Question

In a random sample of 1000 students, 80% were in favor of longer hours at the school library.

What is the standard error of the sample proportion?

Calculate the margin of error. Calculate a 95% confidence interval for proportion of students favoring longer library hours. Interpret.

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = 0.80

1 - = 0.20

Z_/2 = 1.96

standard error = (((0.80 * 0.20) /1000 ) = 0.1265

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

= 1.96 * (((0.80 * 0.20) /1000 )

= 0.025

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.80 - 0.025 < p < 0.80 + 0.025

0.775< p < 0.825