A sample of $40$ observations from a normal distribution $X$ gave $\Sigma x = 24$ and $\Sigma x^2 = 596$. Performing a two-tailed test at the $5 \%$ level, test whether the mean of the distribution is zero.
What I did: Found the population variance: $596.42 = (596 - [(24)^2)/40]*(40/39)$ Used the z statistic $(0.6 - 0)/(24.42/6.324)$ and got the value as $0.155$
Answer is $0.983$. Where am I wrong?
sample standard deviation $= \sqrt{\left(\dfrac{\sum (x-\bar x)^2}{n-1}\right)}$$= \sqrt{\left(\dfrac{\sum x^2 - 2\bar x \sum X + n\bar x ^2}{39}\right)}$
$=\sqrt{\left(\dfrac{\sum x^2 - \frac{(\sum X)^2}{n}}{39}\right)}$
$ = \sqrt{\left(\dfrac{596 - \frac{24^2}{40}}{39}\right)}$
$ s= \sqrt{14.91282}$
the t-statistic$ = \frac{\bar x - 0}{\frac{s}{\sqrt{n}}}$
$ = \dfrac{0.6 - 0}{\frac{\sqrt{14.91282}}{\sqrt{40}}}$
$ = .983$