Question A.2: Suppose that a study of the number of phone calls received a......
Suppose that a study of the number of phone calls received at a certain site is to be made. To do this, the number of phone calls received over a specified 30-minute period is recorded for 25 days. In this study, it is important to not only estimate the average number of calls, but the distribution of such calls as well. For instance, there may be a large number of days where few calls are received, but many days where the line is especially busy. We want to analyze the difference.
The results of an empirical study are recorded and summarized in Table A.3. As can be seen, this is a frequency distribution table, using ξ = 0, δ = 1, and m = 7 classes. Analyze this data statistically and infer an appropriate mass function.
TABLE A.3 Frequency Distribution Table,
Example A.2
\ \begin{array}{c} \hline Number&Score&&\\k&x_k&\phi_k&p_k\\\hline 0&0&2&0.08\\1&1&4&0.16\\2&2&8&0.32\\3&3&5&0.20\\4&4&3&0.12\\5&5&2&0.08\\6&6&1&0.04\\\hline &&25&1\\\hline \end{array}
Since, by the summary description, the variates must be integral, the distribution must be discrete. We therefore consider the mass function to be analogous to the empirical probability, which is shown in Figure A.3. The unbiased mean and variance estimators are computed as follows:
\ \hat{\mu}=\frac{1}{n}\sum\limits_{i=1}^{n}{\phi_ix_i}=2.52,\\ \hat{\sigma}^2=\frac{1}{n-1}\sum\limits_{i=1}^{n}{\phi_i(x_i-\hat{\mu})^2}\\=\frac{1}{n-1}\sum\limits_{i=1}{n \phi_i(x_i-\hat{\mu})^2}\approx 2.343.
In order to determine the actual distribution of this random variable, a slightly artistic approach is called for. For instance, by noting that the distribution has range [0, 6] and is slightly skewed to the right and that there are no “tails”, the binomial distribution as described in Appendix B is a realistic possibility. Conveniently, formulas for the mean and variance of the binomial distribution are also given in Appendix B. They are
\ \mu=np,\\ \sigma^2=np(1-p) (A.7)
Solving Equations (A.7) for n and p and, then substituting the known mean and standard deviation estimators gives values of \ p = 1 – \hat{\sigma}^2/\hat{\mu} \approx 0.07 and n = \hat{\mu}^2/(\hat{\mu} – \hat{\sigma}^2 ) \approx 2.548 , from which the estimated mass function can in principle be computed. The value of\ \hat{p} \approx 0.07 makes good sense, since p is a probability. However, there is no integral value for\ \hat{n} that is obvious (which should it be: n = 2 or n = 3?). Also, it should be noted that the binomial distribution has but a finite range between 0 and n (in this computation, [0.0, 2.548]). From the data, this is certainly not the case.
On the other hand, it can be argued that the true distribution should have a right tail, since theoretically one can receive an unlimited number of calls over a 30-minute period. Also, the mean and variance are nearly equal. Thus, a Poisson distribution may be a better bet! Since the Poisson parameter n equals the mean, an estimated mass function is
\ p(x)=e^{-\hat{\mu}}\frac{\hat{\mu}^x}{x!}\\=(0.08)\frac{(2.52)^x}{x!}, x=0, 1, 2, …However, since the mean and variance are equal for Poisson distributions, couldn’t we use μ = σ² too? The reason that we do this is reflected in Equation (B.13) of Appendix B. The so-called maximum likelihood estimator is\ \hat{\lambda} = n^{-1} \sum\limits_{i=1}^{n}{x_i} , which is the mean and not the variance. We verify this result and the fact that μ = 2.52 is a better estimator in the next section.
\ \hat{\lambda}=\frac{1}{n}\sum\limits_{i=1}^{n}{x_i}. (B.13)
Also, by recalling that the original data in Table A.3 were the result of incoming telephone calls, the Poisson hypothesis of Chapter 8 is reasonable. Therefore, it should be argued that we will use the Poisson if it is good enough. This, too, will be considered in a later section. The graph of the Poisson estimator is sketched along with the empirical mass function in Figure A.3. Clearly, the results are reasonable.
