Question

The bank manager wants to show that the new system reduces typical customer waiting times to less than 6 minutes. One way to do this is to demonstrate that the mean of the population of all customer waiting times is less than 6. Letting this mean be µ, in this exercise we wish to investigate whether the sample of 93 waiting times provides evidence to support the claim that µ is less than 6.

      For the sake of argument, we will begin by assuming that µ equals 6, and we will then attempt to use the sample to contradict this assumption in favor of the conclusion that µ is less than 6. Recall that the mean of the sample of 93 waiting times is = 5.48 and assume that σ, the standard deviation of the population of all customer waiting times, is known to be 2.24.

(a)	Consider the population of all possible sample means obtained from random samples of 93 waiting times. What is the shape of this population of sample means? That is, what is the shape of the sampling distribution of ?

  Normal because the sample is

.

(b)	Find the mean and standard deviation of the population of all possible sample means when we assume that µ equals 6. (Round your answer to 4 decimal places.)

  µ = 6, σ =

(c)	The sample mean that we have actually observed is = 5.48. Assuming that µ equals 6, find the probability of observing a sample mean that is less than or equal to = 5.48. (Round your answer to 4 decimal places.)

P( < 5.48)

(d)	If µ equals 6, what percentage of all possible sample means are less than or equal to 5.48? What do you conclude about whether the new system has reduced the typical customer waiting time to less than 6 minutes? (Round your answer to 2 decimal places.)

  %; conclude that µ is

than 6.

Answer 1

(a)  

Normal because the sample is larger than 30.
.
(b)  

µ = 6,

σ =2.24 /sqrt(93)

=0.2323

(c)  
µ =    6      
σ =    0.23227718      
          
P( X ≤    5.48   ) = P( (X-µ)/σ ≤ (5.48-6) /0.232277179619615)  
=P(Z ≤   -2.239   ) =   0.0126

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