A 1989 sample of 130 college women who visited a gynecologist at a particular un

Question

A 1989 sample of 130 college women who visited a gynecologist at a particular university in a northeastern U.S. indicated that 113 were sexually experienced. Assuming that these women were a simple random sample from the population of all women at that university, calculate a 95% confidence interval for the proportion of the population who are sexually active.

1. the interval and

a sentence describing the information that the intervals gives you, in context of the problem.

2.In regards to the previous problem, do you think it is reasonable to assume that these women form a random sample?

Explain your answer.

3.Would the interval have been wider, narrower, or the same width if 520 women had been samples? You don

Explanation / Answer

(1) Given a=1-0.95=0.05, Z(0.025) = 1.96 (from standard normal table)

p=113/130 = 0.8692308

So the lower bound is

p -Z*sqrt(p*(1-p)/n) =0.8692308- 1.96*sqrt(0.8692308*(1-0.8692308)/130) =0.811274

So the upper bound is

p +Z*sqrt(p*(1-p)/n) =0.8692308+ 1.96*sqrt(0.8692308*(1-0.8692308)/130) =0.9271876

We have 95% confident that proportion of the population who are sexually active will be within this interval.

--------------------------------------------------------------------------------------------------

(2) yes, because it collects 130 college women who visited a gynecologist at a particular university in a northeastern U.S

--------------------------------------------------------------------------------------------------

(3) the interval have been narrower because the sample size becomes large and the standard error becomse narrower.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.