Question

The Pew Research Center reported that in the U.S. in 2012, 16% of stay-at-home parents were fathers. https://www.pewsocialtrends.org/2014/06/05/growing-number-of-dads-home-with-the-kids/ Suppose the population proportion of stay-at-home parents that are fathers is .16. Further, suppose that we have a random sample of the population with 900 observations. What is the probability that the sample proportion from this random sample is within ±.01 of the population proportion?

Answer #1

Solution

Given that,

p = 0.16

1 - p = 1 - 0.16 = 0.84

n = 900

_{}
= p = 0.16

_{}
= [p
( 1 - p ) / n] =
[(0.16 * 0.84) / 900 ] = 0.0122

P( 0.15 < < 0.17)

= P[(0.15 - 0.16) /0.0122 < ( -
_{}
) /
_{}
< (0.17 - 0.16) / 0.0122]

= P(-0.82 < z < 0.82)

= P(z < 0.82) - P(z < -0.82)

Using z table,

= 0.7939 - 0.2061

= 0.5878

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