Out of 100 people sampled, 36 had kids. Based on this, construct a 95% confidence interval for the true
population proportion of people with kids.
Give your answers as decimals, to four places

Respuesta :

Answer:   (0.2659, 0.4541)

This is the same as saying 0.2659 < p < 0.4541

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Work Shown:

At 95% confidence, the z critical value is about z = 1.96 which is determined through a table or calculator.

x = number of people who had kids = 36

n = sample size = 100

phat = sample proportion of those who had kids

phat = x/n

phat = 36/100

phat = 0.36

E = margin of error

E = z*sqrt(phat*(1-phat)/n)

E = 1.96*sqrt(0.36*(1-0.36)/100)

E = 0.09408

This value of E is approximate because z = 1.96 is approximate.

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L = lower bound

L = phat - E

L = 0.36 - 0.09408

L = 0.26592

L = 0.2659

U = upper bound

U = phat + E

U = 0.36 + 0.09408

U = 0.45408

U = 0.4541

The values of L and U are approximate since E is approximate.

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The confidence interval in the format (L, U) is (0.2659, 0.4541)

This is equivalent to writing 0.2659 < p < 0.4541

which is the format L < p < U

We're 95% confident that the true population proportion (p) is somewhere between 0.2659 and 0.4541; i.e. we're 95% confident the true percentage of people who had kids is somewhere between 26.59% and 45.41%