Question

Your professor wishes to estimate the proportion of ALL high school students enrolled in college-level courses each school year. A sample of 1500 students revealed that 18.3% were enrolled in college-level courses. Find the margin of error for a 90% confidence interval for a proportion.

Answer 1

Solution :

Given that,

n = 1500

Point estimate = sample proportion = =18.3%=0.183

1 - = 1-0.183=0.817

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

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

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.183*0.817) / 1500)

= 0.016