Question

In a sample of 270 adults, 216 had children. Construct a 99% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to three places

Answer 1

Solution :

Given that

n = 270

x = 225

= x / n = 216 / 270 = 0.800

1 - = 1 - 0.800 = 0.200

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.800 * 0.200) / 270 )

= 0.063

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.800 - 0.063 < p < 0.800 + 0.063

0.737 < p < 0.863