Question

Out of 300 people sampled, 132 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to three places I am 90% confident that the proportion of people who have kids is between ___and___

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 132 / 300 = 0.440

1 - = 1 - 0.440 = 0.56

Z/2 = 1.645

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

= 1.645 * (((0.440 * 0.56) / 300)

Margin of error = E = 0.047

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.440 - 0.047 < p < 0.440 + 0.047

0.393 < p < 0.487

90% confident that the proportion of people who have kids is
between **0.393 and 0.487**

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