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A 95% confidence interval for percent of american adults who believe in aliens: (0.6578, 0.7822)
In this question we have been given that in a survey conducted on an srs of 200 american adults, 72% of them said they believed in aliens
We need to find the 95% confidence interval for percent of american adults who believe in aliens.
95% confidence interval = (p ± z√[p(1 - p)/n])
Here, n = 200
p = 72%
p = 0.72
And the z-score for 95% confidence interval is 1.960
The upper limit of interval would be,
(p + z√[p(1 - p)/n])
= 0.72 + 1.960 √[0.72(1 - 0.72)/200]
= 0.72 + 1.960 √[(0.72 * 0.28)/200]
= 0.72 + 1.960 √0.001008
= 0.72 + 0.0622
= 0.7822
The lower limit of interval would be,
(p - z√[p(1 - p)/n])
= 0.72 - 1.960 √[0.72(1 - 0.72)/200]
= 0.72 - 1.960 √[(0.72 * 0.28)/200]
= 0.72 - 1.960 √0.001008
= 0.72 - 0.0622
= 0.6578
Therefore, a 95% confidence interval = (0.6578, 0.7822)
Learn more about 95% confidence interval here:
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