Question

Estimate the minimum sample size needed to achieve the margin of
error E= 0.023 for a 95% confidence interval

Answer #1

Solution :

Given that,

= 0.5 ( use 0.5 when not given )

1 - = 1 - 0.5= 0.5

margin of error = E = 0.023

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 ( Using z table ( see the 0.025 value in
standard normal (z) table corresponding z value is 1.96 )

Sample size = n = (Z_{/2}
/ E)^{2} *
* (1 -
)

= (1.96 / 0.023)^{2} * 0.5 * 0.5

= 1816

Sample size = 1816

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