I am trying to solve home made problem, but i am having a hard time solving it..
A Appleseller wants to advertise the average weight of his apple, but since he sells so many it isn't possible to do so. He wants to know the confidence interval of the average apple weight lies in between 150 and 200 g.
His sample of the population consist of 10 apples.
So i know that what i should do is calculate P(150<= µ <= 200) as it was a T-dist. But what is the lower and upper boundary in the T-scores?...
T-scores i've calculated T_lower = $\frac{150-152}{\frac{34.251}{\sqrt{10}}} = -0.0184637$ T_upper = $\frac{200-152}{\frac{34.251}{\sqrt{10}}} = 4.43129$ P(t_lower <= T <= T_upper) = 0.570374
The interval is
$\left[ \overline X-t_{(n-1,1-\alpha_2)}\cdot \frac{s}{\sqrt n}; \overline X+t_{(n-1,1-\alpha_1)}\cdot \frac{s}{\sqrt n}\right] $
with $\hat \mu=\overline X=152$
$\alpha_1+\alpha_2=\alpha$
$\alpha$ is the probability, that the sample is not drawn from a population with $\mu=152$.
It is obvious, that $ \overline X-t_{(n-1,1-\alpha_2)}\cdot \frac{s}{\sqrt n}=152-2=150$
and with $ \overline X+t_{(n-1,1-\alpha_1)}\cdot \frac{s}{\sqrt n}=152+48=200$
Thus $t_{(9,1-\alpha_2)}\cdot \frac{s}{\sqrt{10}}=2$ and $t_{(9,1-\alpha_1)}\cdot \frac{s}{\sqrt{10}}=48$
Do you know how to calculate s ?