Question

In a certain region, 15% of people over the age of 50 didn’t graduate from high school. We would like to know if this percentage is the same among the 25-50 year age group. What is 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% of the true parameter with 99% confidence?

Answer 1

n = 236

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:

`n = ((Z_\alpha_/_2)^2 * p* ( 1 - p ))/(E^2)`

- 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:

`n = ((2.58)^2 * 0.15* ( 1 - 0.15 ))/(0.06^2)\\\\n = (6.6564 * 0.1275)/(0.0036)\\\\n = (0.848691)/(0.0036)\\\\n = 235.7475\\`

Answer: The minimum sample size would be next whole number integer, n = 236.

Answer 2

235 people

Explanation:

Given:

P' = 15% = 0.15

1 - P' = 1 - 0.15 = 0.85

At 99% confidence leve, Z will be:

`\alpha` = 1 - 99%

= 1 - 0.99 = 0.01

`\alpha /2 = (0.01)/(2) = 0.005`

`Z\alpha/2 = 0.005`

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 =
`((Z\alpha /2)/(E))^2 * P* (1 - P)`

`= ((2.576)/(0.06)) ^2 * 0.15 * 0.85`

= 235.02 ≈ 235

A number of 235 people between 25-30 years should be surveyed .