Homework 5 SCM 200 (1)
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School
Pennsylvania State University *
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Course
200
Subject
Industrial Engineering
Date
Feb 20, 2024
Type
Pages
9
Uploaded by ChefCrownCat20
Homework 5 (non-Excel portion)
Fall 2022
Must be uploaded into Canvas by 11:55 pm on Monday, November 7, 2022.
Upon submitting this homework, I affirm that I have not given or received unauthorized aid on this
homework, and I completed this work honestly and according to the instructor’s guidelines.
Answers may be typed or handwritten.
Homeworks are graded on completion, not accuracy. Work must be shown on the homework to receive
full credit. Credit for completion will not be awarded if adequate effort has not been shown. Each
homework is worth 2.5 points. So, for example, if a student gets a score of 2.5, it means that they did all
the work, it does not mean the answers are all correct.
In order to promote professional development of students in learning the importance of appearance and
presentation of submitted work, 0.1 point will be deducted from homework scores for each the following
conditions that are not met (students could lose a total of 0.2 points for not meeting these conditions):
●
Questions numbered and done in order (all work and answers for each question in one
place)
●
Increased space between each question if the answers are included on the posted homework
assignment
If you take any definitions/concepts directly from the packet, please cite the page number in the
packet.
YOU MUST SHOW WORK TO GET CREDIT.
1.A soft drink company was planning to buy up a smaller company. The company was
interested in two different smaller companies, Company A and Company B. The soft drink
company wanted to estimate the percentage of people who preferred Company A’s soft drink
over Company B’s soft drink.
A poll revealed that 296 out of 370 randomly selected people
indicated a preference for Company A’s soft drink. Find a 95.96% confidence interval for the
proportion of people who favored Company A’s soft drink. You must conclude your work by
stating in words what this confidence interval means in the context of people preferring
Company A.
Confidence Interval: P ± z √P (1-p)
n
296 = 0.8
370
1- .9596 = .0202
2
.0202 z table = -2.05
.8 ± 2.05 √.8 (1-.8)
370
.8 ± 2.05 √.16
370
.8 ± 2.05 (.020795)
.8 ± 0.0426 = (.7574 , 8.43)
We are 95.96% confidence that the proportion of people who favor Company A, ranges between 75.74%
to 84.3%
(Page 146 of SCM 200 Course Packet)
2.A promoter is deciding whether to book a band. The promoter decides to do a survey to try
to estimate the true proportion of individuals in the area who will attend the concert. What
should sample size be to estimate the proportion within 2% with a 98.22% confidence level.
Assume maximum error possible. Include the units in your answer.
1 - .9822
= .0089
2
Z = 2.37
n =
π (1 - π) [ z ] ^2
E
n = .5 ( 1 - .5) [2.37] ^2
.02
n = .5 (.5) [2.37] ^2
.02
n = .25 / 14042.25
n = 3510.5625
3511 individuals in the Sample Size
(Page 149 of SCM 200 Course Packet)
(Questions 1-2 can be completed after Lecture 23.)
3.A computer manufacturer claims that at most 8% of computers made will become defective in the
first two years. (200) Two hundred computers were randomly selected and checked for defects. If
12 defects were reported, what conclusion should be reached about the company’s claim if alpha is
.05. Show all steps of hypothesis testing. You must state your conclusion in terms of the
proportion of defective computers.
H
0
: μ ≤ .08
H
A
: μ > .08
∝
(alpha)
=
.05
p = 12 = .06
200
z = .06 - .08
√ .08 (.92)
=
-.02
= -.104
200
0.191833
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P-value = (z < -.104) = .4602
Conclusion: Accept
H
0
(the null) because the p-value is larger in value, compared to the
alpha. Concluding that the percentage of defective products is no more than 8%
(Page 150-151 of SCM 200 Course Packet)
(Question 3 can be completed after Lecture 24.)
4.What are the degrees of freedom for a paired t-test when n
1
= 24 and n
2
= 24?
df = n - 1
24 - 1 = 23
23 degrees of freedom
(Page 156 of SCM 200 Course Packet)
5.Write a scenario of a problem that would be appropriate for a paired t-test. Do not use the
scenarios that are included in the course packet. If you base your idea off of another source,
you must state the source.
