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Project #4 Due 12/10/15 by 5:00pm
Upload one PowerPoint and one Excel file into Blackboard by
the deadline above (e-mailed copies will not be
accepted).
For this project you select a Fortune 500 company of your
choice and assess performance data via the SEC
website EDGAR; here are the minimal requirements:
1. (PowerPoint ) Describe the company including their ticker
symbol, fiscal year, major line(s) of business,
history, financial projections for the future, and any affiliations
currently existing. Good sources include
Yahoo finance and EDGAR but you are free to use others as
well.
2. (Excel and PowerPoint) Using the 10-K reports create an
Excel spreadsheet (label 2 to indicate the sheet
corresponds to question 2) with a column for Fiscal Years (i.e.
2014, 2013, etc.), Fiscal Year Sales.
3. (Excel and PowerPoint) Using the 10-Q reports create an
Excel spreadsheet (label 3 to indicate the sheet
corresponds to question 3) with a column for Quarters, (i.e. Q1-
2015, Q2-2015, etc.) and Quarterly Sales.
Complete a total of 40 quarters (10 years of data) with the last
quarter of 2014 representing quarter 40.
Since the 10-Q reports are not released for the 4
th
quarter of each year you will need to use the
information from question 2 to derive 4
th
quarter data (e.g. if you find that for fiscal year 2012 ABC Inc.
made 1.5 billion in sales and you calculate that the sum of
quarters 1, 2, & 3 net income equals 1.2
billion, then it stands to reason that the 4
th
quarter net income must be 300 million).
4. (Excel and PowerPoint) Using Excel’s scatterplot function,
estimate a linear trend equations [Y = α + β t]
for Sales. Plot the time series data on the X-axis (Quarters from
lowest to highest) and the sales on the Y-
Axis (in millions of dollars). In the graph make sure to show the
trend line and regression metrics
(equation & R-squared). Using the equation forecast Sales for
Q= 41 (Q1-2015).
5. (Excel and PowerPoint) Using Excel’s Data Analysis
Regression function create a spreadsheet that
provides the computer-generated statistical details (label 5 to
indicate the sheet corresponds to question
5). The resulting coefficients should mirror those calculated in
question 4 with the scatterplot. If they
don’t then you may have reversed your dependent and
independent variables so you will need to correct
this before proceeding.
6. (PowerPoint) Summarize and explain your forecasted and
statistical results (i.e. R Square, Coefficients,
T-stat, P-value, and the 95% confidence interval).
WHAT ARE STANDARDS?1
Standards are published documents that establish specifications
and procedures designed to maximize the reliability of the
materials, products, methods, and/or services people use every
day. Standards address a range of issues, including but not
limited to various protocols to help maximize product
functionality and compatibility, facilitate interoperability and
support consumer safety and public health.
Standards form the fundamental building blocks for product
development by establishing consistent protocols that can be
universally understood and adopted. This helps fuel
compatibility and interoperability and simplifies product
development, and speeds time-to-market. Standards also make it
easier to understand and compare competing products. As
standards are globally adopted and applied in many markets,
they also fuel international trade.
It is only through the use of standards that the requirements of
interconnectivity and interoperability can be assured. It is only
through the application of standards that the credibility of new
products and new markets can be verified. In summary
standards fuel the development and implementation of
technologies that influence and transform the way we live, work
and communicate.
Role of IEEE Standards IEEE-SA develops and publishes
standards that include but are not limited to definitions and
terminology; methods of measurement and test; systems;
products; technology; ratings structures; temperature limits and
application guides; recommended practices; and safety. Rating
and dimensional information included in an IEEE standard
result from technical considerations. Rating information
developed by other competent organizations may be included
provided it is consistent with good engineering practice.
The approval and publication of an IEEE standard implies that
the document represents a consensus of the parties who have
participated in its development and review. Since every attempt
is made to involve all interests in the activity, it can be
presumed that the document represents a consensus of interests
concerned with the scope of the standard. Consensus is
established when, in the judgment of the IEEE-SA Standards
Board, substantial agreement has been reached by directly and
materially affected interest categories. Substantial agreement
means much more than a simple majority, but not necessarily
unanimity. Consensus requires that all views and objections be
considered, and that a concerted effort be made toward their
resolution.
