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In: Statistics and Probability

The manager of the new Super Target in West Covina wants to learn how much money...

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.

Solutions

Expert Solution

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

orchestra answered 1 year ago
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