We are given that a local company wants to evaluate their quality of service by surveying their customers. Their budget limits the number of surveys to 100.
Also, Confidence level = 95%
Standard deviation, = 5
Sample size, n = 100
Now, the Margin of error formula is given by =
Here, = value of z score at 2.5% significance level which is = 1.96 {using z table}
So, Margin of error =
= 1.96 * 0.5 = 0.98
Therefore, maximum error of the estimated mean quality for a 95% level of confidence and an estimated standard deviation of 5 is 0.98 .