A pollster wants to construct a 95% confidence interval for the proportion of adults who believe that economic conditions are getting better. A Gallup poll taken in July 2010 estimates this proportion to be 0.33. Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.010 ? Write only an integer as your answer.
Given that,
= 0.33
1 -
= 1 - 0.33= 0.67
margin of error = E = 0.010
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
Sample size = n = (Z/2
/ E)2 *
* (1 -
)
= (1.96 / 0.010)2 * 0.33 * 0.67
= 8494
Sample size = 8494
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