Question

Minnesota Public Radio wants to duplicate a survey conducted in 2016 that found that 65% of adults living in Minnesota felt that the country was going in the wrong direction. How many people would need to be surveyed for a 90% confidence interval to ensure the margin of error would be less than 3%? (You must show the formula and your algebraic work to receive credit for this problem.)

Answer 1

Answer:

The number of people needed  is $n =684$  

Step-by-step explanation:

From the question we are told that

  The population proportion is  $p = 0.65$

   The margin of error is  $E = 0.03$

From the question we are told the confidence level is  90% , hence the level of significance is    

      $\alpha = (100 - 90 ) \%$

=>   $\alpha = 0.10$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } =  1.645$

Generally the sample size is mathematically represented as

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

=>    $n =[ \frac{1.645 }{0.03} ]^2 * 0.65(1-0.65)$

=>    $n =684$