Among 250 employees, of the local office of an international insurance company, 182 are whites, 51 are blacks and the rest are Orientals. If 5% margin of error is used, 5. How many employees will be chosen for a case study as samples? 6. How many Orientals will be chosen? 7. How many whites will be chosen?

Margin of error = 5%

e= 0.05

Confidence level = 100-5= 95%

Population size, N = 250

Since here we don't know the population proportion so, p= 0.5 is assumed.

Now we find Z at 95% confidence level using the normal distribution table and that is

Z= 1.96 at 95% confidence level

5. The formula for sample size:

$n= \frac{\frac{Z^2p(1-p)}{e^2}}{1+\frac{Z^2p(1-p)}{Ne^2}} \\ n= \frac{\frac{1.96^2 \times 0.5(1-0.5)}{0.5^2}}{1+\frac{1.96^2 \times 0.5(1-0.5)}{2050 \times 0.05^2}} \\ n=151.44 = 152$

6. In the population , the number of orientals = 250-182-51=17

The proportion of orientals

$p= \frac{17}{250} = 0.068$

Substituting this value of p in the above formula we get the number of orientals in the sample size of 152.

$n_{Oriental} = \frac{\frac{1.96^2 \times 0.068(1-0.068)}{0.05^2}}{1 + \frac{1.96^2 \times 0.068(1-0.068)}{250 \times 0.05^2}} \\ = 70.08 = 71$

7. Number of whites in the population = 182

Proportion of whites

$p= \frac{182}{250}= 0.728$

On substituting this value of p in the above formula we get the number of whites in the sample.

$n_{Whites} = \frac{\frac{1.96^2 \times 0.728(1-0.728)}{0.05^2}}{1 + \frac{1.96^2 \times 0.728(1-0.728)}{250 \times 0.05^2}} \\ = 137.24 = 138$