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

Answer:

The minimum sample size 'n' = 862.538

Step-by-step 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

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

$0.02=1.645 X \sqrt{\frac{0.85(0.15)}{n} }$

Cross Multiplication, we get

 

         $\sqrt{n} =1.645 X {\frac{\sqrt{0.125} )}{0.02} }$

         √n = 29.369

squaring on both sides, we get

         n = 862.538

Final answer:-

Sample size 'n' = 862.538