Respuesta :
Answer:
235 people
Step-by-step explanation:
Given:
P' = 15% = 0.15
1 - P' = 1 - 0.15 = 0.85
At 99% confidence leve, Z will be:
[tex] \alpha [/tex] = 1 - 99%
= 1 - 0.99 = 0.01
[tex] \alpha /2 = \frac{0.01}{2} = 0.005 [/tex]
[tex] Z\alpha/2 = 0.005 [/tex]
Z0.005 = 2.576
For the minimum number of 25-50 year old people who must be surveyed in order to estimate the proportion of non-grads to within 6%, we have:
Margin of error, E = 6% = 0.06
sample size = n = [tex] (\frac{Z\alpha /2}{E})^2 * P* (1 - P) [/tex]
[tex] = (\frac{2.576}{0.06}) ^2 * 0.15 * 0.85 [/tex]
= 235.02 ≈ 235
A number of 235 people between 25-30 years should be surveyed .
Answer:
n = 236
Step-by-step explanation:
Solution:-
- The proportion of people over the age of 50 who didn't graduate from high school are, p = 0.15 - ( 15 % )
- We are to evaluate the minimum sample size " n " from the age group of 25-50 year in order to estimate the proportion of non-grads within a standard error E = 6% of the true proportion p within 99% confidence.
- The minimum required sample size " n " for the standard error " E " for the original proportion p relation is given below:
[tex]n = \frac{(Z_\alpha_/_2)^2 * p* ( 1 - p )}{E^2}[/tex]
- The critical value of standard normal is a function of significance level ( α ), evaluated as follows:
significance level ( α ) = ( 1 - CI/100 )
= ( 1 - 99/100 )
= 0.01
- The Z-critical value is defined as such:
P ( Z < Z-critical ) = α / 2
P ( Z < Z-critical ) = 0.01 / 2 = 0.005
Z-critical = Z_α/2 = 2.58
- Therefore the required sample size " n " is computed as follows:
[tex]n = \frac{(2.58)^2 * 0.15* ( 1 - 0.15 )}{0.06^2}\\\\n = \frac{6.6564 * 0.1275}{0.0036}\\\\n = \frac{0.848691}{0.0036}\\\\n = 235.7475\\[/tex]
Answer: The minimum sample size would be next whole number integer, n = 236.
