Question

The Nipah Virus. During one year, Malaysia was the site of an encephalitis outbreak caused by the Nipah virus, a paramyxovirus that appears to spread from pigs to workers on pig farms. As reported y K. Goh et al. in the paper "Clinical Features of Nipah Virus Ėncephalitis among Pig Farmers in Malaysia" (New England Journal of Medicine, Vol. 342, No. 17, pp. 1229-1235), neurologists from the University of Malaysia found that, among $94$ patients infected with the Nipah virus, $30$died from encephalitis. Find and interpret a $90\%$ confidence interval for the percentage of Malaysians infected with the Nipah virus who would die from encephalitis.

Short answer

Between $24.1\%$ and $39.9\%$ of Malaysians infected with the Nipah virus will die of encephalitis, according to a $90\%$ confidence interval.

Step-by-step solution

Given Information

The sample size is $n=94$.

A number of successes is $x=30$.

The sample proportion would be $\hat{p}=\frac{x}{n}$

$=\frac{30}{94}$

$=0.32$

Explanation

An interval of confidence of$1-\alpha$,

To find $z_{\alpha /2}$using table II.

The amount of confidence is $90\%$.

i.e. $\alpha =0.10$.

Using table II,

$z_{\alpha /2}=z_{0.10/2}$

$=z_{0.05}$

$=1.645$

The confidence interval for $p$is given as-

$\hat{p}-z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\text{to}\hat{p}+z_{\alpha /2}·\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$\Rightarrow 0.32-1.645\sqrt{\frac{0.32(1-0.32)}{94}}\text{to}0.32+1.645\sqrt{\frac{0.32(1-0.32)}{94}}$$\Rightarrow (0.32-0.079)\text{to}(0.32+0.079)$

$\Rightarrow 0.241\text{to}0.399$

$\Rightarrow 24.1\%\text{to}39.9\%$