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Chapter 5 .

Memory capacity of 9 students was tested before and after a course of medication for a month. State whether the course was effective or not from the data below:

Before 10 15 9 3 7 12 16 17 4
After 12 17 8 5 6 11 18 20 3
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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.

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