BIA 654 STATISTICS HW
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Stevens Institute Of Technology *
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654
Subject
Industrial Engineering
Date
Dec 6, 2023
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docx
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6
Uploaded by MateRaven3704
1.
Motorola used the normal distribution to determine the
probability of defects and the number of defects expected in a
production process. Assume a production process produces
items with a mean weight of 10 ounces. Calculate the
probability of a defect and the expected number of defects for
a 1000-unit production run in the following situations.
Motorola defect probability and expected number of defects
a.
The process standard deviation is .15, and the process control
is set at plus or minus one standard deviation. Units with
weights less than 9.85 or greater than 10.15 ounces will be
classified as defects.
a. Probability of defects with a standard deviation of 0.15:
Z-score for 9.85 ounces:
Z = (9.85 - 10) / 0.15 approximately -1.00
Z-score for 10.15 ounces:
Z = (10.15 - 10) / 0.15 approximately 1.00
Using a standard normal distribution table or calculator:
The probability for Z < -1.00 is approximately 0.1587.
The probability for Z > 1.00 is also approximately 0.1587.
The total probability of defects is 2 * 0.1587 = 0.3174
Expected number of defects = P(defect) = 0.3174*1000 = 317
b. Through process design improvements, the process standard
deviation can be reduced to .05. Assume the process control
remains the same, with weights less than 9.85 or greater than
10.15 ounces being classified as defects
.
z_lower = (9.85 - 10) / 0.05 = -3
z_upper = (10.15 - 10) / 0.05 = 3
P(defect) = 0.9987 - 0.0013 = 0.9974
The probability of a defect after reduced standard deviation is 0.9974
Expected number of defects with reduced process variation:
Expected number of defects = P(defect) * sample size = 0.9974 * 1000 =
997
c. What is the advantage of reducing process variation, thereby
causing process control limits to be at a greater number of
standard deviations from the mean?
Reducing process variance has the benefit of enabling more items to meet
specification limitations, which lowers the number of faults. Reducing the
standard deviation will reduce
probability of defects in cases (a) and (b).
2.
After deducting grants based on need, the average cost to
attend the University of Southern California (USC) is $27,175
(U.S. News & World Report, America’s Best Colleges, 2009 ed.).
Assume the population standard deviation is $7400. Suppose
that a random sample of 60 USC students will be taken from
this population.
a. What is the value of the standard error of the mean?
a. Standard Error (SE):
SE = σ / √n
SE = $7,400 / √60 approximately $955.75
b. What is the probability that the sample mean will be more than
$27,175?
b. Calculate the Z-score:
Z = (27175 - 27175) / SEM = 0
P(Z > calculated Z) = 0.5
c. What is the probability that the sample mean will be within
$1000 of the population mean?
c. Calculate the Z-scores for the lower and upper limits:
Lower limit: 27175 - 27175 - 1000 = -1.046
Upper limit: 27175 - 27175 + 1000 = 1.046
P(lower limit < X < upper limit) = 0.704
d. How would the probability in part (c) change if the sample size
were increased to 100?
With a larger sample size (n = 100), SE = $7,400 / √100 = $740.
A smaller SE means a smaller margin of error and a more precise estimate
of the sample mean.
3.
A production process is checked periodically by a quality
control inspector. The inspector selects simple random samples
of 30 finished products and computes the sample mean
product weights
𝑥̅
. If test results over a long period of time
show that 5% of the
𝑥̅
values are over 2.1 pounds and 5% are
under 1.9 pounds, what are the mean and the standard
deviation for the population of products produced with this
process?
Given n=30
To calculate mean which is the average of the two given boundaries
M = 2.1+2.9/2 = 2.0
Z-score with a probability of 5% or 0.05
Z-score approximately -1.645
Solving the standard error using Z-score
z = (x-μ)/σ
-1.645 = (1.9-2.0)/
σ
σ
approximately equals to 0.0608
Standard deviation is the product of standard error and the square root of
the sample size
σ
= 0.0608
√30 approximately 0.3330
4.
