Sampling Distributions

Basic Ideas

• An important goal of data analysis is to distinguish between features of the data that reflect real biological facts and features that may reflect only chance effects.
• The random sampling model provides a framework for making this distinction: Chance effects are regarded as sampling error. That is, discrepancy between the sample and the population.
• In this chapter we develop the theoretical background that will enable us to place specific limits on the degree of sampling error to be expected in a study.

Sampling variability

• The variability among random samples from the same population is called sampling variability.
• A probability distribution that characterizes some aspect of sampling variability is termed a sampling distribution.
• We have to expect a certain amount of discrepancy between the sample and the population due to the sampling error.

The meta-study

A meta-study consists of indefinitely many repetitions, or replications, of the same study. If the study consists of drawing a random sample of size $n$ from some population, the corresponding meta-study involves drawing repeated random samples of size $n$ from the same population.

The Sample Mean

• The sample mean $\bar{y}$ can be used, not only as a description of the data in the sample, but also as an estimate of the population mean $\mu$. It is natural to ask, "How close to $\mu$ is $\bar{y}$?"
• We cannot answer this question for the mean $\bar{y}$ of a particular sample due to the randomness of the sample. Regarding the sample mean as a random variable $\bar{Y}$, the question then becomes: "How close to $\mu$ is $\bar{Y}$ likely to be?"
• To characterize such randomness, we resort to the sampling distribution of the sample mean $\bar{Y}$, the probability distribution that describes sampling variability in $\bar{Y}$.

To visualize the sampling distribution of $\bar{Y}$, imagine the meta-study as follows:

• Random samples of size $n$ are repeatedly drawn from a fixed population with mean $\mu$ and standard deviation $\sigma$; each sample has its own mean $\bar{y}$.
• The variation of the $\bar{y}$'s among the samples is specified by the sampling distribution of $\bar{Y}$.

When we think of $\bar{Y}$ as a random variable, we need to be aware of two basic facts

• On average, the sample mean equals to the population mean. That is, the average of the sampling distribution of $\bar{Y}$ is $\mu$.
• As the sample size increases, the standard deviation of $\bar{Y}$ decreases. That is, for larger samples, $\bar{Y}$ will tend to be closer to the population mean.

Theorem

• Consider the random sample $Y_1, \ldots, Y_n$, drawn from a population with mean $\mu$ and standard deviation $\sigma$. The sample mean is denoted as $\bar{Y}=\frac{1}{n}\sum_{i=1}^nY_i$. Try to derive Parts 1 and 2 of the above theorem.

• The Central Limit Theorem states that, no matter what distribution $Y$ may have in the population, if the sample size is large enough, then the sampling distribution of $\bar{Y}$ will be approximately a normal distribution.
• It is because of the Central Limit Theorem (and other similar theorems) that the normal distribution plays such a central role in statistics.
• It is natural to ask how "large" a sample size is required by the Central Limit Theorem.
  • If the shape is normal, any n will do.
  • If the shape is moderately nonnormal, a moderate n is adequate.
  • If the shape is highly nonnormal, then a rather large n will be required.

Example: weights of seeds

A large population of seeds of the princess bean Phaseotus vulgaris is to be sampled. The weights of the seeds in the population follow a normal distribution with mean $\mu=500$ mg and standard deviation $\sigma=120$ mg. Suppose now that a random sample of four seeds is to be weighed, and let $\bar{Y}$ represent the mean weight of the four seeds. What is the sampling distribution of $\bar{Y}$? $N(500, 3600)$

Dependence of sample size

• Larger $n$ gives a smaller value of $\sigma_{\bar{Y}}$ and consequently a smaller expected sampling error if $\bar{y}$ is used as an estimate of $\mu$.
• If the population distribution is not normal, then the shape of the sampling distribution of $\bar{Y}$ depends on $n$, being more nearly normal for larger $n$.
• The mean of a larger sample is not necessarily closer to $\mu$ than the mean of a smaller sample, but it has a greater probability of being close. It is in this sense that a larger sample provides more information about the population mean than a smaller sample.

Populations, samples, and sampling distributions

It is important to distinguish clearly among three different distributions related to a quantitative variable $Y$:

• the distribution of $Y$ in the population;
• the distribution of $Y$ in a sample of data, and
• the sampling distribution of $\bar{Y}$.

