Recall that bank manger has developed 8 new system to reduce the time customer s

Question

Recall that bank manger has developed 8 new system to reduce the time customer spend waiting to be served by the tellers during peak business hours. The mean walking time during peak business hours under the current system is strongly is roughly 90 to 10 minutes. The bank manger hopes that the new system will have a mean waiting time that is less than six minutes. The mean of the sample of 98 bank customers waiting times is x = 5.49. If we let mu denote the mean of all possible bank customer waiting times using the new system and assume that sigma equals 2.41: (a) Calculate 95 percent and 99 percent confidence intervals for mu. (Round your answers to 3 decimal places.) (b) Using the 95 confidence interval, can the bank manager to 95 percent confident that mu is less than six minutes? Explain. ______, 95 percent interval is ______ 6. (c) Using the 99 percent confidence interval, can the bank manager be 99 percent confident that mu is less than the six minutes? Explain. ______, 99 percent interval extends ______ 6. (d) Based on your answers to parts b and c, how convinced are you that the new mean waiting time is less than six minutes? ______ confident, since the 95 percent Cl ______ 6 while the 99 percent Cl contains 6.

Explanation / Answer

here we assume that the population standard deviation is known as 2.41 thus we need to used z interval for estimating confidence interval for mean.

Using minitab

Step 1) Click on Stat>>>Basic Statistics >>1 sample Z...

Click on summarized data

Sample size = 98

Mean = 5.49

Standard deviation = 2.41

Step 2) then click on Option

Confidence level = 95.0

Alternative "not equal "

then click on OK

again click on OK

So we get the following output:

One-Sample Z

The assumed standard deviation = 2.41

N Mean SE Mean 95% CI
98 5.490 0.243 (5.013, 5.967)

Now, let's find 99% confidence interval for population mean .

Do the same steps as above just replace confidence leval = 99

The minitab output is

One-Sample Z

The assumed standard deviation = 2.41

N Mean SE Mean 99% CI
98 5.490 0.243 (4.863, 6.117)

b) See we find 95% confidence interval for true mean as (5.013, 5.967)

The upper limit of this interval < 6 so we can say that the mean is less than 6.

C) See we find 99% confidence interval for true mean as  (4.863, 6.117)

Here 6 is included in this interval so there does not have sufficient evidence to say the mean is less than 6

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