The observations given below are related with duration between two patients that come to ER...

Step 1
Under null hypothesis ${\displaystyle H_0}$: The distributions between two patients follow exponential distribution with mean 1.778
${\displaystyle \mu =\frac{1}{\lambda }=1.778}$
Under alternate hypothesis ${\displaystyle H_1}$: The distributions between two patients do not follow exponential distribution of which mean is 1.778
${\displaystyle \mu =\frac{1}{\lambda }\ne 1.778}$
Step 2
Sample Mean:
${\displaystyle \bar{x}=\frac{\sum x_i}{n}}$
${\displaystyle =\frac{35.244}{22}}$
${\displaystyle =1.602}$
Exponential distribution has mean and standard deviation ${\displaystyle \mu =\frac{1}{\lambda }\text{and}{\sigma }_x=\frac{1}{{\lambda }^2}}$
Using central limit theorem,
${\displaystyle \bar{X}\sim N(\frac{1}{\lambda },\frac{1}{{\lambda }^2})}$
Step 3
Test statistic:
${\displaystyle z=\frac{\bar{X}-\frac{1}{\lambda }}{\sqrt{\frac{1}{{\lambda }^2}}}}$
${\displaystyle =\frac{1.602-1.778}{\sqrt{\frac{1}{\frac{1}{{1.778}^2}}}}}$
${\displaystyle =\frac{1.602-1.778}{\sqrt{{1.778}^2}}}$
${\displaystyle -0.09899}$
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 ${\displaystyle p-value\le \alpha }$, 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.