0
2.8kviews
A certain drug administered to 12 patients resulted in the following change pressure.5, 2,8, -1,3, 0,6, -2 ,1, 5, 0 ,4. Can we conclude that the drug increases the blood pressure.
1 Answer
0
108views

$\text{First we calculate $\bar{X}$ and s$^2$}$ $\text{$\bar{X}$=2.58}$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline X & 5 & 2 & 8 & -1 & 3 & 0 & 6 & -2 & 1 & 5 & 0 & 4 & SUM \\ \hline 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 \\ \hline (X-\bar{X})^2 & 5.86 & 0.34 & 29.38 & 12.82 & 0.18 & 6.66 & 11.70 & 20.98 & 2.50 & 5.86 & 6.66 & 2.02 & 56.07 \\ \hline \end{array} $\text{s$^2$ = (∑ $\frac{(X – \bar{X})^2 }{9}$ = $\frac{104.96}{12}$ = 8.74}$ $\text{(i) The null hypothesis H$_0$ : μ = 0.}$ $\text{Alternative hypothesis H$_1$ : μ ≠ 0.}$ $\text{(ii) Calculation of test statistic: Since the sample size is small , we use t – distribution}$ $\text{t = $\frac{(\bar{X} - μ)}{(\frac{s}{\sqrt{(n-1)}})}$ = $\frac{(2.58-0)}{( \frac{\sqrt{8.74}}{\sqrt{(12-1)}})}$ = 2.89}$ $\text{(iii) Level of significance : α =0.05}$ $\text{(iv) Critical value : the value of t$_α$ at 5% level of significance for ν= 12-1 =11 degrees of freedom is 2.201}$ $\text{(v) Decision : since the calculated value of |t| =2.89 is greater than the table value t$_α$=2.201, the null hypothesis is rejected.}$

Please log in to add an answer.

Continue reading...

The best way to discover useful content is by searching it.

Search