Answer:
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.
