Question

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.

Answer #1

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}
= Z_{0.025} = 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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