written 8.3 years ago by | modified 2.3 years ago by |

Can it be concluded that the injection will be in general accompanied by an increase in blood pressure?

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A certain injection administered to 12 patients resulted in the following changes of blood pressure: 5, 2, 8, -1, 3, 0, 6, -2, 1, 5, 0, 4. Also check the following.

written 8.3 years ago by | modified 2.3 years ago by |

Can it be concluded that the injection will be in general accompanied by an increase in blood pressure?

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written 8.3 years ago by | • modified 8.3 years ago |

$$N = 12 ( < 30, 20 \ it \ is \ a \ small \ sample)$$

Step 1 :

Null hypothesis ($H_0)\ :\ {\mu }_1=0\ (\ i.e.\ the\ injection\ is\ not\ accompanied\ by\ an\ increase\ in\ blood\ pressure)$

Alternative hypothesis : ${\mu }_1\lt0\ (\ i.e.\ the\ injection\ is\ \ accompanied\ by\ an\ increase\ in\ blood\ pressure)$

(One test failed)

Step 2:

L.O.S = 5\% (Two failed test)

L.O.S = 10\% (One failed test)

Degree of freedom = n-1 = 12-1 = 11

Critical value ($t_d)=1.796$

Step 3:

Since sample is small: $\overline{d}=\ \frac{\sum{d_i}}{n}=\ \frac{31}{12}=2.5833$

s = $\sqrt{\frac{{\sum{d_i}}^2}{n}-\ {(\frac{\sum{d_i}}{n})}^2}$ = $\sqrt{\left(\frac{185}{12}\right)-\ {(\frac{31}{12})}^2}$ = 2.9569

S.E = $\frac{s}{\sqrt{n-1}\ }=\ \frac{2.9569}{\sqrt{11}}=0.3915$

Step 4:

Test statistic $t_{cal}=\ \frac{\overline{d_i}-\mu }{S.E} = \frac{2.5833-0}{0.8915}=2.8976$

Step 5:

Decision

Since |$t_{cal}|\gt t_{\alpha }$ , $H_0\ $is rejected.

$\mathrm{\therefore }$ The injection will be in general accompanied by an increase in blood pressure.

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