Question

In a sample of 210 adults, 195 had children. Construct a 90% confidence interval for the true population proportion of adults with children.

Give your answers as decimals, to three places

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Answer 1

Solution :

Given that,

n = 210

x = 195

= x / n =195 / 210 = 0.929

1 - = 1 - 0.929 = 0.071

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * (((0.929 * 0.071) / 210) = 0.029

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.929 - 0.029 < p < 0.929 + 0.029

0.900 < p < 0.958

The 90% confidence interval for the population proportion p is : ( 0.900 , 0.958)