Question

A newspaper article states that 90% of drivers in Iowa wear their seatbelt. Suppose we are interested in estimating the proportion of drivers in Iowa City who wear their seatbelt when driving. We will observe drivers at different times of the day at different locations around town and record whether they are wearing their seatbelt. How large of a sample would we need to estimate this proportion to within 1% with 95% confidence?

Answer 1

Answer:

The sample size is  $n = 3457$

Step-by-step explanation:

From the we are told that

  The population proportion is  p =  0.90

  The margin of error is  E =  0.01  

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally the sample size is mathematically represented as

       $n =[ \frac{ Z_{\frac{\alpha }{2} } }{E} ]^2 * p(1-p)$

=>  $n =[ \frac{ 1.96 }{0.01} ]^2 * 0.90(1-0.90)$

=>   $n = 3457$