Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $43,000 and $61,600. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. Required:
a. What is the planning value for the population standard deviation (to the nearest whole number)?
b. How large a sample should be taken if the desired margin of error is as shown below (to the nearest whole number)?

1. $500?
2. $200?
3. $100?

a. 4650

b.

1. 332

2. 2076

3. 8306

Step-by-step explanation:

a. The planning value for population standard deviation is,

s = (maximum - minimum) / 4

s = (61600 - 43000) / 4

s = 4650

that is, it would be 4650, the planning value for population standard deviation

b. we have to:

n = (z * s / E) ^ 2

z for confidence level 95% is 1.96, E = 500; 200; 100

replacing:

1.

n = (1.96 * 4650/500) ^ 2

n = 332.2 = 332

2.

n = (1.96 * 4650/200) ^ 2

n = 2076.6 = 2076

3.

n = (1.96 * 4650/100) ^ 2

n = 8306.4 = 8306

898.1666667

step-by-step explanation:

5389/6 = 898.1666667

      2

  2 is to go somewhat to the top right of a because it is squared    

v= a h/3