Question

Beth wants to determine a 99 percent confidence interval for the true proportion ? of high school students in the area who attend their home basketball games. Out of ? randomly selected students she finds that that exactly half attend their home basketball games. About how large would ?nhave to be to get a margin of error less than 0.01 for ??
[Use the values for z* from a z-table or t-table, and round to the smallest integer that works.]

?≈

Answer 1

Solution,

Given that,

= 0.51 - =  

margin of error = E = 0.01

At 99% confidence level

= 1 - 99%  

= 1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005  = 2.576

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

= (2.576 / 0.01)2 * 0.5 * 0.5

= 16589.44

sample size = n = 16590