Question

A graduate student in the School of Education is interested in whether families of students in the Chicago Public Schools are for or against the new legislation proposing school uniform requirements. She surveys 600 students and finds that 480 are against the new legislation. Compute a 90 and 98 percent confidence interval for the true proportion who are for the new legislation.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 480 / 600 = 0.80

1 - = 1 - 0.80 = 0.20

a) Z/2 = Z0.05 = 1.645

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

= 1.645 (((0.80 * 0.20) / 600 )

= 0.027

A 90% confidence interval for population proportion p is ,

± E

= 0.80  ±0.027

= ( 0.773, 0.827 )

b) Z/2 = Z0.01 = 2.33

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

= 2.33 (((0.80 * 0.20) / 600 )

= 0.038

A 98% confidence interval for population proportion p is ,

± E

= 0.80  ±0.038

= ( 0.762, 0.838 )