Question #97574
2. Newsweek performed a poll in which 567 American parents were asked the question and found that only 12% agreed with the question , “Would you prefer to have your child taught by a male for grades K-12?”. Find the 95% C.I. for the true proportion of Americans agreeing with their child being taught by a male for grades K-12.
1
Expert's answer
2019-11-01T10:45:06-0400

Let p^\hat{p} be the sample proportion. When nn  is large and pp is not close to zero or one, we can use the normal distribution to approximate the binomial.

The sampling distribution of p^\hat{p} is normal.

The mean of the sampling distribution is p.p.

The standard deviation of the sampling distribution (called the standard error) is


SE=p(1p)nSE=\sqrt{{p(1-p) \over n}}

Given that n=576,p=0.12.n=576, p=0.12.


SE=0.12(10.12)5760.013540SE=\sqrt{{0.12(1-0.12) \over 576}}\approx0.013540

Confidence Level 95%: zvalue=1.96z^*-value=1.96

Confidence interval for pp is


p^±z×SE\hat{p}\pm z^*\times SE

0.12±1.96×0.0135400.12±0.02650.12\pm 1.96\times 0.013540\approx0.12\pm0.0265

95% CI=(0.0935,0.1465)95\%\ CI=(0.0935, 0.1465)

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