sample_week_3__.xlsx

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Data
4.65
5.21
4.84
5.64
5.1
5.29
5.25
4.73
5.28
4.68
5.31
5.22
5.32
5.87
5.1
4.65
5.23
4.8
4.79
5.06
4.78
4.87
4.63
4.72
5.41
5.53
5.1
4.65
5.21
4.91
3.99
5.91
5.73
4.95
5.38
5.22
4.67
4.64
5.06
4.92
Mike is the manager of a flour factory in Michigan. Recently, he often receives complaints that the factory
flour packages often contain less than 5 pounds. Mike decides to do some test before calibrating the pack
He takes a sample of 40 packages and measure how many pounds of flower each package contain. He get
shown in Column A.
Question 1. Construct a 95% confidence interval on the average weight of packaged flour. Explain very ca
and to the packaging workers what the 95% confidence interval numbers mean.
Question 2. Mike’s factory has been supplying large amounts of flour to a supermarket chain. Due to the
chain decides to audit the flour packages by taking a sample of 40 packages from time to time. Mike want
that 95% of the time, the chain will find a sample mean of at least 5.0 pounds. Meanwhile, he doesn’t wan
machines to fill packages with any more product than needed.
Before the chain conducts any test, Mike took multiple samples of 40 in 2 weeks and eventually determin
mean of all packages is 5.03 pounds with a standard deviation the same as in Question 1.
What advice would you give to Mike as to how to calibrate their machines? In particular, should they incre
decrease the amount they are putting in the packages? Are there any other steps they might be able to ta
the situation?
Question 3. Mike calibrated the machines and provided training to workers. One month later, he determ
mean of all packages is 5.02 pounds with a standard deviation of 0.07 pound.
The supermarket chain is still auditing the flour packages by taking a sample of 40 packages from time to
going to find a sample mean of at least 5.0 pounds 95% of the time after the calibration and training?
ceives complaints that the factory’s 5 pounds
me test before calibrating the packaging machines.
wer each package contain. He gets the data
of packaged flour. Explain very carefully to Mike
a supermarket chain. Due to the complaints, the
ages from time to time. Mike wants to make sure
ounds. Meanwhile, he doesn’t want to adjust the
2 weeks and eventually determined that the true
as in Question 1.
es? In particular, should they increase or
ther steps they might be able to take to improve
kers. One month later, he determined the true
mple of 40 packages from time to time. Are they
r the calibration and training?
Data
4.65
5.21
4.84
5.64
5.1
5.29
5.25
4.73
5.28
4.68
5.31
5.22
5.32
5.87
5.1
4.65
5.23
4.8
4.79
5.06
4.78
4.87
4.63
4.72
5.41
5.53
5.1
4.65
5.21
4.91
3.99
5.91
5.73
4.95
5.38
5.22
4.67
4.64
5.06
4.92
Question 1:
Data
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Confidence Level(95.0%)
5.0575
0.061477
5.08
4.65
0.388816
0.151178
0.56402
0.040889
1.92
3.99
5.91
202.3
40
0.12435
Then continue to calculate the 95% confidence interval as below:
lower end = mean – Confidence Level (95.0%) =
4.93
upper end = mean + Confidence Level (95.0%) =
5.18
The population is all the 5 pounds flour packages produced in the factory
recently. The sample mean is 5.06 pounds (rounded from 5.0575). It is a point
estimate of the population mean. Is this point estimate accurate? How much
confidence do we have in it?
The confidence interval means that we can reasonably expect that 95% of the
means of all similar samples of 40 can fall between 4.93 to 5.18 pounds.
other words, if we take many samples of 40 packages again and again, we will
get many sample means. These means may differ from each other slightly. The
95% confidence interval tells us that 95% of these sample means are expected
to fall between 4.93 to 5.18 pounds. Therefore, we can conclude that
95% confident that the MEAN of all flour packages would fall between 4.93 to
5.18 pounds. It doesn’t mean that 95% of all flour packages fall between 4.93
– 5.18. We can see that 95% of the data do not necessarily fall in this interval.
