Question

Recall that a bank manager has developed a new system to reduce the time customers spend waiting to be served by tellers during peak business hours. The mean waiting time during peak business hours under the current system is roughly 9 to 10 minutes. The bank manager hopes that the new system will have a mean waiting time that is less than six minutes. The mean of the sample of 90 bank customer waiting times is x⎯⎯ x ¯ = 5.43. If we let µ denote the mean of all possible bank customer waiting times using the new system and assume that σ equals 2.49:

(a) Calculate 95 percent and 99 percent confidence intervals for µ. (Round your answers to 3 decimal places.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 5.43

Population standard deviation = = 2.49

Sample size = n = 90

a)

At 95% confidence level the z is ,

Z_/2 = Z_0.05 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * ( 2.49 / 90 )

= 0.514

At 95% confidence interval estimate of the population mean is,

- E < < + E

5.43 - 0.514 < < 5.43 + 0.514

4.916 < < 5.944

( 4.916 , 5.944 )

At 99% confidence level the z is ,

Z_/2 = Z_0.025 = 2.576

Margin of error = E = Z_/2* ( /n)

= 2.576* ( 2.49/ 90 )

= 0.676

At 99% confidence interval estimate of the population mean is,

- E < < + E

5.43 - 0.676 < < 5.43 + 0.676

4.754 < < 6.106

( 4.754 , 6.106 )