Purpose of IEEE standardization IEEE standards provide a
common ground for communication in some specific area of
electrotechnology. They also provide criteria for measuring the
acceptable performance of equipment or materials pertinent to
the field of electrotechnology. The purpose of the review by the
IEEE-SA Standards Board is to ensure that IEEE standards
represent a consensus of interests from those that are materially
affected by these standards and that proper procedures have
been followed during the development of these standards. An
active IEEE standard gives an authoritative reference that is
kept up to date through review at least every ten years by the
Sponsor responsible for its preparation.
1. http://standards.ieee.org/
-
intervals is:
Point Estimate ± Margin of Error
Point & Confidence Interval Estimates
Population Mean Estimation (σ Known)
e:
ecreased, (1 –
n
σ
zx
α/2
n
σ
zxUCL
α/2
n
σ
zxLCL
α/2
Application
1. A random sample of 75 students taking the quantitative
analysis test at SDSU
showed an average score of 18.3 with a population standard
deviation of 5.0.
What would be a 95% confidence interval for all scores on the
quant?
2. A sample of 120 randomly selected cars showed highway
speeds in a certain
location that averaged 66.2 MPH with a population standard
deviation of 13.8
MPH. Find a 99% confidence interval for the average highway
speed of the cars
at this location.
3. A recent study of 38 college students found they spent an
average $275 per
semester for books with a population standard deviation of $28.
Find a 90%
confidence interval for the average amount spent per semester
by college
students.
freedom:
en σ is unknown and n<30
d.f. = n - 1
Population Mean Estimation (σ Unknown)
sample to sample so we
use the t-distribution instead of the normal distribution.
n
s
tx
α/21,-n
n
σ
txUCL
α/21,-n
n
σ
txLCL
α/21,-n
Student’s t-Table
Application
1. A medical researcher would like to estimate the average
resting heart rate of
subjects that are being treated with a new medication. A sample
of size n=18
gives an average resting rate of 87.2 with a sample standard
deviation of 11.31.
Find a 90% confidence interval for the true resting rate.
2. The average deposit of 20 customers selected at random from
the depositors of
a local bank is $83.60 with a sample standard deviation of
$12.41. What would
be a 95% confidence interval for the mean deposit of all bank
depositors?
3. A researcher would like to estimate the average weight loss
of people using a
new high protein diet. A random sample of 12 people using the
diet showed an
average weight loss of 12.3 pounds with s = 2.5 pounds. Find a
90%
confidence interval for the true mean weight loss.
Application
wing identify the appropriate distribution
to use, t or Z, in inferring sample
statistics to population parameters.
1. The average time it took a sample of 45 selected mice to
complete a certain
maze was 3.2 minutes. The population standard deviation is
unknown but the
sample standard deviation is 0.4 minutes.
2. A sample of 20 sport fishing boats working out of San Diego
ports showed an
average catch of 80 tuna on three day trips during tuna season.
The population
standard deviation is known to be 12.8 fish.
3. A random sample of 18 families found that they sent an
average of 44.7 greeting
cards during the holiday season with a sample standard
deviation of 8.1.
Population Proportion Estimation
le proportion is approximately
normal if the sample size is large.
and n(1−P) > 5
val are:
n
)p̂ (1p̂
n
P)P(1
σ
P
n
)p̂ (1p̂
zp̂
α/2
n
)p̂ (1p̂
z-p̂ LCL
α/2
n
)p̂ (1p̂
zp̂ UCL
α/2
→
→ Ep̂ M
Application
1. A Cardiologist is doing a study of damage to the heart muscle
due to heart
attack. As part of the study, the doctor would like to know the
proportion of
patients who suffer a second heart attack within one year of
their first attack. A
random sample of 300 patients finds that 64 suffer a second
heart attack within
one year. Find a 98% confidence interval for the true proportion
of patients who
suffer a second heart attack within one year of their first.
2. To estimate the number of accidents that involve alcohol, 80
past accidents
were chosen at random. Sixty three were found to involve
alcohol. Find a 90%
confidence interval for the true proportion of accidents that
involve alcohol.
3. A survey asked college students the following question. Do
you consider
yourself a social liberal? The survey questioned 220 students
and 168
answered yes to the question. Find a 94% confidence interval
for the true
proportion of students who consider themselves social liberals.
Sampling and the
Distribution
of Sample Means
Sampling
ting, presenting, and describing data.
population based
only on sample data.
sufficiently precise.
