Hey
service station advertises that customers will have to wait no more
than 30 minutes for...
Hey
service station advertises that customers will have to wait no more
than 30 minutes for an oriole change a sample of 28 oil change has
a standard deviation of five. Two minutes find the 95% confidence
interval of the population variance and standard deviation of the
time spent waiting for an oil change
Solutions
Expert Solution
95% confidence interval for population variance
is
[ 15.627 , 46.3173 ]
95 % confidence interval for standard deviation
is
A service station advertises that customers will have to wait no
more than 30 miniutes for an oil change. A sample of 25 oil changes
has a standard deviation of 5.2 minutes.
a) Find the 95% confidence interval of the population variance
of the time spent waiting for oil change.
b) Find the 95% confidence interval of the population standard
deviation of the time spent waiting for oil change.
A drugstore considers a wait of more than 5
minutes to be a defect. Each week 100 customers are randomly
selected and timed at the checkout line. The numbers of defects for
20 consecutive weeks are given below.
4 4 5 5 5 5 5 6 6 6 6 12 6 6 6 7 6 7 8 7
Find the correct lower control limit
El toro grande restaurant advertises that customers will have
their orders taken with three minutes of being seated. Management
wants to monitor average times, as it is such an important
guarantee for business. Construct x? - and s-charts for the data
given in the worksheet pro07-05 in the ch07data.xlsx file for this
chapter.
a. compute the sample means and the average standard deviation,
calculate the control limits, and plot them on control charts.
b. does the process appear to be...
Research by First Bank
of Illinois revealed that 8 percent of its customers wait more than
five minutes to do their banking when not using the drive-through
facility. Management considers this reasonable and will not add
more tellers unless the proportion becomes larger than 8 percent.
The branch manager at the Litchfield Branch believes that the wait
is longer than the standard at her branch and requested additional
part-time tellers. To support her request she found that, in a
sample...
Suppose the mean wait time for a bus is 30 minutes and the
standard deviation is 10 minutes. Take a sample of size n =
100.
Find the probability that the sample mean wait time is more than
31 minutes.
Suppose the mean wait time for a bus is 30 minutes and the
standard deviation is 10 minutes. Take a sample of size n =
100.
Find the 95th percentile for the sample mean wait time.
Suppose the mean wait time for a bus is 30 minutes and the
standard deviation is 10 minutes. Take a sample of size n =
100.
Find the probability that the sample mean wait time is between
29 minutes and 31 minutes.
An insurance company has an online help service for its
customers. Customer queries that take more than 5 minutes to
resolve are categorized as "unsatisfactory" experiences. To
evaluate the quality of its service, the company takes 10 samples
of 100 calls each while the process is under control. The control
chart that developed is described as follows: p = 0.1, UCL,,,, =
0.226 and LCL„ = 0. A new randomly selected sample shows that 30
queries took longer than 5...
A local hospital claims that you will not wait longer than 45
minutes in the emergency room. You go out to test this claim and
find a sample of 81 patients chosen at random from their emergency
room, you find a sample mean of 52 minutes with a sample standard
deviation of 23 minutes. Can you publicly debunk their claim (call
their claim fraudulent) at the 1% significance level? Use a test
stat and then a p-value to determine if...
A local hospital claims that you will not wait longer than 45
minutes in the emergency room. You go out to test this claim and
find a sample of 81 patients chosen at random from their emergency
room, you find a sample mean of 52 minutes with a sample standard
deviation of 23 minutes. Can you publicly debunk their claim (call
their claim fraudulent) at the 1% significance level? Use a test
stat and then a p-value to determine if...