Question

A study reported that 44% of Internet users have searched for information about themselves online. The 44% figure was based on a random sample of Internet users. Suppose that the sample size was

n = 300.

Construct a 90% confidence interval for the proportion of Internet users who have searched online for information about themselves. (Round your answers to three decimal places.)

( , )

Interpret the interval.

We are confident that the proportion of all Internet users who have searched online for information about themselves is within this interval 90% of the time.We are 90% confident that the proportion of all Internet users who have searched online is within this interval. We are 90% confident that the proportion of all Internet users who have searched online for information about themselves is within this interval.We are confident that 90% of the proportion of all Internet users who have searched online for information about themselves is within this interval.

Answer 1

$\hat{p}$ = 0.44

$CI = \hat{p} \pm z * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$CI = 0.44 \pm 1.645 * \sqrt{\frac{0.44(1-0.44)}{300}}$

CI = 0.44 + /- 0.0973

CI = (0.343 , 0.537)

We are 90% confident that the proportion of all Internet users who have searched online for information about themselves is within this interval