Unit S2: Statistics 2

Binomial and Poisson distributions both model counts, but their random processes differ. Choose the model from the conditions before calculating a probability, and keep the count interval and event inequality explicit.

Translate words into integer events before using a formula or cumulative table: 'fewer than 3' is $P(X\le2)$, 'at least 6' is $1-P(X\le5)$, and $4<X<10$ is $P(X\le9)-P(X\le4)$. For $X\sim Po(4.5)$, the last probability is $0.9829-0.5321=0.4508$.

Poisson means scale with exposure. If complaints occur at mean 3 per day, then a 7-day total has distribution $Po(21)$. More generally, independent Poisson counts add: if $X\sim Po(\lambda_1)$ and $Y\sim Po(\lambda_2)$ independently, then $X+Y\sim Po(\lambda_1+\lambda_2)$.

The mean gives the long-run expected count and the variance measures its spread. Binomial and Poisson models have characteristic mean–variance relationships that can also identify unknown parameters or test plausibility.

If a binomial variable has mean 90 and variance 36, then$$np=90,\qquad np(1-p)=36.$$Dividing the second equation by the first gives $1-p=36/90=0.4$, so $p=0.6$ and $n=90/0.6=150$. Check that $n$ is a positive integer and $0\le p\le1$.

For $Y\sim B(20,0.3)$, $E(Y)=6$ and $\operatorname{Var}(Y)=4.2$. For $W\sim Po(4)$, both the mean and variance are 4. An observed count dataset whose sample mean is far from its sample variance may therefore cast doubt on a Poisson model, although closeness alone does not prove all Poisson assumptions.

When $X\sim B(n,p)$ has many trials and a small success probability, the rare-success count can be approximated by a Poisson variable with the same mean.

First check that the binomial conditions hold and that the context really describes many independent opportunities for a rare event. Then calculate $\lambda=np$, preserve the original integer event exactly, and evaluate it with the Poisson formula or table.

If $X\sim B(200,0.012)$, use $Y\sim Po(2.4)$. Then$$P(X\le1)\approx P(Y\le1)=e^{-2.4}(1+2.4)=0.3084.$$For 'at least 2', use the complementary Poisson event $1-P(Y\le1)$.

A continuous random variable can take any real value in an interval, so it models a measurement rather than a count. Waiting time, mass and length are typical examples; the number of calls is discrete even when calls arrive through time.

Probability is assigned to intervals, not isolated values. For every particular number $a$, $P(X=a)=0$. Consequently, endpoint choices do not change a continuous probability:$$P(a<X<b)=P(a\le X<b)=P(a<X\le b)=P(a\le X\le b).$$

This does not say that observing a particular rounded value is impossible. A recorded value such as 2.3 seconds represents a measurement interval determined by the instrument's precision; the exact mathematical point still has probability zero.

A probability density function $f$ describes probability through area. It must satisfy $f(x)\ge0$ on its support and have total area 1. For a continuous random variable,$$P(a<X\le b)=\int_a^b f(x)\,dx.$$The height $f(x)$ is not itself the probability $P(X=x)$, which is zero.

A cumulative distribution function is non-decreasing, remains between 0 and 1, and runs from 0 below the support to 1 above it. Once $F$ is known, interval probabilities are differences:$$P(a<X\le b)=F(b)-F(a),\qquad P(X>b)=1-F(b).$$

For a simple piecewise polynomial density$$f(x)=\begin{cases}x,&0\le x\le1,\\2-x,&1<x\le2,\\0,&\text{otherwise},\end{cases}$$both pieces are non-negative and their two triangular areas total 1. Accumulating area from the left gives$$F(x)=\begin{cases}0,&x<0,\\x^2/2,&0\le x\le1,\\2x-x^2/2-1,&1<x\le2,\\1,&x>2.\end{cases}$$The second middle expression starts with the area already accumulated up to 1; integrating that piece as though the accumulation restarted at zero would make the CDF jump or finish at the wrong value.

Where the cumulative distribution function is differentiable, its rate of increase is the density:$$f(x)=\frac{dF(x)}{dx}.$$In the other direction, integrating the density from the lower end of the support reconstructs $F$.

Suppose$$F(x)=\begin{cases}0,&x<0,\\x^2/4,&0\le x<2,\\1,&x\ge2.\end{cases}$$Differentiating each interval gives$$f(x)=\begin{cases}x/2,&0<x<2,\\0,&\text{otherwise}.\end{cases}$$Its integral over $0<x<2$ is 1, so the result is a valid density.

