Question

A study shows that 47% of young adults prefer to go out-of-state for college. How many young adults would you survey to state with 90% confidence that the estimated proportion is within 5% of the true proportion?

Group of answer choices

272

270

271

269

Answer 1

Solution :

Given that,

= 0.47

1 - = 1 - 0.47 = 0.53

margin of error = E = 5% = 0.05

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.645 / 0.05)2 * 0.47 * 0.43

=270

Sample size = 270