Sample: each element has the same probability of
being selected and
is independent of other selections.
Methods of Sampling
• Simple Random
• Systematic
• Cluster (Area)
• Stratified
ion of all of
the possible values of
a statistic for a given size sample randomly selected from a
population
Sampling Distribution
1
st
2
nd
Observation
Obs 18 20 22 24
18 18,18 18,20 18,22 18,24
20 20,18 20,20 20,22 20,24
22 22,18 22,20 22,22 22,24
24 24,18 24,20 24,22 24,24
Now consider all possible samples of size n = 2
16 possible samples
(sampling with replacement)
1st 2nd Observation
Obs 18 20 22 24
18 18 19 20 21
20 19 20 21 22
22 20 21 22 23
24 21 22 23 24
Sampling Distribution
16 Sample
Means
Sample Mean
Different samples of the same size from the same population
will yield
different sample means.
sample is given
by the Standard Error of the Mean:
n decreases as the
sample size
increases
n
1i
i
X
n
1
X
n
σ
σ
X
Summary Measures for the Population Distribution:
Summary Measures of the Sampling Distribution:
Population vs. Sampling Distribution
ormal with mean μ and standard deviation
σ, the sampling distribution is also normally distributed with:
replacement use a
Finite Population Correction Factor
to adjust the standard error to
μμ
X
n
σ
σ
X
1N
nN
n
σ
σ
X
If it is unknown whether the population is
distributed normal, assume such when n ≥ 30.
N - n
N -1
Finite Population Correction
Factor
Approaches 1 as N move toward infinity.
Population vs. Sampling Distribution
Unbiased Estimator: mean of the sampling distribution =
parameter being estimated.
Consistent Estimator: when the difference between the unbiased
estimator and the
parameter becomes smaller as the sample size becomes larger.
Application
For each of the following find the mean of the sample means
and the
standard error of the mean. Determine what the shape of the
sample
distribution is.
1. All samples of size 12 are taken, without replacement, from a
population
of size 90 with μ=127 and σ=28.
2. Calculate the above but this time with replacement.
3. All samples of size 21 are taken, without replacement, from a
normal
population of size 125 with μ=22.4 and σ=4.52.
4. All samples of size 53 are taken, with replacement, from a
population of
size 805 with μ=36.2 and σ=6.20.
Standard Normal Distribution for the
Sample Means
-value for the sampling distribution of:
where: = sample mean
= population mean
= standard error of the mean
Z is a standardized normal random variable with mean of 0
and a variance of 1
X
μ
n
σ
μX
σ
μX
Z
X
X
x
σ
Sampling Distribution Properties
(i.e. is unbiased )
Normal Population
Distribution
Normal Sampling
Distribution
x
x
x
μ
x
μ
(both distributions have the same mean)
n
σ
σ
x
(the distribution of has a reduced standard deviation) x
Central Limit Theorem
mple means from the
population will be
approximately normal as long as the sample size is large
enough.
variables having identical
distributions with mean µ, variance σ2, and X as the mean of
these random
variables.
distribution of
approaches the standard normal distribution
X
x
σ
μX
Z
If the Population is Not Normal
Population Distribution
Sampling Distribution
(becomes normal as n increases)
Central Tendency
Variation
x
x
Larger
sample
size
Smaller
sample size
Sampling distribution
properties:
μμ
x
n
σ
σ
x
x
μ
μ
ng
distribution
that is nearly normal.
of the mean is always normally distributed.
Example
deviation σ = 3.
Suppose a random sample of size n = 36 is selected.
and 8.2?
0.5
36
3
n
σ
σ
x
x
μ = 8
n ≥ 30 justifying central limit theorem.
Application
In the following exercises assume that the sample is taken from
a large
population.
1. The mean weight of 10-year-old girls is 80 pounds, and the
standard deviation is
6 pounds. If a sample of 31 girls is selected, what is the
probability that the mean
of the sample is between 78.5 and 80.5 pounds.
2. The average purchase by a customer in a large department
store the week
before Christmas is $120 with a standard deviation of $15.20. If
50 customers
are selected at random, what is the probability that their average
purchase will be
more than $124?
