The life of a cell can be measured by $X\sim Exp(0.1)$. After a procedure "S" the life of the cell is given by $X\sim Exp(\lambda_1).$ From a sample of $200$ we know $\sum_{i=1}^{200} x_i=1182.5$ and $\sum_{i=1}^{200} x_i^2=13239.14$ with $c=95\%$. Determine if the cell lives longer.
I can calculate the mean and deviation
$\bar{x}=\frac{1}{200}\sum_{i=1}^{200}x_i=1182,25/200=5,91.$
$\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(x_i-\bar{x})^2}=\sqrt{\frac{1}{200}\sum_{i=1}^{200}(x_i^2-2x_i\bar{x}+\bar{x}^2)}\\ = \sqrt{\frac{1}{200}\left(\sum_{i=1}^{200}x_i^2-2\sum_{i=1}^{200}x_i\bar{x}+\sum_{i=1}^{200}\bar{x}^2\right)}\\= \sqrt{\frac{1}{200}\left(13239,14-2(5,91)\sum_{i=1}^{200}x_i+\sum_{i=1}^{200}(5,91)^2\right)} \\ =\sqrt{\frac{1}{200}\left(13239,14-2(5,91)(1182,25)+(200)(5,91)^2\right)}\\ =5.59. $
Because both before and after are exponential distribution I consider the hypothesis may be
$ \begin{cases} H_0: & \lambda_1\leq 0.1 \\ H_1: & \lambda_1 > 0.1 \end{cases} $
Using z-test
$z=\displaystyle \frac{\bar{x}-0.1}{\sigma\sqrt{n}}= \frac{5.91-0.1}{(5.59)\sqrt{200}}=0.074$
Because $c=95\%$ then $\alpha= 0.05$ and according to the $z-table$ the z-score is 1.645. The result doesn't fall into the rejection region then according to the test the cell do not lives longer.
Is it correct? I am particularly unsure about the hypothesis I've chosen.