written 6.1 years ago by
teamques10
★ 64k
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modified 6.1 years ago
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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
$t=\frac{\bar X-\mu}{s/\sqrt n-1}=\frac{2.58-0}{\sqrt 8.74 / \sqrt 12-1}=2.89$
(iii)Level of significance : $\alpha=0.05$
(iv)Critical value : the value of $t_\alpha$ at 5% level of significance for ν= 12-1 =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.