Assume a hair-dryer manufacturer company is testing a new assembling system that will result in
two different processing rates. The blow-drying company wants to know if the new assembling
system is as equally different as the current system, by looking at the differences between the
processing rates we can come to a conclusion. With the mean difference of the two-tailed
alternative hypotheses, the number of paired differences range to the sample size of 24.
(Page 153 of SCM 200 Course Packet)
6.A doctor claims that the average person is at least 12 pounds overweight. To test the claim, the
difference between actual and ideal weight of 36 randomly selected people was calculated. The
sample mean and sample standard deviation were 10 pounds and 2 pounds respectively. At alpha of
1%, what conclusion would you reach? Show all steps of hypothesis testing. You must state your
conclusion in terms of the mean difference of the weight of the average person.
H
0
: μ
≥
12
H
A
: μ < 12
t = (D-bar) -
μ
D
S
D
/ √ n
t = 10 - 12
= - 2
= -6
2 / √36
2/6
α (alpha) = .01
p value= -6
P value < .00001
The average person is less than 12 pounds overweight, concluding that the claim is true, there is a
difference between the actual and ideal rate of 36 people in the sample size. Reject the null (H
0
) and
accept the claim as true.
(Page 158 of SCM 200 Course Packet)
(Questions 4-6 can be completed after Lecture 25.)
7.What are the degrees of freedom for an independent t-test when n
1
= 30 and n
2
= 44?
n
1
+ n
2
= 2
30 + 44 -2 = 72
74 degrees of freedom
(Page 158 of SCM 200 Course Packet)
8.The following information is given about a hypothesis test of the difference between two
means based on independent random samples. (Assume normal distributions with equal
variances.) What conclusion should be reached if alpha = .05? Show all steps of hypothesis
testing.
H
0
: μ
1
– μ
2
= 0
H
a
: μ
1
– μ
2
> 0
x
1
= 18, x
2
= 17, s
1
= 3, s
2
= 2
n
1
= 22, n
2
= 25
n
1
= 22 - 1 = 21
n
2
= 25 -1 = 24
df = 45
Sp
=
√21 (3)^2 + 24 (2)^2
= √21 (9) + 24 (4)
= 2.5166
21 + 24 -2
45
t = (18 - 17) - 0
2.5166 √1/22 + 1/25
=
1
=
1
2.5166 √.045 + .04
2.5166 (.29154)
=
1
= 1.3631
.7336
.05 < p value < .10
Accept the null H
0
. The p-value of 1.3631 is greater than the alpha of .05 therefore we are accepting the
null.
(Page 163 of SCM 200 Course Packet)
9.Write a scenario of a problem that would be appropriate for an independent test. Do not
use any scenarios that are in the course packet. If you base your idea off of another source,
you must state the source.
Apple, the manufacturing company, wants to test if there is an increase in product functionality when
workers receive a sufficient amount of sleep the night prior to work. As opposed to receiving little or no
sleep the night before constructing the products. Assume two independent random samples of a week
consisting of 5 days yielded the product functionally in the 000s of units.
(Page 161 of SCM 200 Course Packet)
10. In a test of the difference of two proportions, the z-value was calculated to be 1.69.
Compute an upper tail, lower tail, and two tail p-values for this test statistic.
Z-value - 1.69
Upper Tail - shade to right
1 - 9.545 = -8.545
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Lower Tail - shade to left
= 9.545
Two-Tailed Test - shade both ends
-8.545 x 2 = 17.09
(Page 167 of SCM 200 Course Packet)
(Questions 7-10 can be completed after Lecture 26.)
Excel Homework 5
Fall 2022
All Excel tutorials have a red box titled “Are you in the right tutorial?” Read the box before you
start – it will help you make sure that you are using the correct tutorial for your Excel version.
Using the Homework 5 data set in Canvas and the appropriate Excel Homework 5 Tutorial or any other
sources, answer all of the questions below.
Problem: A large company is considering paying for employees to take part in a weight loss program.
The company decides that if the average weight loss of a sample group is at least 20 pounds, the company
will contract with the weight loss program and pay all expenses for the future employees that participate.
