Question A.6: Using the same data as in Example A.5, estimate the density ......
Using the same data as in Example A.5, estimate the density function using an exponential distribution. Test the hypothesis that this is a reasonable estimate.
The mean of the data is μ = 7. Since the mean is also the MLE for the
exponential distribution, the estimated density function is
\ \hat{f}(x)=\begin{cases} \frac{1}{7}e^{-x/7},&x≥0,\\0,&x<0. \end{cases}
Notice that for an exponential distribution, one must use an additional interval x ≥ 15 since the exponential is right-tailed. The expected frequencies are calculated as follows:
\ e_1=n\int_{0}^{2.5}{\hat{f}(x) dx}\\= 10\int_{0}^{2.5}{\frac{1}{7}e^{-x/7} dx}\\=-10e^{-2.5/7}|^{2.5}_0\\=10(1-e^{-2.5/7})\approx 3.00,\ e_2= 10\int_{2.5}^{5.0}{\frac{1}{7}e^{-x/7} dx}\approx 2.101,
\ e_3= 10\int_{5.0}^{7.5}{\frac{1}{7}e^{-x/7} dx}\approx 1.470,
\ e_4= 10\int_{7.5}^{10.0}{\frac{1}{7}e^{-x/7} dx}\approx 1.029,
\ e_5= 10\int_{10.0}^{12.5}{\frac{1}{7}e^{-x/7} dx}\approx 0.720,
\ e_6= 10\int_{12.5}^{15.0}{\frac{1}{7}e^{-x/7} dx}\approx 0.506,
\ e_7= 10\int_{15.0}^{\infty}{\frac{1}{7}e^{-x/7} dx}\approx 1.172.
Again calculating\ \chi^2,
\ \chi^2=\sum\limits_{k=1}^{7}{\frac{(\phi_k – e_k)^2}{e_k}}\\[0.3cm] = \frac{(3-3.000)^2}{3.000}+\frac{(2-2.101)^2}{2.101}+\frac{(1-1.470)^2}{1.470}+\frac{(0-1.029)^2}{1.029} \\+\frac{(2-0.720)^2}{0.720} +\frac{(2-0.506)^2}{0.506}+\frac{(0-1.172)^2}{1.172}\approx 9.043.Since there is but one parameter determined\ (l = 1) and 7 classes (m = 7), this time the number of degrees of freedom is\ \nu = 7 – 1 – 1 = 5 . Checking the critical values,\ \chi^2_{0.05} = 11.070 and \chi^2_{0.01} = 15.086 . Since\ \chi^2 < \chi^2_{0.05} and \chi^2 < \chi^2_{0.01} , we will accept this hypothesis at either the 5% or 1% significance level.