3. The average mileage between oil changes of cars taken to a
local quick oil
change store is 4200 miles with a standard deviation of 680
miles. What is the
probability that a randomly selected sample of 45 cars have a
mean time
between oil changes of more than 4000 miles?
Copyright © 2013 Pearson Education, Inc. Publishing as
Prentice Hall
Regression
Analysis
Population Linear Regression Model
in Y are assumed to be influenced by changes in X.
House Price
in $1000s
(Y)
Square
Feet
(X)
245 1400
312 1600
279 1700
308 1875
199 1100
219 1550
405 2350
324 2450
319 1425
255 1700
0
50
100
150
200
250
300
350
400
450
0 500 1000 1500 2000 2500 3000
Square Feet
H
o
u
s
e
P
ri
c
e
(
$
1
0
0
0
s
)
Slope
=0.10977
Intercept
=98.248
Sample Linear Regression Model
Total variation is made up of two parts:
Sum of Squares
Total
Sum of Squares
Regression
Sum of Squares
Error (residual)
SST = (y
i
- y)
2
ii
2
i
)yy(SSR ˆ
Sample Linear Regression Model
their mean, y.
linear relationship between x and y.
factors other than the linear
relationship between x and y.
Coefficient of Determination (R2)
variable that is explained by variation in the
independent variable.
1R0
2
squares of sum regression
SST
SSR
R
2
Excel Output
Regression Statistics
Multiple R 0.76211
R Square 0.58082
Adjusted R Square 0.52842
Standard Error 41.33032
Observations 10
ANOVA
df SS MS F Significance F
Regression 1 18934.9348 18934.9348 11.0848 0.01039
Residual 8 13665.5652 1708.1957
Total 9 32600.5000
Coefficients Standard Error t Stat P-value Lower 95% Upper
95%
Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720
232.07386
Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580
58.08% of the variation in
house prices is explained by
variation in square feet
0.58082
32600.5000
18934.9348
SST
SSR
R
2
Standard Error
Y Y
X X
e
s small
e
s large
se is a measure of the variation of observed y values from
the regression line
The magnitude of se should always be judged relative to the
size
of the y values in the sample data
i.e., se = $41.33K is large relative to house prices in the $200 -
$300K range
Excel Output
Regression Statistics
Multiple R 0.76211
R Square 0.58082
Adjusted R Square 0.52842
Standard Error 41.33032
Observations 10
ANOVA
df SS MS F Significance F
Regression 1 18934.9348 18934.9348 11.0848 0.01039
Residual 8 13665.5652 1708.1957
Total 9 32600.5000
Coefficients Standard Error t Stat P-value Lower 95% Upper
95%
Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720
232.07386
Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580
41.33032s
e
Standard Error
Y
X
Y
X
1b
S small
1b
S large
is a measure of the variation in the slope of regression
lines from different possible samples
1b
S
Regression Statistics
Multiple R 0.76211
R Square 0.58082
Adjusted R Square 0.52842
Standard Error 41.33032
Observations 10
ANOVA
df SS MS F Significance F
Regression 1 18934.9348 18934.9348 11.0848 0.01039
Residual 8 13665.5652 1708.1957
Total 9 32600.5000
Coefficients Standard Error t Stat P-value Lower 95% Upper
95%
Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720
232.07386
Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580
0.03297s
1b
Excel Output
H0: β1 = 0 (no linear relationship)
Inference about the Slope: t Test
1b
11
s
βb
t
where:
b1 = regression slope
coefficient
β1 = hypothesized slope
sb1 = standard
error of the slope
H0: β1 = 0
Test Statistic: t = 3.329
From Excel output:
Coefficients Standard Error t Stat P-value
Intercept 98.24833 58.03348 1.69296 0.12892
Square Feet 0.10977 0.03297 3.32938 0.01039
1b
s t b1
Decision: Reject H0
Conclusion: There is sufficient
evidence that square footage affects a
homes price.
Reject H0 Reject H0
a/2=.025
-tn-2,α/2
Do not reject H0
0
a/2=.025
-2.3060 2.3060 3.329
d.f. = 10-2 = 8
t8,.025 = 2.3060
tn-2,α/2
Inference about the Slope: t Test
H0: β1 = 0
P-value = 0.01039
From Excel output:
Coefficients Standard Error t Stat P-value
Intercept 98.24833 58.03348 1.69296 0.12892
Square Feet 0.10977 0.03297 3.32938 0.01039
P-value
Decision: P-value < α so Reject H0
Conclusion: There is sufficient evidence
that square footage affects house price
Inference about the Slope: P value
Confidence Interval Estimate
11 bα/22,n11bα/22,n1
stbβstb
Coefficients Standard Error t Stat P-value Lower 95% Upper
95%
Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720
232.07386
Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580
d.f. = n - 2
This 95% confidence interval does not include 0.
