Question A.5: Consider the data {2, 2, 2, 4, 4, 7, 10, 12, 13, 14}. Estima......
Consider the data {2, 2, 2, 4, 4, 7, 10, 12, 13, 14}. Estimate an appropriate continuous uniform density function and test to see if such a hypothesis is reasonable.
The frequency distribution table and empirical density functions are shown in Table A.4 and Figure A.4 respectively. The partition defined by ξ = 0 and δ = 2.5 seems reasonable, but is relatively arbitrary.
TABLE A.4 Frequency Distribution Table, Example A.5
\ \begin{array}{c} \hline Number&Interval&\phi _k&p_k&f_k\\\hline 0& 0.0≤x<2.5&3&0.3&0.12\\1& 2.5≤x<5.0&2&0.2&0.08\\2& 5.0≤x<7.5&1&0.1&0.04\\3& 7.5≤x<10.0&0&0.0&0.00\\4& 10.0≤x<12.5&2&0.2&0.08\\5& 12.5≤x<15.0&2&0.1&0.08\\\hline &&10&1.0&0.40\\\hline \end{array}
The uniform density function is defined by parameters a and b as given in
Appendix C. The MLEs for a and b are 2 and 14 respectively. Therefore, the density function is
\ f(x)=\begin{cases} \frac{1}{b-a},&a≤x≤b,\\0,&otherwise. \end{cases} (C.1)
\ \hat{f}(x)=\begin{cases} \frac{1}{12},&2≤x≤14,\\0,&otherwise. \end{cases} (A.17)
Using this estimator, we proceed to calculate the expected frequency for each interval, being careful to integrate only over the interval defined by Equation (A.17):
\ I_1: e_1=n\int_{0}^{2.5}{\hat{f}(x) dx}= 10\int_{2}^{2.5}{\frac{1}{12} dx}\approx 0.417;\ I_2: e_2=n\int_{2.5}^{5.0}{\hat{f}(x) dx}= 10\int_{2.5}^{5.0}{\frac{1}{12} dx}\approx 2.083;
similarly, it may be shown that\ e_3 = e_4 = e_5 \approx 2.083;
\ I_6: e_6=n\int_{12.5}^{15.0}{\hat{f}(x) dx}= 10\int_{12.5}^{14.0}{\frac{1}{12} dx}= 1.25Comparing these calculated frequencies with the actual ones listed in Table 3.4, we next compute\ \chi^2:
TABLE 3.4 Frequency Distribution Table for Example 3.7
\ \begin{array}{c} \hline Interval & Empirical \space frequency & Expected \space frequency \\ \hline [0.0,\space \space 0.5] & 8 &17.52 \\ [0.5,\space \space 1.0] & 37 &44.03\\ [1.0,\space \space 1.5] & 56 &60.35 \\ [1.5,\space \space 2.0] & 64 &69.31\\ [2.0,\space \space 2.5] & 76 &73.03\\ [2.5,\space \space 3.0] & 64 &73.13 \\ [3.0,\space \space 3.5] & 77 &70.79\\ [3.5,\space \space 4.0] & 78 &66.90\\ [4.0,\space \space 4.5] &64 &62.09\\ [4.5,\space \space 5.0] & 49 &56.83\\ [5.0,\space \space 5.5] & 53 &51.44\\ [5.5,\space \space 6.0] & 46 &46.13\\ [6.0,\space \space 6.5] & 50 &41.06\\ [6.5,\space \space 7.0] & 35 &36.31\\ [7.0,\space \space 7.5] & 27 &31.93\\ [7.5,\space \space 8.0] & 29 &27.95\\ [8.0,\space \space 8.5] & 21 &24.36\\ [8.5,\space \space 9.0] & 22 &21.15\\ [9.0,\space \space 9.5] & 9 &18.31\\ [9.5,\space \space 10.0] & 25 &15.80\\ [10.0,\space \space 10.5] & 21 &13.60\\ [10.5,\space \space 11.0] & 9 &11.68\\ [11.0,\space \space 11.5] & 4 &10.01\\ [11.5,\space \space 12.0] & 6 &8.56\\ [12.0,\space \space 12.5] & 3 &7.30\\ [12.5,\space \space 13.0] & 5 &6.22\\ [13.0,\space \space 13.5] & 10 &5.30\\ [13.5,\space \space 14.0] & 10 &4.50\\ [14.0,\space \space 14.5] & 5 &3.82\\ [14.5,\space \space 15.0] & 5 &3.24\\ \hline \end{array}
Since there were two parameters estimated in the density function,\ l = 2 and \nu = 6 – 2 – 1 = 3 . Checking Appendix F, the critical\ \chi^2 -values at the 5% and 1% significance levels are\ \chi^2_{0.05} = 7.815 and \chi^2_{0.01} = 11.341 . In both cases,\ \chi^2 exceeds the critical values. Therefore, the estimated uniform density function defined by Equation (A.17) is rejected.
Appendix F THE CHI-SQUARE DISTRIBUTION FUNCTION
Values of x for given 1 – F(x) with ν degrees of freedom
\ \begin{array}{c cccc} \hline \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \rlap{\text{Complemented distribution, 1 – F(x)}}\\ \hline Degrees of freedom \nu &&0.10&0.05&0.02&0.01\\\hline 1&&2.706&3.841&5.412&6.635\\2&&4.605&5.991&7.824&9.210\\3&&6.251&7.815&9.837&11.341\\4&&7.779&9.488&11.688&13.277\\5&&9.236&11.070&13.388&15.086\\6&&10.645&12.592&15.033&16.812\\7&&12.017&14.067&16.622&18.475\\8&&13.362&15.507&18.168&20.090\\9&&14.684&16.919&19.679&21.666\\10&&15.987&18.307&21.161&23.209\\\\11&&17.275&19.675&22.618&24.725\\12&&18.549&21.026&24.054&26.217\\13&&19.812&22.362&25.472&27.688\\14&&21.064&23.685&26.873&29.141\\15&&22.307&24.996&28.259&30.578\\16&&23.542&26.296&29.633&32.000\\17&&24.769&27.587&30.995&33.409\\18&&25.989&28.869&32.346&34.805\\19&&27.204&30.144&33.687&36.191\\20&&28.412&31.410&35.020&37.566\\\\21&&29.615&32.671&36.343&38.932\\22&&30.813&22.924&37.659&40.289\\23&&32.007&35.172&38.968&45.638\\24&&33.196&36.415&40.270&42.980\\25&&34.382&37.652&41.566&44.314\\26&&35.563&38.885&42.856&45.642\\27&&36.741&40.113&44.140&46.963\\28&&37.916&41.337&45.419&48.278\\29&&39.087&42.557&46.693&49.588\\30&&40.256&43.773&47.962&50.892\\ \hline \end{array}
