Statistics based question

Question

A local company wants to evaluate their quality of service by surveying their customers. Their budget limits the number of surveys to 200. What is their maximum error of the estimated population mean for a 95% level of confidence and an estimated standard deviation of 6?

 

Answer 1

The margin of error is 

 

m = (z^*) σ / sqrt(n)

 

For 95% confidence level z^* = 1.96 so 

 

m = (1.96)*6 / sqrt(200) = 11.76 / 14.142 = 0.832