91Ó°ÊÓ

A random sample of eight observations X_1,….,X_s is taken from the normal distribution.

The sample data are such that \begin{aligned}\sum_{i-1}^{8}X_{i}=-11.2\end{aligned} and \begin{aligned}\sum_{i-1}^{8}X_{i}^{2}=43.7\end{aligned}

The symmetric t-test rejects the null hypothesis at level 0.1 if \begin{aligned}\left|T\right|> T_{n-1}^{-1}(0.95)\end{aligned} where

\begin{aligned}\T=\frac{\sqrt{n}\overline{X}}{s}\end{aligned}

Where s is the sample standard deviation.

Here n =8

\begin{aligned}\overline{X}=\frac{-11.2}{8}\end{aligned}

= -1.4

Sample standard deviation is:

\begin{aligned}s=\sqrt{\frac{1}{7}}\sum_{i-1}^{n}X_{i}^{2}-8\overline{X^{2}}\end{aligned}

\begin{aligned}=\sqrt{4}\end{aligned}

\begin{aligned}=2\end{aligned}

Therefore, the value of the T statistic turns out to be

\begin{aligned}T=\frac{-1.4\sqrt{7}}{2}\end{aligned}

= -1.852

That is, T = -1.852

The cut off value t table is \begin{aligned}\left|T\right|>T_{n-1}^{-1}(0.95)\end{aligned}

Since 1.852<1.89, the null hypothesis fails to be rejected.