the interval uses $\hat{p}$ (exercise 4): the equivalence is then only approximate.

$\overline{x}\pm 2.576\times 0.01301=9.774\pm 0.0335$, that is, $[9.740,\,9.808]$ m/s².

$9.81>9.808$: the reference value lies outside the $99\%$ interval, so we reject $H_{0}$ at the $1\%$ level, with no other calculation. This is consistent with $p=0.0056<0.01$.

The margin is thin: $9.81$ is only $0.002$ m/s² from the bound. Rounding the standard error too early to $0.013$ would be enough to blur the reading.

Conclusion: at the $1\%$ level, the data from the $50$ measurements allow us to conclude that the setup gives, on average, a value of $g$ different from $9.81$ m/s², and lower.

$g=\frac{4\pi^{2}L}{T^{2}}$ underestimates $g$ if $T$ is too large. The formula is valid only for small oscillations: at large amplitude, the real period lengthens, which lowers $g$ in every measurement, in the same direction.

A random timing error spreads the measurements without shifting them all to the same side: it increases $s$, not the mean gap. Measuring $L$ to the centre of the mass is the correct practice.

Pitfall: concluding that $g$ equals $9.774$ m/s² in the laboratory. The test judges the SETUP, not the value of $g$.