A researcher at a major clinic wishes to estimate the proportion of the adult po

Question

A researcher at a major clinic wishes to estimate the proportion of the adult population that has sleep deprivation. How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 6% A researcher at a major clinic wishes to estimate the proportion of the adult population that has sleep deprivation. How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 6%

Explanation / Answer

Solution:

Given that a = 1 – 0.90 = 0.10 and E = 0.06

Since population proportion is not known, we assume p = 0.5

Using Z-tables, the critical value is

Z (0.10/2) = Z (0.05) = 1.645

The sample size is given by:-

N = [Z/E]^2*p*(1 – p)

N = [1.645/0.06]^2*0.5*0.5

N = 751.6736*0.5*0.5

N = 187.9 ~ 188

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