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

6,250

step-by-step explanation:

to get the area, multiply 360 by 160.(57,600)

to find out how many bags of fertilizer it will take, first figure out how many square feet one bag can hold. divide 1000 by 8(125) to get the unit rate(how many square feet one pound of fertilizer can hold)and multiply it by 50(6250). if i haven't made a mistake, that is the answer.