The mode, mean and median are not close. The range is huge. These indicate a
relatively wide distribution of products and the producing process may not be
in control. The manager may need to bring the process into control by
calibrating the machines.
Go to Data tab, click Data Analysis, then to Descriptive Statistics, then
OK. For the input range, cursor over the entire data set (Cells A1:A41).
Check Label in First Row. The data need to be grouped by columns.
Select a blank cell as the starting cell of the Output Range. Check
Summary Statistics, and make sure that the Confidence Level for the
Mean is 95%, then click OK. The output should look something like the
left.
ced in the factory
m 5.0575). It is a point
accurate? How much
expect that 95% of the
3 to 5.18 pounds. In
gain and again, we will
each other slightly. The
le means are expected
conclude that we are
uld fall between 4.93 to
kages fall between 4.93
ily fall in this interval.
huge. These indicate a
process may not be
nto control by
Question 2:
According to the given information,
40
sample size
n=
5.03
population mean (mean of sample means)
m=m =
s=
0.388816
Standard deviation of population
Calculate Standard deviation of sample means (standard error of the mean)
σ = σ/√n = 0.388816/√40 =
0.0614772
Calculate probability:
P(drawn sample mean < 5.0) = P(drawn sample mean >= 5.0) =
0.31278
1 – 0.313 =
See textbook Black (2010) Pages 230
68.7%
Since the sample size n is 40 (big enough), the central limit theorem may be applied. As a
result, it may be assumed that the sample means are normally distributed. This normal
distribution has a mean of 5.03 and a standard deviation of 0.061477. Based on these data,
we can calculate the probability of getting a sample mean no less than 5.0 (cutoff point X)
on this normal distribution. It can be completed by using the NORM.DIST formula in Excel
shown below:
Use Excel Formula NORM.DIST (enter the values and text shown in bold):
Formulas – Insert Function – NORM.DIST – OK – X 5.0
Mean 5.03
Standard_dev 0.061477
Cumulative true
The result is 0.312. It means that P(drawn sample mean < 5.0) = 0.313. Then P(drawn sample mean >= 5.0) = 1 – P(drawn sample mean < 5.0) = 1 – 0.313 = 0.687 = 68.7%. This result indicates that only 68.7% of randomly drawn samples will have a mean equal to or greater than 5.0 pounds. That is, only for 68.7% of the time, the supermarket chain will find a sample mean of at least 5.0 pounds. This doesn't meet Mike's expectation, which is to make sure that 95% of the time, the chain will find a sample mean of at least 5.0 pounds. Short Term Solution: The short term solution is to calibrate the machines to add a bit more flour to each package so that the whole distribution can be pushed to the right. This way, we will ensure that 95% of the time, the chain will find a sample mean of at least 5.0 pounds. Long Term Solution: Given the relatively wide distribution of product, the process may not be in control. The larger issue is to bring the process into control. 0.313 0.687 or 68.7% x=5.0 µ =5.03 σ =0.06 Question 3: According to the given information, 40 sample size n= 5.02 population mean (mean of sample means) m=m = s= 0.07 Standard deviation of population Calculate Standard deviation of sample means (standard error of the mean) σ = σ/√n = 0.388816/√40 = 0.011068 Calculate probability: P(drawn sample mean < 5.0) = P(drawn sample mean >= 5.0) =
See textbook Black (2010) P
1.0003
0.035380
1 – 0.035 =
96.5%
Since the sample size n is 40 (big enough), the central limit theorem may be applied. As a result, it
may be assumed that the sample means is normally distributed. This normal distribution has a
mean of 5.02 and a stardard deviation of 0.011068. Based on these data, we can calculate the
probablity of getting a sample mean no less than 5.0 (cutoff point X) on this normal distribution.
It can be completed by using the NORM.DIST formula in Excel shown below:
Use Excel Formula NORM.DIST (enter the values and text shown in bold):
Formulas – Insert Function – NORM.DIST – OK – X 5.0
Mean 5.02
Standard_dev 0.011068
Cumulative true