Because a CDF cannot decrease, a candidate derivative must not be negative. Corners at piece boundaries may make the derivative undefined at isolated points, but changing a density at finitely many individual points changes no continuous probability.

For a continuous random variable, moments are density-weighted integrals across the whole support. Split the integral wherever the density formula changes.

If $f(x)=x/2$ for $0<x<2$ and is zero otherwise, then$$E(X)=\int_0^2\frac{x^2}{2}\,dx=\frac43,$$$$E(X^2)=\int_0^2\frac{x^3}{2}\,dx=2,$$so$$\operatorname{Var}(X)=2-\left(\frac43\right)^2=\frac29.$$The variance is non-negative and has squared units.

For constants $a$ and $b$, use linear-transformation rules rather than reintegrating:$$E(aX+b)=aE(X)+b,\qquad \operatorname{Var}(aX+b)=a^2\operatorname{Var}(X).$$A shift changes the mean but not the variance.

Mode, median and quartiles describe different features of a continuous distribution. The mode is located from density height; the median and quartiles are located from cumulative probability.

For $f(x)=x/2$ on $0<x<2$, $F(x)=x^2/4$. Hence$$Q_1=1,\qquad m=\sqrt2,\qquad Q_3=\sqrt3,$$and the interquartile range is $\sqrt3-1$. Since the density increases throughout its support, its greatest height is at the upper endpoint, so the modal location is 2.

First decide which piece of a piecewise CDF contains 0.25, 0.5 or 0.75, then solve only within that piece and check the answer lies in its interval. For a mode, differentiating the density can locate interior candidates, but endpoints and any corners must also be compared.

A continuous uniform variable gives equal probability to intervals of equal length. If $X\sim U(a,b)$ with $a<b$, its density is a rectangle of width $b-a$. Unit total area fixes its height:

Probability is therefore a length ratio. Intersect the requested event with the support first: for $a\le c<d\le b$,$$P(c<X<d)=\int_c^d\frac{1}{b-a}\,dx=\frac{d-c}{b-a}.$$For $X\sim U(-5,19)$, $P(|X|>3.5)=[1.5+15.5]/24=17/24$. Endpoints make no difference for a continuous variable.

Accumulating rectangular area from the left derives the cumulative distribution function:$$F(x)=\begin{cases}0,&x<a,\\\dfrac{x-a}{b-a},&a\le x\le b,\\1,&x>b.\end{cases}$$The middle piece is linear because each extra unit of $x$ adds the same area $1/(b-a)$.

The mean follows from$$E(X)=\int_a^b\frac{x}{b-a}\,dx=\frac{b^2-a^2}{2(b-a)}=\frac{a+b}{2}.$$Likewise,$$E(X^2)=\frac{b^3-a^3}{3(b-a)}=\frac{a^2+ab+b^2}{3},$$so$$\operatorname{Var}(X)=E(X^2)-[E(X)]^2=\frac{(b-a)^2}{12}.$$Thus the mean is the rectangle's midpoint, while variance depends only on its width.

A Normal distribution can approximate a Binomial or Poisson count when the count distribution is sufficiently spread out and not strongly skewed. Match the discrete mean and variance before converting the integer event to a continuous boundary.

Each integer count represents a unit-width bar extending 0.5 on either side of its centre. Moving the Normal boundary to the outer edge of the last included bar preserves approximately the same area. First rewrite words such as 'fewer than 32' as the integer event X ≤ 31; then its corrected boundary is 31.5.

Suppose $X\sim Po(36)$ and we need $P(X<32)$. Use $Y\sim N(36,36)$, so its standard deviation is 6. With the correction,$$P(X<32)\approx P(Y<31.5)=P\left(Z<\frac{31.5-36}{6}\right)=P(Z<-0.75)=0.2266.$$The value 32 is not the boundary because the count 32 is excluded.

For a Binomial example, $X\sim B(200,0.4)$ gives $Y\sim N(80,48)$. Then$$P(X\ge90)\approx P(Y>89.5)=P\left(Z>\frac{89.5-80}{\sqrt{48}}\right).$$Keep the second Normal parameter as the variance, but divide by its square root when standardising.

A statistical investigation begins by defining exactly what is being studied. The population is the complete set of units of interest; a census seeks data from every population unit, while a sample survey collects data from only a subset.

To study the fill mass of all cans produced during one shift, the population is every can from that shift, one can is a sampling unit, and the production list or numbered stream may provide the frame. Measuring 100 selected cans is a sample survey; measuring every can is a census.

