Question

Most people can roll their tongues, but many can’t. The ability to roll the tongue is

genetically determined. Suppose we are interested in determining what proportion of students can roll their tongues. We test a simple random sample of $400$students and find that $317$can roll their tongues. The margin of error for a $95\%$confidence interval for the true proportion of tongue rollers among students is closest to which of the following?

a. $0.0008$

b. $0.02$

c. $0.03$

d. $0.04$

e. $0.05$

Short answer

Option (d) is correct.

Step-by-step solution

Given Information

It is given that $\left(n\right)=400$

$\left(\hat{p}\right)=\frac{317}{400}=0.7925$

Confidence level$=95\%=0.95$

Concept Used

Formula to be used is:

$E=z_{\alpha /2}\times \sqrt{\frac{\hat{p}\times (1-\hat{p})}{n}}$

Calculating Margin of Error

Corresponding to $95\%$confidence level, $z_{\frac{\alpha }{2}}=1.96$

Hence, margin of error,

$E=z_{a/2}\times \sqrt{\frac{\hat{p}\times (1-\hat{p})}{n}}$

$=1.96\times \sqrt{\frac{(0.7925\times (1-0.7925\left)\right)}{400}}$

$=0.0397\approx 0.04$

Option (d) is correct.