In a survey, the planning value for the population proportion is p* = 0.26. How large a sample should be taken to provide a 95% confidence interval with a margin of error of 0.05? (Round your answer up to nearest whole number.)

n = 296

Sample size n = 296

Step-by-step explanation:

Confidence interval can be defined as a range of values so defined that there is a specified probability that the value of a parameter lies within it.

The confidence interval of a statistical data can be written as.

p+/-z√(p(1-p)/n)

x+/-M.E

M.E = z√(p(1-p)/n)

Making n the subject of formula;

n = (p(1-p)/(M.E/z)^2) 1

Given that;

Proportion p = 0.26

Number of samples n = ?

Confidence interval = 95%

z(at 95% confidence) = 1.96

Substituting the values into equation 1;

n = (0.26(1-0.26))/((0.05/1.96)^2)

n = 295.649536

n = 296

Sample size n = 296

ok

step-by-step explanation:

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