 vetrila10

2021-11-26

The observations given below are related with duration between two patients that come to ER service of certain hospital. There is a claim that the distribution of durations is Exponential with $V\left(X\right)=1.778$. Evaluate the claim.
$\begin{array}{|ccccccccccc|}\hline 3.264& 4.62& 0.439& 0.13& 1.975& 1.271& 0.82& 0.42& 1.145& 0.61& 2.408\\ 0.87& 1.942& 0.32& 0.62& 0.375& 0.383& 2.512& 1.92& 3.19& 2.12& 3.89\\ \hline\end{array}$ Steven Arredondo

Step 1
Under null hypothesis ${H}_{0}$: The distributions between two patients follow exponential distribution with mean 1.778
$\mu =\frac{1}{\lambda }=1.778$
Under alternate hypothesis ${H}_{1}$: The distributions between two patients do not follow exponential distribution of which mean is 1.778
$\mu =\frac{1}{\lambda }\ne 1.778$
Step 2
Sample Mean:
$\stackrel{―}{x}=\frac{\sum {x}_{i}}{n}$
$=\frac{35.244}{22}$
$=1.602$
Exponential distribution has mean and standard deviation
Using central limit theorem,
$\stackrel{―}{X}\sim N\left(\frac{1}{\lambda },\frac{1}{{\lambda }^{2}}\right)$
Step 3
Test statistic:
$z=\frac{\stackrel{―}{X}-\frac{1}{\lambda }}{\sqrt{\frac{1}{{\lambda }^{2}}}}$
$=\frac{1.602-1.778}{\sqrt{\frac{1}{\frac{1}{{1.778}^{2}}}}}$
$=\frac{1.602-1.778}{\sqrt{{1.778}^{2}}}$
$-0.09899$
Computation of P-value:
The P-value for the two tailed z-value test is 0.9211 which can be obtained using the excel formula, "=2*NORM.S.DIST(-0.09899,TRUE)".
Step 4
Decision rule:
If $p-value\le \alpha$, then reject the null hypothesis.
Otherwise, do not reject the null hypothesis.
Conclusion:
Consider level of significance as 0.05.
The p-value is greater than level of significance.
Hence, we have insufficient evidence to reject null hypothesis and thus conclude that the duration between two patients that come to ER service of certain hospital follow exponential distribution with mean 1.778.

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