Conclusion: There is a significant relationship between house
price and
square feet at the .05 level.
we are 95% confident that the average impact on sales price
is between $33.70 and $185.80 per square foot of house size
Forecasting with Regression Analysis
The predicted price for a house with 2000 square feet is
98.24833+2000(.10977)=317,788
Multiple Regression
εXβXβXββY
KK22110
Population Multiple Regression Equation with K Independent
Variables:
Y-intercept Population slopes Random Error
KiK2i21i10i
Estimated
(or predicted)
value of y
Estimated slope coefficients
Sample Multiple Regression Equation with K Independent
Variables:
Estimated
intercept
Example: Julian Pie Company
Week
Pie
Sales
Price
($)
Advertising
($100s)
1 350 5.50 3.3
2 460 7.50 3.3
3 350 8.00 3.0
4 430 8.00 4.5
5 350 6.80 3.0
6 380 7.50 4.0
7 430 4.50 3.0
8 470 6.40 3.7
9 450 7.00 3.5
10 490 5.00 4.0
11 340 7.20 3.5
12 300 7.90 3.2
13 440 5.90 4.0
14 450 5.00 3.5
15 300 7.00 2.7
Sales = b0 + b1(Price) + b2(Advertising)
Excel Output
Regression Statistics
Multiple R 0.72213
R Square 0.52148
Adjusted R Square 0.44172
Standard Error 47.46341
Observations 15
ANOVA df SS MS F Significance F
Regression 2 29460.027 14730.013 6.53861 0.01201
Residual 12 27033.306 2252.776
Total 14 56493.333
Coefficients Standard Error t Stat P-value Lower 95% Upper
95%
Intercept 306.52619 114.25389 2.68285 0.01993 57.58835
555.46404
Price -24.97509 10.83213 -2.30565 0.03979 -48.57626 -1.37392
Advertising 74.13096 25.96732 2.85478 0.01449 17.55303
130.70888
44.2% of the variation in pie sales is explained by the variation
in price
and advertising, taking into account the sample size and number
of
independent variables
Adjusted Coefficient of
Determination,
added to the model, even if the new variable is
not an important predictor variable
e a disadvantage when comparing
models
variable is added
explanatory power to offset the loss of one
degree of freedom?
2
R
Hypothesis
Testing
Concepts of Hypothesis Testing
Hypothesis: a claim about a population parameter.
The proportion of adults in this city with cell phones is P =
88%
Null Hypothesis
ue (e.g. H0: P=.88).
Alternative Hypothesis
the status quo
Hypothesis Testing Process
Level of Significance and Rejection Region
sample
statistic if the null hypothesis is true.
ion region of the sampling distribution.
Testing the Mean of a Normal Distribution
(σ Known)
ample result ( ) to a computed Z value Zc
The decision rule is:
α
0
c0
z
n
σ
μx
σ Known σ Unknown
Hypothesis
Consider the test
00
01
(Assume the population is normal)
x
Hypothesis Testing Steps
Reject H0 Do not reject H0
zα 0
μ0
H0: μ ≤ μ0
H1: μ > μ0
Critical value
n
σ
zμ
α0
α
0
c0
z
n
σ
μx
1. State the null and alternative hypotheses.
2. Specify the desired level of significance (.05, .10, .01).
3. Choose a sample size (n=?).
4. Determine statistics.
5. Determine Critical Value.
6. Calculate Computed Value.
7. Decision Rule.
1. A phone industry manager thinks that customer monthly cell
phone bills have increased, and
now average over $52 per month. The company wishes to test
Suppose a sample is taken with the following results: n = 64, x
= 53.1. Calculate the critical
and computed value at a .05 level of significance and determine
whether to accept or reject
the hypothesis.
2. Test the hypothesis that the average number of T.Vs in U.S.
households is 3. Your sample
consists of 100 households with a mean of 2.84 T.Vs. You know
the population standard
deviation to be .8. