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Bus waiting times are distributed like this (they are independent)

independent

I know the average time is 8 minutes. I need to find the pivotal quantity of Theta parameter and after it of P. (P is the probability that waiting time will take more than 5 minutes )

I don't know How to treat each of them separately ? (P and $\theta$) ?

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1 Answer 1

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If $T_1, T_2, \dots, T_n$ are exponentially distributed with mean $\theta,$ then one can show (e.g., using moment generating functions( that the sample mean $\bar T$ has $\frac{\bar T}{\theta} \sim \mathsf{Gamma}(\mathrm{shape}=n, \mathrm{rate}=n).$

Then one can find values $L$ and $U$ that cut probability $0.025,$ respectively, from the lower and upper tails of $\mathsf{Gamma}(n,n):$

$$0.95 = P\left(L \le \frac{\bar T}{\theta} \le U\right) = P\left(\frac{\bar T}{U} \le \theta \le \frac{\bar T}{L}\right),$$ so that a 95% confidence interval for $\theta$ is of the form $\left(\frac{\bar T}{U},\;\frac{\bar T}{L}\right).$

Here is an example in R with thirty observations from an exponential distribution with rate $\lambda = 1/8$ and mean $\theta = 8.$

The resulting 95% CI is $(5.52, 11.35),$ which does cover the population mean $8,$ as should happen for 95% of such datasets. [In R, probability functions for exponential distribution use the rate $\lambda = 1/\theta$ as the parameter.]

set.seed(101)
t = rexp(30, 1/8)   # data: 30 0bservations
summary(t)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.373   2.631   4.737   7.660  10.455  30.051 

ci = mean(t)/qgamma(c(.975,.025), 30, 30);  ci 
[1]  5.517638 11.353423

Finally, $P(T_i > 5) = e^{-5/\theta} = e^{-5/8} = 0.5353.$

1 - pexp(5, 1/8)
[1] 0.5352614

Then the CI for the probability is $(0.4040, 0.6438).$

exp(-5/ci)
[1] 0.4040628 0.6437815

Note: If you are not familiar with gamma distributions or computations in R, then you can look at information on the gamma and chi-squared distributions (perhaps in your text or the relevant Wikipedia pages) to see how to use printed tables of the chi-squared distribution to get the 95% CI for $\theta.$

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  • $\begingroup$ I can't use R, and I know Gamma. I need to use Pivotal Quantities, and to get an numeric answer for theta and for P. but I dont succeed undertsand in which Pivotal Quantities I need to use with my data $\endgroup$
    – user123
    Jan 2, 2021 at 10:14
  • $\begingroup$ The pivotal quantity is $\bar T/\theta.$ The 'pivot' takes place at the last member of my displayed equation. $\endgroup$
    – BruceET
    Jan 2, 2021 at 10:36
  • $\begingroup$ ok and if I have a chi-square with 60 df, how can I find it in the table? the table ends in 30 $\endgroup$
    – user123
    Jan 2, 2021 at 11:51
  • $\begingroup$ Get a different table, use a statistical calculator, learn to use R (if only for probability look-up), or google for chi-square tables online (of which one example is from NIST).) $\endgroup$
    – BruceET
    Jan 2, 2021 at 18:26
  • $\begingroup$ You are correct that $\mathsf{Chisq}(\nu=k)\equiv\mathsf{Gamma}(\mathrm{shape}=k/2,\mathrm{rate}=1/2),$ so in R: qchisq(.975,60) and qgamma(.975,30,1/2) both return $83.29767.$ Note that qchisq and qgamma are quantile functions (inverse CDFs) of chi-squared and gamma distributions, respectively. $\endgroup$
    – BruceET
    Jan 2, 2021 at 18:36

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