Question

Can someone check my answer?! (How to solve is already answered) (:

Answer 1

a) sum the values, divide by 10. (empirical mean \(\frac{1}{n}\sum_{i=1}^{n} x_i\))

Answer 2

doesnt that just give me the average @reemii

Answer 3

that's right. its' fine. a "point" estimate is a real number computed from the data.
usually people choose the mean \(\bar x\).

Answer 4

(b)
from http://en.wikipedia.org/wiki/Student's_t-distribution#Confidence_intervals
This :
\[\overline{X}_n - A\frac{S_n}{\sqrt{n}}, \overline{X}_n + A\frac{S_n}{\sqrt{n}}\]
is the confidence interval, where
\(\bar X_n\) is the mean computed in (a)
\(S_n\) is the empirical standard deviation
\(A\) is the 99.5% quantile of the student distribution.