## Example

Let k1 = 20 and k2 = 120. The 5 percent critical F value for these df is 1.48. Therefore, k1 F = (20)(1.48) = 29.6. From the chi-square distribution for 20 df, the 5 percent critical chi-square value is about 31.41.

In passing, note that since for large df the t, chi-square, and F distributions approach the normal distribution, these three distributions are known as the distributions related to the normal distribution.

### The Bernoulli Binomial Distribution

A random variable X is said to follow a distribution named after Bernoulli (a Swiss mathematician) if its probability density (or mass) function (PDF) is:

P(X = 0) = 1 - p P (X = 1) = p where p, 0 < p < 1, is the probability that some event is a "success," such as the probability of obtaining a head in a toss of a coin. For such a variable,

E(X) = [1 x p(X = 1) + 0 x p(X = 0)] = p var (X) = pq where q = (1 - p), that is, the probability of a "failure."

### Binomial Distribution

The binomial distribution is the generalization of the Bernoulli distribution. Let n denote the number of independent trials, each of which results in a

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"success" with probability p and a "failure" with a probability q = (i — p). If X represents the number of successes in the n trials, then X is said to follow the binomial distribution whose PDF is:

f ( x)=( n)px (i—p)n—x where x represents the number of successes in n trials and where n n!

where n!, read as n factorial, means n(n — i)(n — 2) ■ i.

The binomial is a two-parameter distribution, n and p. For this distribution,

For example, if you toss a coin i00 times and want to find out the probability of obtaining 60 heads, you put p = 0.5, n = i00 and x = 60 in the above formula. Computer routines exist to evaluate such probabilities.

You can see how the binomial distribution is a generalization of the Bernoulli distribution.

The Poisson Distribution

A random X variable is said to have the Poisson distribution if its PDF is:

The Poisson distribution depends on a single parameter, X. A distinguishing feature of the Poisson distribution is that its variance is equal to its expected value, which is X. That is,

The Poisson model, as we saw in the chapter on nonlinear regression models, is used to model rare or infrequent phenomena, such as the number of phone calls received in a span of, say, 5 minutes, or the number of speeding tickets received in a span of an hour, or the number of patents received by a firm, say, in a year.

A.7 STATISTICAL INFERENCE: ESTIMATION

In Section A.6 we considered several theoretical probability distributions. Very often we know or are willing to assume that a random variable X follows a particular probability distribution but do not know the value(s) of the parameter(s) of the distribution. For example, if X follows the normal

896 APPENDIX A: A REVIEW OF SOME STATISTICAL CONCEPTS

distribution, we may want to know the value of its two parameters, namely, the mean and the variance. To estimate the unknowns, the usual procedure is to assume that we have a random sample of size n from the known probability distribution and use the sample data to estimate the unknown para-meters.5 This is known as the problem of estimation. In this section, we take a closer look at this problem. The problem of estimation can be broken down into two categories: point estimation and interval estimation.

### Point Estimation

To fix the ideas, let X be a random variable with PDF f (x; 6), where 6 is the parameter of the distribution (for simplicity of discussion only, we are assuming that there is only one unknown parameter; our discussion can be readily generalized). Assume that we know the functional form—that is, we know the theoretical PDF, such as the t distribution—but do not know the value of 6. Therefore, we draw a random sample of size n from this known PDF and then develop a function of the sample values such that

provides us an estimate of the true 6.6) is known as a statistic, or an estimator, and a particular numerical value taken by the estimator is known as an estimate. Note that 6 can be treated as a random variable because it is a function of the sample data. 6 provides us with a rule, or formula, that tells us how we may estimate the true 6. Thus, if we let

n where X is the sample mean, then X is an estimator of the true mean value, say, If in a specific case X = 50, this provides an estimate of The estimator 6 obtained previously is known as a point estimator because it provides only a single (point) estimate of 6.

### Interval Estimation

Instead of obtaining only a single estimate of 6, suppose we obtain two estimates of 6 by constructing two estimators 61(x1, x2,..., xn) and 62(x1, x2,..., xn), and say with some confidence (i.e., probability) that the interval between 61 and 62 includes the true 6. Thus, in interval estimation, in contrast with point estimation, we provide a range of possible values within which the true 6 may lie.

5Let X\, X2,..., Xn be n random variables with joint PDF f (x\, x2,..., xn). If we can write f (x1, x2, ..., xn) = f (x\) f fe) ••• f (xn)

where f (x) is the common PDF of each X, then x1, x2,..., xn are said to constitute a random sample of size n from a population with PDF f (xn).

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The key concept underlying interval estimation is the notion of the sampling, or probability distribution, of an estimator. For example, it can be shown that if a variable X is normally distributed, then the sample mean X is also normally distributed with mean = \x (the true mean) and variance = a2/n, where n is the sample size. In other words, the sampling, or probability, distribution of the estimator X is X ~ N(f, a2/n). As a result, if we construct the interval x± 2 a 