Find the minimum sample size required to estimate the population proportion. if the required margin of error is 0.05, confidence level is 95% and p=0.15

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The margin of error is given by `ME = z_(alpha/2)*sqrt((p*(1-p))/n)`

For a confidence level of 95%, `z_(alpha/2) = 1.96`

`p = 0.15, 1-p = 0.85, ME = 0.05`

Substituting the values in the formula gives:

`0.05 = 1.96*sqrt((0.15*0.85)/n)`

=> `n = (1.96/0.05)^2*(0.15*0.85)`

=> `n ~~196`

**The minimum sample size is 196**

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