Q.11.54 The Nipah Virus. During one year... [FREE SOLUTION]

Chapter 11: Q.11.54 (page 461)

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 臇ncephalitis 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

Expert verified

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 $1 - 伪$ ,

To find $z_{伪 / 2}$ using table II.

The amount of confidence is $90 %$ .

i.e. $伪 = 0.10$ .

Using table II,

$z_{伪 / 2} = z_{0.10 / 2}$

$= z_{0.05}$

$= 1.645$

The confidence interval for $p$ is given as-

$\hat{p} - z_{伪 / 2} 路 \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}} to \hat{p} + z_{伪 / 2} 路 \sqrt{\frac{\hat{p} (1 - \hat{p})}{n}}$ $鈬� 0.32 - 1.645 \sqrt{\frac{0.32 (1 - 0.32)}{94}} to 0.32 + 1.645 \sqrt{\frac{0.32 (1 - 0.32)}{94}}$ $鈬� (0.32 - 0.079) to (0.32 + 0.079)$

$鈬� 0.241 to 0.399$

$鈬� 24.1 % to 39.9 %$

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