A bank officer wants to determine the amount of the average total monthly deposits per customer at the bank.

Question

He believes an estimate of this average amount using a confidence interval is sufficient. He assumes the standard deviation of total monthly deposits for all customers is about $1,475. How large a sample should he take to be within $200 of the actual average with 95% confidence?

Answer 1

The expected value of the sample mean, E(m), is the population mean, μ.

 

The standard deviation of the sample mean (standard error), s_m, is the population standard deviation, s, (1475, in this case) divided by the square root of the sample size, i.e., s/√n.

 

(m-μ)/s_m is a standard normal variate, z.

 

From a table of the standard normal variate, you find that there is a 95% probability that z will be between -1.96 and +1.96.

 

Therefore, you want 1.96(s_m)<=200.

 

1.96(s/√n)<=200

 

√n>=1.96s/200=1.96(1475)/200=14.455

 

n>=14.455²=209