Question

The Nipah Virus. Refer to Exercise 11.54.

a. Find the margin of error for the estimate of the percentage.

b. Obtain a sample size that will ensure a margin of error of at most five percentage points for a $90\%$confidence interval without making a guess for the observed value of $\hat{p}$.

c. Find a $90\%$ confidence interval for $p$ if, for a sample of the size determined in part (b), $28.8\%$ of the sampled Malaysians infected with the Nipah virus die from encephalitis.

d. Find the margin of error for the estimate in part (c) and compare it to the margin of error specified in part (b).

e. Repeat parts (b)-(d) if you can reasonably presume that the percentage of sampled Malaysians infected with the Nipah virus who would die from encephalitis would be between $25\%$ and $40\%$.

Short answer

(a) The margin of error is $7.91\%$

(b) Required Sample Size is $271$.

(c) $90\%$Confidence Interval of population is $(0.2428,0.3332)$

(d) The margin of error in (c) is less than that of (b)

(e) The margin of error in this case is$0.0462$.

Step-by-step solution

Given Information

It is given that $\hat{p}=0.3191,n=94$, confidence interval is $90\%$

Value of$z_{\frac{\alpha }{2}}$is$1.645$

(a) Margin of Error

It is calculated as $E=z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=1.645\sqrt{\frac{0.3191(1-0.3191)}{94}}$

$=1.645(0.0481)~7.91\%$

Margin of Error is$7.91\%$

(b) Sample Size when Error is 5%

It is obtained as $n=\hat{p}(1-\hat{p}){\left(\frac{\frac{z_{\alpha }}{2}}{E}\right)}^2$

$=0.5(1-0.5){\left(\frac{1.645}{0.05}\right)}^2$

$=(0.25)(1,082.41)=270.6$

The sample size is$271$(approx)

(c) 90% confidence interval when n=271 and p^=0.288

It is calculates as $\hat{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=0.288\pm 1.645\sqrt{\frac{0.288(1-0.288)}{271}}$

$=0.288\pm 1.645(0.0275)$

$=(0.2428,0.3332)$

The required confidence interval is$(0.2428,0.3332)$

(d) Comparison of Margin of Error

The margin of error of (c) is $0.0452$

Hence, margin of error of (c) is less than that of (b).

Margin of error when n=260 and p^=0.288. 

As per the information, $90\%$confidence interval is calculated as

$\hat{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.288\pm 1.645\sqrt{\frac{0.288(1-0.288)}{260}}$

$=0.288\pm 1.645(0.0281)=(0.2418,0.3342)$

Above result gives margin of error as $0.0462$.

It is less than that in part (b).

(f) Comparison

From above results, sample size is reduced by $11$in (e).

The margin of error is increased$0.045\text{to}0.046$