First we calculate $\bar X $&$s^2$
$\bar X=2.58$
X 
5 
2 
8 
1 
3 
0 
6 
2 
1 
5 
0 
4 
SUM 
$ X \bar X $ 
2.42 
0.58 
5.42 
3.58 
0.42 
2.58 
3.42 
4.58 
1.58 
2.42 
2.58 
1.42 

$ (X \bar X)^2 $ 
5.86 
.34 
29.38 
12.82 
0.18 
6.66 
11.70 
20.98 
2.50 
5.86 
6.66 
2.02 
56.07 
$ s^2 = \frac {\sum (x\bar x)^2}{9}=\frac{104.96}{12}=8.74$
(i)The null hypothesis$ H_0 : \mu= 0 $
Alternative hypothesis $H_1 : \mu ≠ 0$
(ii) Calculation of test statistic: Since the sample size is small , we use t – distribution
$t=\frac{\bar X\mu}{s/\sqrt n1}=\frac{2.580}{\sqrt 8.74 / \sqrt 121}=2.89$
(iii)Level of significance : $\alpha=0.05$
(iv)Critical value : the value of $t_\alpha$ at 5% level of significance for ν= 121 =11 degrees of freedom is 2.201
(v)Decision : since the calculated value of t =2.89 is greater than the table value $t_\alpha=2.201$ ,the null hypothesis is rejected.