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1
Many manufacturing problems involve the accurate matching of
machine parts such as shafts that fit into a valve hole. A particular
design requires a shaft with a diameter of 22.000 mm, but shafts with
diameters between 21.900 mm and 22.010 mm are acceptable. Suppose
that the manufacturing process yields shafts with diameters normally
distributed with a mean of 22.002 mm and a standard deviation of .005
mm.
What is the probability of producing an acceptable shaft?
(22.01 – 22.002)/0.005 = 1.6 = z-value
22.01 is 1.6 standard deviations above the mean
(21.9 – 22.002)/0.005 = -20.4 = z-value
21.9 is 20.4 standard deviations below the mean
Probability of being less than 1.6 σ above the mean is 0.9452
Probability of being less than 20.4 σ below the mean is 0.0000
Probability of being between the two points is 0.9452 – 0.0000 = 0.9452
What diameter will be exceeded by 95% of the shafts?
If 95% of the shafts are greater than this value, then 5% are less.
So, look up probability 0.05 in normal table and get z-value of -1.645.
So, the diameter is 1.645 standard deviations below the mean.
22.002 – 1.645 * 0.005 = 21.99378
If a random sample of 36 shafts is chosen from the above process, what
is the probability of the sample mean being above 20.003 mm?
(22.003 – 22.002)/(0.005/√36) = 1.2 = z-value
22.003 is 1.2 standard deviations above the mean
Probability of being less than 1.2 σ above the mean is 0.8849
Probability of being greater than 1.2 σ above the mean is 1.0000 - 0.8849
= 0.1151
2
McDonalds claims that their quarter-pounders have an average weight
of at least .25 lbs. The weight standard deviation is known to be .05 lbs.
State the null and alternative hypotheses.
Null: μ ≥ 0.25 pounds
Alternative: μ < 0.25 pounds
What test statistic will be used to test the null hypothesis?
Sample mean
A random sample of 36 quarter-pounders indicates a sample weight of
.2475 lbs.
At the 0.01 level of significance, is there evidence that the mean weight is
less than .25 lbs?
This is a one-tail test with a lower critical value, look up 0.01 in the
normal table and find a z-value of -2.33. The lower critical value is 2.33
standard deviations below the mean.
Lower critical value = 0.25 – 2.33 * (0.05/√36) = 0.230583
Since the sample mean of 0.2475 is greater than the lower critical value,
the sample statistic is in the region of nonrejection. Therefore, we accept
the null and conclude the average weight is at least a quarter pound.
There is not evidence that the mean weight is less than 0.25 pounds.
3
The quality-control manager at a light bulb factory needs to determine
whether the average life of a large shipment of light bulbs from a new
process is equal to the specified value of 375 hours. The process
standard deviation is 95 hours. State the null and alternative hypothesis.
Null: μ = 375 hours
Alternative: μ ≠ 375 hours
What test statistic will be used to test the null hypothesis?
Sample mean
A random sample of 64 light bulbs indicates a sample mean life of 360
hours.
At the 0.05 level of significance is there evidence that the mean life is
different from 375 hours?
This is a two-tail test with lower and upper critical values. Divide 0.05
by 2 and get 0.025, look up 0.025 in the normal table and find a z-value
of -1.96. The lower critical value is 1.96 standard deviations below the
mean and the upper critical value is 1.96 standard deviations above the
mean
Lower critical value = 375 – 1.96 * (95/√64) = 351.725
Upper critical value = 375 + 1.96 * (95/√64) = 398.275
Since the sample mean of 360 is between the two critical values, the
sample statistic is in the region of nonrejection. Therefore, we accept the
null and conclude the average life of a light bulb is equal to 375 hours.
There is not evidence that the mean life is different from 375 hours.
4
A local farmer asked me for help with his feed operation. The
manufacturer set up the feed dispenser to dispense 500 lbs of chop. The
farmer fears that his young son has been playing with the feed
dispensing dial and the setting has been changed. Thirty chop weights
are measured and the data is contained in the Feed Weights column of
the StatProblems Minitab worksheet.
Using Minitab, determine the 99% confidence interval for the true
mean weight.
508.27 to 516.40
Interpret the 99% confidence interval.
We are 99% confident that the average chop weight is between 508.27
and 516.40.
At the 99% confidence level, would I conclude that the young son has
changed the setting? Why?
Yes, because we are 99% confident that the average weight is greater
than 500 pounds.
Use Minitab to construct the following three graphs of the chop weights:
Histogram
Dot Plot
Run Chart
5
I charted the points that my favorite Wolfpack player scored over 15
games last year. The data is contained in the Points column. Using
Minitab determine the following measures:
Mean
22.33
Median
22.00
Mode
First Quartile
18.00
Third Quartile
28.00
Range
24.00
Variance
50.38
Standard Deviation
7.10
A company has been having a large amount of late deliveries to
customers. In order to improve deliveries, the company has been
tracking reasons for late deliveries. The reasons and data are contained
in the Late Reason and Quantity columns. Use Minitab to construct the
following graphs:
Bar Chart
Pie Chart
Pareto Diagram
6
Pepsi has a bottling plant that is dispensing pepsi in 20 ounce bottles.
They are concerned about the variation among two dispensing
machines, Machine 1 and Machine 2. They measured fifteen bottles
from the two machines and the data is in the Machine 1 and Machine 2
columns. Use Minitab to run an F Test for the Ratio of Two Variances
to determine if the amount of variation is equal between the two
machines with a level of significance of 0.01.
The p-value is 0.000 which is less than 0.01 so we reject the null and
conclude that the variation among the machines is not equal.
If the probability of a certain employee producing a defective part is
0.07 and the employee produces 20 parts per day, use Minitab to
compute the following probabilities (Assume independent parts):
Producing two defects
0.252141
Producing less than 3 defects
0.839
Producing at least four defects
1.00000 – 0.95287 = 0.04713
A call center receives an average of 12.35 calls per hour. Assume the
calls follow the Poisson distribution. Use Minitab to determine the
following probabilities:
Receiving exactly 11 calls
0.110572
Receiving no more than 14 calls
0.739476
Receiving more than 12 calls
1.000000 – 0.536004 = 0.463996