Question 440970
In a survey of 5100 T.V. viewers, 40% said they watch network news programs. Construct a 95% confidence interval for our estimate of the percentage of T.V. viewers who watch network news programs
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p-hat = 0.4
ME = z*sqrt(pq/n)
ME = 1.96*sqrt(0.4*0.6/5100) = 0.0134
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95% CI: 0.4-0.0134 < p < 0.4+0.0134
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Cheers,
Stan H.