A sample survey of 54 discount brokers showed that the mean
price charged for a trade of 100 shares at $50 per share was
$33.77 (AAII Journal, February 2006). The survey is conducted
annually. With the historical data available, assume a known
population standard deviation of $15.
Given data:
(x̄) = $33.77
population standard deviation (σ) = $15
(n) = 54
Confidence level = 95%
a.
Using the sample data, what is the margin of error associated
with a 95% confidence interval?
Calculate the Margin of Error (ME):
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The formula for the margin of error is:
ME = Z * (σ / √n)
margin of error: 4.0007
b.
Develop a 95% confidence interval for the mean price charged
by discount brokers for a trade of 100 shares at $50 per share.
Develop a 95% Confidence Interval:
Confidence Interval = (x̄ - ME, x̄ + ME)
Lower Limit of Confidence Interval:
=33.77 - 4.0007 = $29.77
Upper Limit of Confidence Interval:
=33.77 + 4.0007 = $37.77
5.
The 92 million Americans of age 50 and over control 50 percent
of all discretionary income (AARP Bulletin, March 2008). AARP
estimated that the average annual expenditure on restaurants
and carryout food was $1873 for individuals in this age group.
Suppose this estimate is based on a sample of 80 persons and
that the sample standard deviation is $550.
Given data:
(x̄) = $1,873
(s) = $550
(n) = 80
Confidence level = 95%
a.
At 95% confidence, what is the margin of error?
Margin of Error (ME):
The formula for the margin of error is:
ME(margin of error)
= Z * (s / √n)
ME = 120.5
b. What is the 95% confidence interval for the population mean
amount spent on restaurants and carryout food?
Develop a 95% Confidence Interval:
Lower Limit of Confidence Interval:
=1873 - ME = 1752.5
Upper Limit of Confidence Interval:
=1873 + ME = 1993.5
c. What is your estimate of the total amount spent by the 92
million Americans of age 50 and over on restaurants and carryout
food?
Total Amount Spent by Americans Aged 50 and Over:
Total Estimate = Sample Mean * Population Size
total expenditure =1873 * 92,000,000 = 172,316,000,000
d. If the amount spent on restaurants and carryout food is
skewed to the right, would you expect the median amount spent
to be greater or less than $1873?
If the amount spent on restaurants and carryout food is skewed to the
right, you would expect the median amount spent to be less than $1,873.
This is because in a right-skewed distribution, the tail on the right side is
longer, indicating that there are some individuals who spend significantly
more than the mean, which pulls the mean to the right, away from the
median.
6. The management of Regional Hospital has made substantial
improvements in their hospital and would like to test and
determine whether there has been a significant decrease in the
average length of stay of their patients in their hospital. The
following data has been accumulated from before and after the
improvements. At 95% confidence (5% significance), test to
determine if there has been a significant reduction in the average
length of stay.
After
Before
Sample size
45
58
Mean (in days)
4.6
4.9
Standard Deviation (σ)
0.5
0.6
a.
Formulate the hypothesis.
Null Hypothesis(H0): μ_after ≥ μ_before
Alternative Hypothesis : μ_after < μ_before
b. Compute the Test Statistic:
Test Statistic (t)
t = (X
‾
- μ0) / (s / √n)
Given
n
after
= Sample size after improvements = 45
n
before
= Sample size before improvements = 58
S
After = Standard deviation after improvements = 0.5
S
Before = Standard deviation before improvements = 0.6
Mean
After = Mean after improvements = 4.6
Mean
Before = Mean before improvements = 4.9
Using the values given and the pooled standard deviation S = 0.5074
Calculating T value using the formula above gives T= -2.77
c. Using the p-value approach, test to see if the average length of
stay in RH “after”
improvements is significantly less than the average length of stay
“before” improvements at RH. Let α = 0.05.
p-value = P(TS | H0 is true) = cdf(ts) = 0.0034
Comparing p-value to α (0.05):
IF p-value < 0.05, Reject H0
IF p-value < 0.05, Fail to reject H0
Since the p-value is less than 0.05, we can reject the null hypothesis and
conclude that there has been a significant reduction in the average length
of stay at the Regional Hospital after the improvements.
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