In: Statistics and Probability
The manager of the new Super Target in West Covina wants to learn how much money the average West Covina family spends in similar stores yearly within a $100 margin of error. Past research has shown that the average amount spent is $2500 per year, with a standard deviation of $300. What sample size will he need for an estimate with a 99% level of confidence?
Please show all your work (Formulas and Math) for full points.
Solution :
Given that,
standard deviation = =$300
Margin of error = E = $100
At 99% confidence level the z is,
= 1 - 99%
= 1 - 0.99 = 0.01
/2 = 0.005
Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )
sample size = n = [Z/2* / E] 2
n = ( 2.58* 300/ 100 )2
n =59.90
Sample size = n =60