Problem: the lifespan of some items follows the normal distribution N$ (\mu, \sigma^2) $, with $\mu $ and $\sigma$ both unknown. Now we have $N=16$ of the items whose lifespans have the mean of $\bar{X}=241.5$ (hours) and the variance of $s^2=98.7259$. Is it reasonable to believe that the average lifespan $\mu$ is more than $\mu_0=225$ (hours)? ($\alpha=0.05, t_{0.05}(16-1)=1.7531$)
My Answer is:
suppose:
$H_0:\mu\le225$
$H_1:\mu>225 $
and we have: $$t = \frac{\bar{X}-\mu_0}{s/\sqrt{N}} = \frac{241.5-225}{(98.7259)^{0.5}/\sqrt{16}} \approx 6.6424 > t_{0.05}(16-1)=1.7531$$ So, I refuse $H_0$, accept $H_1$, which means that the average lifespan of items $\mu$ is more than $\mu_0=225$ (hours).
Does everybody agree with my answer?