Distribution	Mean	Standard deviation
$Y$ in population	$\mu$	$\sigma$
$Y$ in sample	$\bar{y}$	$s$
$\bar{Y}$ (in meta-study)	$\mu_{\bar{Y}}=\mu$	$\sigma_{\bar{Y}}=\sigma/\sqrt{n}$

Example

Recall the weights of seeds example, the population mean and standard deviation are $\mu=500$ mg and $\sigma=120$ mg. Suppose we weigh a random sample of $n=25$ seeds from the population and obtain the data in the table below

• The population distribution of $Y=$ weights is represented in (a)
• the sample mean is $\bar{y}=526.1$ mg and the sample standard deviation is $s=113.7$ mg. (b) shows a histogram of the data; this histogram represents the distribution of $Y$ in the sample.
• The sampling distribution of $\bar{Y}$ as shown in (c) is a theoretical distribution which relates, not to the particular sample shown in the histogram, but rather to the meta-study of infinitely repeated samples of size $n=25$. The mean and standard deviation of the sampling distribution are $\mu_{\bar{Y}}=500$ mg and $\sigma_{\bar{Y}}=120/\sqrt{25}=24$ mg.

Notice that the distributions in (a) and (b) are more or less similar; in fact, the distribution in (b) is an estimate of the distribution in (a). By contrast, the distribution in (c) is much narrower, because it represents a distribution of means rather than of individual observations.

The Normal Approximation to the Binomial Distribution

• The binomial random variable $X\sim B(n, p)$ is the sum of $n$ identical Bernoulli random variables, each with expected value $p$ and variance $p(1-p)$. In other words, if $X_1, \ldots, X_n$ are identical (and independent) Bernoulli random variables with parameter $p$, then $X=X_1+\cdots+X_n$.
• Think of $X_1, \ldots, X_n$ as a random sample. Then the sample mean $\hat{P}=\frac{1}{n}\sum_{i=1}^nX_i$ is governed by the Central Limit Theorem.

Theorem

• If $n$ is large, then the binomial distribution of the probability of success, $\hat{P}$, can be approximated by a normal distribution with mean $=p$ and standard deviation $=\sqrt{p(1-p)/n}$.
• If $n$ is large, then the binomial distribution of the number of successes, $Y$, can be approximated by a normal distribution with mean $=np$ and standard deviation $=\sqrt{np(1-p)}$.

Example: normal approximation to binomial

We consider a binomial distribution with $n=50$ and $p=0.3$. (a) shows this binomial distribution, using spikes to represent probabilities; superimposed is a normal curve with mean $=np=15$ and standard deviation $=\sqrt{np(1-p)}=3.24$. (b) shows the sampling distribution of $\hat{P}$; superimposed is a normal curve with mean $=p=0.3$ and standard deviation $=\sqrt{p(1-p)/n}=0.0648$.

To illustrate the use of the normal approximation, let us find the probability that $50$ independent trials result in at least $18$ successes, i.e., $P(Y\geq18)$. The exact calculation using the binomial formula is very tedious, which involves $50-18+1=33$ terms ($0.2178$). If instead the normal approximation is adopted, we only need to find the corresponding area under the normal curve.

The Z score that corresponds to $18$ is $$z=\frac{18-15}{3.2404}=0.93.$$ We find that the area is $1-0.8238=0.1762$ using Z table.

The continuity correction

• What would happen if we want to compute $P(Y=18)$ using the normal approximation, the probability of 18 successes?
• We think of "$18$" as covering the space from $17.5$ to $18.5$ and thus we consider the area under the normal curve between $17.5$ and $18.5$.
• Compute $P(Y\geq18)$ using the continuity correction.
  • The Z score is $$z=\frac{17.5-15}{3.2404}=0.77$$
  • From the Z table, we find that the area above $0.77$ is $1-0.7794=0.2206$.
• What about $P(12\leq Y\leq18)$ and $P(12<Y<18)$?

Summary of continuity correction

• If $P(Y=n)$ use $$P(n-0.5<Y<n+0.5).$$
• If $P(Y>n)$ use $$P(Y>n+0.5).$$
• If $P(Y\leq n)$ use $$P(Y<n+0.5).$$
• If $P(Y<n)$ use $$P(Y<n-0.5).$$
• If $P(Y\geq n)$ use $$P(Y>n-0.5).$$

How large must $n$ be?

The required $n$ depends on the value of $p$.

• If $p=0.5$, then the binomial distribution is symmetric and the normal approximation is quite good even for $n$ as small as $10$.
• However, if $p=0.1$, the binomial distribution for $n=10$ is quite skewed and is poorly fitted by a normal curve; for larger $n$ the skewness is diminished and the normal approximation is better.
• A simple rule of thumb is the following:
  • The normal approximation to the binomial distribution is fairly good if both $np$ and $n(1-p)$ are at least equal to $5$.