Question

In: Statistics and Probability

A poll conducted in 2013 surveyed a random sample of 2003 American adults, and 18% of...

A poll conducted in 2013 surveyed a random sample of 2003 American adults, and 18% of them said that they have seen a ghost. Construct a 95% confidence interval for the proportion of American adults who have seen a ghost.

Expert Solution

Solution :

Given that,

n = 2003

Point estimate = sample proportion = =18%=0.18

1 - = 0.82

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96 ( Using z table )

Margin of error = E = Z/2 * $\sqrt$ ((( * (1 - )) / n)

= 1.96 ( $\sqrt$ ((0.18*0.82) /2003 )

E = 0.017

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.18 - 0.017 < p < 0.18+0.017

(0.163 , 0.197)

orchestra answered 2 years ago

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