Out of 100 people sampled, 51 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
? < p < ?
Ans:
sample proportion=51/100=0.510
90% confidence interval for the true population proportion
=0.510+/-1.645*sqrt(0.510*(1-0.510)/100)
=0.510+/-0.082
=(0.428, 0.592)
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