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.
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.
6 bags × 24 balloons = the amount of round balloons 6× 24 = 144 3 bags × 36 balloons = the amount of long balloons 3 ×36=108
108 long balloons - 144 round balloons = number of more long balloons that round ones 108-144=-36 He has -36 more long balloons that round ones Or 36 more round balloons that long ones
