A Internet provider has implemented a new process for handli

A Internet provider has implemented a new process for handling customer complaints.

Based on a review of customer complaint data for the past 2 years, the mean time for handling a customer complaint was 27 minutes. Three months after implementing the plan, a random sample of the records of 50 customers who had complaints produced the following response times.Use the 50 data values to determine if the new process has reduced the mean time to handle customer complaints.

32.3 26.9 25.4 32.9 27.7 32.2 24.8 20.5 30.4 21.3 25.9 27.1 19.2 28.4 18.0 33.1 31.1 21.9 33.4 24.3 25.5 29.6 32.7 21.3 31.8 27.6 17.4 26.9 18.9 28.6 23.5 21.6 20.1 30.9 26.8 28.7 24.6 21.5 21.9 28.3 24.1 28.9 29.8 27.1 23.8 25.3 30.7 27.2 19.0 30.0

a. Estimate the mean time for handling a customer complaint under the new process using a 95% confidence interval.

b. Is there substantial evidence (a = .05) that the new process has reduced the mean time to handle a customer complaint?

c. What is the population about which inferences from these data can be made?

*please show the calculations on how you got the answers/explanations*

Solution

a)

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    26.218          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    4.46115616          
n = sample size =    50          
              
Thus,              
              
Lower bound =    24.98145332          
Upper bound =    27.45454668          
              
Thus, the confidence interval is              
              
(   24.98145332   ,   27.45454668   )

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b)

Formulating the null and alternative hypotheses,              
              
Ho:   u   >=   27  
Ha:    u   <   27  
              
As we can see, this is a    left   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    -   1.644853627      
              
Getting the test statistic, as              
              
X = sample mean =    26.07708333          
uo = hypothesized mean =    27          
n = sample size =    48          
s = standard deviation =    4.46448108          
              
Thus, z = (X - uo) * sqrt(n) / s =    -1.432227871          
              
Also, the p value is              
              
p =    0.076039309          
              
Comparing z and zcrit (or, p and significance level), we   FAIL TO REJECT THE NULL HYPOTHESIS.          
              
Thus, there is no substantial evidence (a = .05) that the new process has reduced the mean time to handle a customer complaint. [conclusion]

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c)

The population is all the customer complaint handling times of that specific internet provider.