The size of the sample they should take to estimate p with a 2% margin of error and 90% confidence is n = 1691.
In statistics, the margin of error is just the degree of a significant error in the outcomes of random sample surveys.
The formula of margin error is, E = z√((p-vector)(1 - (p-vector)) ÷ n)
E = 2% = 0.02
Confidence level = 90%
Now, the proportion is not given so adopt nominal (p-vector) = 0.05
The critical value at CL of 90% is 1.645.
Thus, making n the subject,
n = z²(((p-vector) × (1 - (p-vector))) ÷ E²)
n = 1.645²((0.5 × 0.5) ÷ 0.02²)
n = 1691.266
n ≈ 1691
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The answer is nine and one third
