Question

STATISTICS QUESTION: If a sample consisting of 64 working people indicates that 20 percent of them skip lunch, what is the 95 percent confidence interval for that proportion?

Answer 1

Solution :

Given that,

n = 64

= 0.200

1 - = 1 - 0.200 = 0.800

At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

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

= 1.960 * (((0.200 * 0.800) / 64)

= 0.098

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.200 - 0.098 < p < 0.200 + 0.098

0.102 < p < 0.298