The company pays for 20 randomly selected participants to participate in the trial run. The weights of the
20 participants are recorded before the trial begins. After eight months of the participants being on the
program, their weights are recorded. The level of significance for the problem is .05.
●
Weight Before Program will be Sample 1. Weight After Program will be Sample 2. Setting the
problem up as Sample 1 – Sample 2, write the null and alternative for the problem. (.1)
H
0
: μ
D
≥
20
H
a
: μ
D
< 20
●
Why is a paired t-test appropriate for this problem? (.1)
The paired t-test is appropriate for this problem because the controlled variable is for the people being
sampled. The same employees are being weighed before and after the program.
In the column to the right of the Weight After Program column, create a difference column.
(This
column does not need to be turned it with your homework.)
Now name the difference column
Difference by using the instructions given in the Excel Homework 5 Tutorial. When you have named a
column, when you click on the arrow to the right of A1 (see the picture below), the word Difference will
be there where you see A1 below. If you click on Formulas and then Name Manager, your named column
will show up in the list if you have named it correctly.
3.
Write in your own words the process you went through to name the difference column. (You may
not copy/paste the instructions from the tutorial.) (.1)
Make a histogram of the difference column to check for normality since our sample size is small. (If Data
Analysis is not loaded in your Excel, follow the instructions in Excel Tutorial 1 to install Data Analysis
Toolpak.) As stated in the tutorial, if you don’t want to fill in Bin values, you can leave the Bin Range
blank and Excel will fill in bins for you.
You do not need to turn in the histogram and frequency table
with your homework.
In order to name the difference column using excel, I had to first create the difference column. I created
this column by typing in the formula =B2-C2 and then proceeded to drag down my cursor to fill in the
corresponding rows. Once I’ve created my difference column it was time to name it; already having these
rows selected and not including the difference label at the top, I went to the box in the top-left hand corner
of the screen, which was previously entitled D2 and renamed the column difference then pressed enter. To
ensure that I correctly named this column, I clicked the little arrow to the right of the box and the drop
down menu had the only selectable option entitled difference. Thus verifying that I correctly named the
difference column.
4.
Looking at the histogram that you just made, describe the shape of the distribution. (.1)
Although the sample size is small, since the distribution of the sample is close to normal, we will assume
the population is also close to normal so that we can use the t-distribution. Perform a paired t-test in
Excel. (Hint: you need to change the hypothesized mean difference from 0 to 20.)
You do not need to
turn in the Excel output with your homework.
When looking at the histogram that I created in excel, the majority of the values are shifted to the left of
the distribution. The shape of the distribution is left skewed because the tail is on the left side of the
distribution.
5.
Read the Excel Homework 5 Tutorial carefully and pay careful attention to how to calculate the
p-value from the output. Using the Excel output from your paired t-test, calculate the p-value and write
down in your own words why you calculated it the way you did. (You cannot copy/paste the explanation
given in the tutorial.) You must use the output to get your p-value. You cannot use tables or your
calculator. (.1 for p-value, .1 for description of the process)
P = 1 - 0.25004175
= 0.74995825
The T statistic is positive meaning P (T<=t). The one-tail value gives the area to the right of the t-value, in
order to calculate the area to the left of the distribution I needed to subtract the p-value given minus one to
identify the p-value from the test. Therefore I subtracted one from the P(T<=t) value of 0.25004175,
giving me the p-value of 0.74995825.
6.
Would you accept or reject the null hypothesis? (.1)
Through comparing my p-value to my level of significance, I can conclude that we are accepting the null
hypothesis of
H
0
. I am accepting the null hypothesis due to the fact that my p-value of 0.74995825 is
greater than my alpha value of 0.05.
7.
Now state your conclusion in terms of the context of the problem. You must include what action
the company will take or not take regarding paying for employees to take part in the weight loss program.
(.1)
In conclusion the average weight loss of a sample group is not less than 20 pounds. Therefore the
company will in fact contract the weight loss program in the future and evidently the company will pay
for all expenses of employees who do choose to participate in the weight loss program.
(Questions 1-7 can be completed after Lecture 25.)
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