A statistic is a quantity calculated entirely from sample observations. It may estimate or test a population feature, but its formula cannot contain an unknown population parameter. Thus $\bar X$, the sample range and the number of successes are statistics; $(X_1-\mu)/\sigma$ is not when $\mu$ and $\sigma$ are unknown.

If the same random-sampling procedure were repeated with a fixed sample size, the statistic would usually change. Its sampling distribution lists every possible value of the statistic together with its probability under the stated population model.

Suppose two independent Bernoulli observations are drawn from a population with success probability $0.5$, and let $T$ be the sample proportion of successes. The four equally likely ordered samples give$$T=0,\ \frac12,\ 1$$with probabilities$$P(T=0)=\frac14,\quad P\left(T=\frac12\right)=\frac12,\quad P(T=1)=\frac14.$$This probability model is the sampling distribution of $T$, not the distribution of a single observation.

A test statistic is a chosen statistic whose sampling distribution is known under the null hypothesis. That distribution lets an observed sample value be judged as ordinary or unusually extreme.

A hypothesis test asks whether sample evidence is sufficiently unusual under a stated model to justify rejecting that model in favour of a specific alternative. It can therefore refine a mathematical model, but it cannot prove either hypothesis.

State both hypotheses before using the data. Assume H0, identify the sampling distribution of a suitable test statistic, and calculate the probability of the observed value or values at least as supportive of H1. If this p-value is no greater than alpha—or the statistic lies in the precomputed critical region—reject H0; otherwise do not reject H0.

A historical breakdown rate is $8$ per week and a refurbishment is claimed to have changed it. Use$$H_0:\lambda=8,\qquad H_1:\lambda\ne8.$$If the observed count is outside the two-tailed critical region, do not reject $H_0$: there is insufficient evidence that the mean breakdown rate changed.

A critical region is the set of test-statistic values that cause rejection of the null hypothesis. It is chosen from the sampling distribution under H0 so that its total probability—the actual significance level—does not exceed the stated level and is as close as the discrete distribution permits.

For an upper-tailed test with $J\sim Po(9)$ under $H_0$, tables give$$P(J\le13)=0.9261,\qquad P(J\le14)=0.9585.$$Hence$$P(J\ge14)=0.0739>0.05,\qquad P(J\ge15)=0.0415\le0.05.$$The 5% critical region is therefore $J\ge15$. An observed value of 15 or more is significant at the 5% level.

Because a discrete tail changes in jumps, the actual significance is often below the nominal level. Moving the boundary one integer further inward here would make the rejection probability too large; moving it outward would be valid but unnecessarily conservative.

The wording of the alternative hypothesis fixes the tail structure before the sample is inspected. A directional claim uses one tail; a claim of any change uses both tails.

For a one-tailed test, the critical-region probability is placed in the stated tail. For a two-tailed test at nominal level alpha, choose lower and upper regions with probabilities as close as practical to alpha/2 while keeping their combined probability no greater than alpha. Discreteness may prevent equal tails or an exact total.

Testing whether a historical proportion $0.35$ has changed requires $H_0:p=0.35$ and $H_1:p\ne0.35$, so both unusually few and unusually many successes oppose $H_0$. Testing whether it has fallen uses $H_1:p<0.35$ and only the lower tail.

For a Binomial test, the parameter is a population success probability p; for a Poisson test, it is the mean event rate lambda for a stated exposure. Under H0, that exact parameter determines the sampling distribution used for the test.

Use this order: (1) define the parameter and state H0 and H1; (2) translate the observed count and direction into the correct tail or tails; (3) calculate an exact tail probability from tables/formulae or construct a critical region; (4) compare with alpha; (5) reject or do not reject H0 and give a contextual conclusion.

Suppose 8 of 40 items fail and the historical failure proportion is $0.35$. To test whether it is lower, use$$H_0:p=0.35,\qquad H_1:p<0.35,\qquad X\sim B(40,0.35).$$The lower-tail probability is$$P(X\le8)=0.0303<0.05.$$Reject $H_0$: there is sufficient evidence at the 5% level that the failure proportion is lower than $0.35$.

Use an approximation only when suitable, preserving the null model: a rare-event Binomial may use $Po(np_0)$, while a sufficiently spread Binomial or Poisson count may use a matched Normal distribution with continuity correction. For example, six weeks at a null Poisson rate of 6 per week gives total mean 36 and may be approximated by $N(36,36)$.

For a discrete two-tailed test, construct both rejection tails and add their null probabilities to obtain the actual significance. Do not automatically double one observed-tail probability unless that rule is justified by the chosen test; discrete tails are often unequal.