**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.