Answer to Question #290373 in Statistics and Probability for dafuq

Question

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?

Answer 1

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.

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