In a random sample of 600 U.S. high school students, 11% (or 66 out of 600)...

In a random sample of 600 U.S. high school students, 11% (or 66 out of 600) said they do not plan on having children.

What sample size would allow us to create a 99% confidence interval for the true proportion of U.S. high school students who do not plan on having children with a margin of error of only 0.02?

Group of answer choices

1083

459

220

1625

882

Solutions

Expert Solution

Solution :

Given that,

= 0.11

1 - = 1 - 0.11 = 0.89

margin of error = E = 0.02

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.576 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (2.576 / 0.02)2 * 0.11 * 0.89

=1625

Sample size = 1625

orchestra answered 1 year ago

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