Question

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level. In an earlier study, the population proportion was estimated to be 0.22. How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 95% confidence level with an error of at most 0.02? Round your answer up to the next integer.

Answer 1

Answer:

The large sample n = 1713.96

Step-by-step explanation:

Explanation:-

given the population proportion was estimated to be 0.22

population proportion (P) = 0.22

The 95 % level of significance = 1.96≅ 2

The margin of  error = 0.02

The formula of margin error

                                   $\frac{2\sqrt{p(1-p)} }{\sqrt{n} } = 0.02$    …  ( i )

Substitute 'p' values in equation (1)

$\frac{2\sqrt{0.22(1-0.22)} }{\sqrt{n} } = 0.02$

cross multiplication and simplification, we get

$2X0.414= 0.02 \sqrt{n}$

$\frac{2X0.414}{0.02} = \sqrt{n}$

$41.4 = \sqrt{n}$

squaring on both sides, we get

the large sample n = 1713.96