Solution:
Before ($x_{1}$) |
After ($x_{2}$) |
Difference $di=x_{1}-x_{2}$ |
$di^{2}$ |
10 |
12 |
-2 |
4 |
15 |
17 |
-2 |
4 |
9 |
8 |
1 |
1 |
3 |
5 |
-2 |
4 |
7 |
6 |
1 |
1 |
12 |
11 |
1 |
1 |
16 |
18 |
-2 |
4 |
17 |
20 |
-3 |
9 |
4 |
3 |
1 |
1 |
|
|
$\sum di=-7$ |
$\sum di^{2}=29$ |
n=9
$\therefore$ n-1=9-1=8
$\bar{X} = \cfrac{\sum di}{n} = -\cfrac{7}{9}$
$\therefore \bar{X} = -0.7778$
$s^{2}=\cfrac{\sum di^{2} - (\sum di)^{2}}{n} = \cfrac{29- (-7)^{2}}{9}$
$s = \sqrt{\cfrac{29-(-7)^{2}}{9}} = 1.6178$
Step 1: The null hypothesis $H_{0}=\mu=0$
Alternative hypothesis $H_{a}=\mu =0$
Step 2: Calculation of the test statistic.
$\Longrightarrow$ Since sample is small,
$t=\cfrac{\bar{X}-\mu}{s/\sqrt{n-1}}= \cfrac{-0.7778-0}{1.6178/\sqrt{9-1}} = \cfrac{-0.7778}{0.5720} = -1.3598$
$\therefore |t| = |-1.3598|=1.3598$
Step 3: Level of significance = 5%
i.e., $\alpha=0.05$
Step 4: Critical Value $\Longrightarrow$ The value of $t_{\alpha}$ for 5% level of significance and degrees of freedom.
$V=9-1=8$ degrees of freedom is = 2.306 (percentage point distribution table)
Step 5: Decision
Since, $|t| \lt t_{a}$, i.e., calculated value of $|t|=1.3598$ is less than the critical value $t_{a}=2.306$, the hypothesis is accepted.
$\therefore$ The medication course was not accepted.