written 4.3 years ago by
teamques10
★ 26k

•
modified 4.1 years ago

(i) The null hypothesis $H_0$: The efficacy of vaccine in preventing tuberculosis.
Alternative hypothesis $H_a$: The efficacy of vaccine is not in preventing tuberculosis.
(ii) Calculation of test statistic: Table of calculated frequencies $(using formula (A×B)N)$
$
\begin{array}{cccc}
\hline & \text{Affected} & \text{Not Affected} & \text{Total} \\ \hline
\text{Inoculated} & 250 & 44 & 294 \\ \hline
\text{Not Inoculated} & 774 & 138 & 912 \\ \hline
\text{Total} & 1024 & 182 & 1206 \\ \hline
\end{array}
$
Calculation of $\frac{(O – E)^2}{E}$ :
$
\begin{array}{cccc}
\hline \text{O} & \text{E} & \text{(O  E)$^2$} & \text{$\frac{(O  E)^2}{E}$} \\ \hline
267 & 250 & 289 & 1.156 \\ \hline
27 & 44 & 289 & 6.568 \\ \hline
757 & 774 & 289 & 0.374 \\ \hline
155 & 138 & 289 & 2.094 \\ \hline
& & \text{Total} & 10.192 \\ \hline
\end{array}
$
(iii) Level of significance : $α = 0.05$
(iv) Degrees of freedom = $(r  1)(c  1)$ = $(2  1)(2  1)$ = 1
(v) Critical value : The table value of $χ^2$ at 5% level of significance for 1 degree of freedom is 3.81
(vi) Decision : Since the calculated value of $χ^2$ = 10.192 is greater than the table value $χ^2$ = $3.81$, the null hypothesis is rejected.