Annual starting salaries for college graduates with degrees in business administration are generally expected to be between $30,000 and $50,000. Assume that a 95% confidence interval estimate of the population mean annual starting salary is desired. Given the information in the Microsoft Excel Online file below, construct a spreadsheet to determine how large a sample should be taken for each desired margin of error.

A. 384.16

B. 2,401

C. 9,604

D. No

Step-by-step explanation:

Calculation to determine how large a sample should be taken for each desired margin of error

First step is to find σ which represent Population Standard deviation

σ=($50,000-$30,000)/4

σ=$20,000/4

σ = 5,000

Now let calculate how large a sample should be taken for each desired margin of error

Using this formula

n = (Za/2*σ/E)^2

Where,

Za/2=1-0.95/2

Za/2=0.05/2

Za/2=0.025

Z-score 0.025=1.96

Za/2=1.96

σ =5,000

E represent Desired margin of error

Let plug in the formula

a. $500

n = (1.96* 5,000/$500)^2

n=(9,800/$500)^2

n=(19.6)^2

n = 384.16

b. $200

n = (1.96*5,000/200)^2

n=(9,800/$200)^2

n=(49)^2

n = 2,401

c. $100

n = (1.96*5,000/$100)^2

n=(9,800/$100)^2

n=(98)^2

n = 9,604

Therefore how large a sample should be taken for each desired margin of error will be :

A. $500= 384.16

B. $200= 2,401

C. $100= 9,604

d.NO, Based on the information calculation i would NOT recommend trying to obtain the $100 margin of error reason been that it is highly costly compare to $500 margin of error and $200 margin of error.

yor answer is 7/6 times 5

step-by-step explanation:

b

step-by-step explanation: