Answer:
See Explanation
Explanation:
Given
[tex]N = 1975[/tex] --- Population
[tex]s = 142[/tex] --- Selected
Required
Determine the sample size (n) using: [tex]n = \frac{z^2\sigma^2}{E^2}[/tex]
This question is incomplete as the margin of error (E) and the z-score are not given.
To solve this question, I'll assume values for z and E.
Calculating [tex]\sigma^2[/tex]
This is calculated as:
[tex]\sigma^2 = N * p * (1 - p)[/tex]
Where
[tex]p = \frac{s}{N}[/tex]
[tex]p = \frac{142}{1975}[/tex]
[tex]p = 0.0719[/tex]
So, we have:
[tex]\sigma^2 = N * p * (1 - p)[/tex]
[tex]\sigma^2 = 1975 * 0.0719 * (1 - 0.0719)[/tex]
[tex]\sigma^2 = 131.79[/tex]
The formula: [tex]n = \frac{z^2\sigma^2}{E^2}[/tex] becomes
[tex]n = \frac{z^2 * 131.79}{E^2}[/tex]
Assumptions:
[tex]z = 1.96[/tex]
[tex]E = 10[/tex]
[tex]n = \frac{1.96^2 * 131.79}{10^2}[/tex]
[tex]n = \frac{3.8416 * 131.79}{100}[/tex]
[tex]n = 5.06[/tex]
Round up: [tex]n = 6[/tex]
