Question

From earlier studies, it is believed that the percentage of students favoring a four-day school week during May and June of every school year is approximately 85%. What is the minimum sample size that will create a margin of error of 2% with 90% confidence

Answer 1

The minimum sample size 'n' = 862.538

Explanation:

Step(i):-

Given the percentage of students favoring a four-day school week during May and June of every school year is approximately 85%

Given the Proportion of students favoring a four-day school week during May and June of every school year

p = 85% = 0.85

q = 1-p = 1-0.85 = 0.15

Given Margin of error = 2 % or 0.02

Level of significance = 0.90 or 0.10

The critical value Z₀.₁₀ = 1.645

Step(ii):-

The Margin of error is determined by

`M.E = Z_(0.10) \sqrt{(p(1-p))/(n) }`

`0.02=1.645 X \sqrt{(0.85(0.15))/(n) }`

Cross Multiplication, we get

`√(n) =1.645 X {(√(0.125) ))/(0.02) }`

√n = 29.369

squaring on both sides, we get

n = 862.538

Final answer:-

Sample size 'n' = 862.538