91Ó°ÊÓ

Another assumption is that the generated samples are taken from the 10 different normal distributions from all possible combinations of means and standard deviations given in the exercise. For each of the 10 combinations, there should be 3 different quantiles estimates, that is,\(3 \times 10 = 30\)in total.

The number of intervals should not exceed 5, so\(5 - 1 = 4\) it will be used in this simulation. The intervals depend on the probability of the distribution under the null hypothesis, which is the mentioned normal distribution. The upper limit of the first interval is

\(U{L_1} = \mu + 0.5 \times {\Phi ^{ - 1}}(0.25) = 3.192 + 0.5 \times ( - 0.674) = 3.575\)

where\(P\left( {Z < - 0.674} \right) = 0.25.\)

This way, the probability that the observation falls in the interval\(( - \infty ,3.575)\)is 0.25. The other 3 intervals shall have the exact probabilities and are computed using\({\Phi ^{ - 1}}(0.5)\)\({\Phi ^{ - 1}}(0.75)\)instead of\({\Phi ^{ - 1}}(0.25)\)

\(\begin{aligned}{c}U{L_2} = \mu + 0.5 \times {\Phi ^{ - 1}}(0.5) = 3.192 + 0.5 \times 0 = 3.912U{L_3} \\ = \mu + 0.5 \times {\Phi ^{ - 1}}(0.75) = 3.192 + 0.5 \times 0.674 = 4.249\end{aligned}\)

Then, the\({\chi ^2}\)statistic value is computed using

\(Q = \sum\limits_{i = 1}^k {\frac{{{{\left( {{N_i} - np_i^0} \right)}^2}}}{{np_i^0}}} \)

Where, in this case,\({N_i},i = 1,2,3,4\)are the number of observations that fall in the\({i^{th }}\)interval, n=23, and\(p_i^0\)will be calculated using the maximum likelihood estimates of the generated samples. For example, if\(\hat \mu \)and\({\hat \sigma ^2}\)are the MLE of the parameters\(\mu \)and\(\sigma ,\)then the probabilities are given/estimated by

\(\begin{aligned}{l}\hat p_1^0 = \Phi \;\;\;{\kern 1pt} \frac{{U{L_1} - \hat \mu }}{{\hat \sigma }}\\\hat p_2^0 = \Phi \;\;\;{\kern 1pt} \frac{{U{L_2} - \hat \mu }}{{\hat \sigma }} - \Phi \frac{{U{L_1} - \hat \mu }}{{\hat \sigma }}\\\hat p_3^0 = \Phi \;\;\;{\kern 1pt} \frac{{U{L_3} - \hat \mu }}{{\hat \sigma }} - \Phi \frac{{U{L_2} - \hat \mu }}{{\hat \sigma }}\\\hat p_4^0 = 1 - \Phi \;\;\;{\kern 1pt} \frac{{U{L_3} - \hat \mu }}{{\hat \sigma }}\end{aligned}\)

Then, use the\({\chi ^2}\)statistic given above. After computing the Q statistic values, order them and take the\(n \times {p^{th }}\)value where\(p \in \{ 0.9,0.95,0.99\} \)to obtain the estimate of the quantile of the distribution.

The resulting estimates are given in the following table. As mentioned, there are a total of 30 estimates of the quantiles. The "Quantile Matrix" from the code gives the desired result. The code in RStudio is given below.

There are changes. However, they do not seem to be huge.

Generate a student t distribution with 5 degrees of freedom, as given in the exercise. The estimated probability using the code below is around \(0.0357.\)