Question A.1: Consider the following sorted data: {0.1, 0.3, 0.3, 0.5, 0.6......
Consider the following sorted data: {0.1, 0.3, 0.3, 0.5, 0.6, 0.6, 0.7, 0.8, 0.8, 0.8, 0.9, 1.2, 1.3, 1.4, 1.7, 1.9, 2.2, 2.6, 2.9, 3.4}. This data might be the recorded length of long-distance telephone calls (measured in minutes) at a residence over a one-month period. It is desired to plot the histogram for this data and infer an appropriate density function.
The unbiased estimates for the population mean and standard deviation are calculated to be
\ \hat{\mu}=\frac{1}{n}\sum\limits_{i=1}^{n}{x_i}=1.25, (A.4)
\ \hat{\sigma}=\sqrt{\frac{1}{n-1}\sum\limits_{i=1}^{n}{(x_i-\hat{\mu})^2}}\\=\sqrt{\frac{1}{n-1}(\sum\limits_{i=1}^{n}{x_i^2-n\hat{\mu}^2)}}=0.9259. (A.5)
Using ξ = 0 and δ = 0.5 along with m = 7 intervals, the frequency distribution table is given in Table A.2. The corresponding empirical probability and density functions are plotted in Figure A.2. Since the graph resembles a Gamma distribution, this may be assumed to be a fair picture of the underlying (actual) distribution. Assuming this to be the case, it remains to estimate the parameters that characterize the Gamma distribution: α and β.
TABLE A.2 Frequency Distribution Table, Example A. 1
\ \begin{array}{c} \hline Number&Interval&\phi _k&p_k&f_k\\\hline 0& 0.0≤x<0.5&3&0.15&0.30\\1& 0.5≤x<1.0&8&0.40&0.80\\2& 1.0≤x<1.5&3&0.15&0.30\\3& 1.5≤x<2.0&2&0.10&0.20\\4& 2.0≤x<2.5&1&0.05&0.10\\5& 2.5≤x<3.0&2&0.10&0.20\\6& 3.0≤x<3.5&1&0.05&0.10\\\hline &&20&1.00&2.00\\\hline \end{array}
One method to estimate the parameters of the Gamma distribution is to simply note (using Appendix C) that its mean and variance are given by\ \mu = αβ and \sigma^2 = αβ^2 , respectively. Solving for α and β in terms of μ and σ, then using the unbiased estimators found by Equations (A.4) and (A.5), the estimators for α and β are found to be\ α = \hat{\mu}^2/\hat{\sigma}^2 \approx 1.8224 and β = \hat{\sigma}^2/\hat{\mu}\approx 0.6859.
Again referring to appendix C, the density function estimator is
\ \hat{f}(x)=\frac{\beta^{-\alpha}}{\Gamma(\alpha)}x^{\alpha -1 }e^{-x/\beta}\\=\frac{(0.6859)^{-1.8224}}{\Gamma(1.8224)}x^{0.8224}e^{-x/0.6859}\\[0.3cm]\\=2.122\frac{x^{0.8224}}{e^{1.4579x}}, x≥0. (A.6)
This function is graphed and compared against the empirical density function in Figure A.2. It should be noticed that the artistic comparison (smoothness / sharpness eyeball test) agrees fairly well with the inferred density function, especially considering the small sample size.
