Question

1. 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 100 bank customer waiting times in Table 1.9 is 5.46. If we let u denote the mean of all possible bank customer waiting times using the new system and assume that the population standard deviation equals 2.47: a. Calculate 95 percent and 99 percent confidence intervals for u. b. Using the 95 percent confidence interval, can the bank manager be 95 percent confident that u is less than six minutes? Explain. c. Using the 99 percent confidence interval, can the bank manager be 99 percent confident that u is less than six minutes? Explain.

Answer 1

n = 100 X = 5.46 20 - 2.47. a) for pe 95% confidence Internals [20.05/2 J 5.46 + .96 2.471 vioo E [ 5.46 + 0.48412 ] pe E [ 4.9758 , 5.94412 ]. 99 % Confidence Interuals for E [ at 20.0/2 on 6 [ 5.46 2.5758 2.47 7 V100 E [ 5.46 $ 0.6362] 6 [ 4.8238 , 6.0962).

Yes do usering 95% confidence intowal, the nk manager came be 95% confident that re es less than 6 minutes. Because the gry. c. I is less than 6 miss 9 No, ureng 99 % c. I, the bank manager cannot be ggy. Confident that pe is less than 6 memuts because the ggy.c. contains 6 mess