Question

You wish to estimate the proportion of adults that have ever used alternative medicine. Obtain a sample size that will ensure a margin of error of at most 0.01 for a 95% confidence interval. It is deemed reasonable to presume that of those sampled, the proportion that have used alternative medicine will be at least 0.64? Round to a whole number.

Answer #1

Solution :

Given that,

= 0.64

1 - = 1 - 0.64= 0.36

margin of error = E =0.01

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 )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.96 / 0.01)2 * 0.64 * 0.36

= 8851.0464

Sample size =8851

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