Math Help Forum - View Single Post

chromium · #1 June 30th, 2009, 12:08 PM

Hello all! I would really appreciate some help. I kind of used an equation that seemed to work.

If I have 9 observations, specifically 240,216,265,284,229,232,250,225,252, and need to find a 95% confidence interval for the mean $\mu$ assuming that these are observations of a $N(\mu,\sigma^{2})$ distribution, can I do the following?:

( $\bar{x}-t_{\alpha/2}\frac{s}{\sqrt{n}},$ $\bar{x}+t_{0.025,8}\frac{s}{\sqrt{n}})$
*I think $\alpha/2$ becomes 0.025 with 8 degrees of freedom and therefore $t_{\alpha/2}=2.306$

$(243.66667-2.306\frac{21.418449999}{\sqrt{9}}, 243.66667+2.306\frac{21.418449999}{\sqrt{9}})$
(227.20, 260.13)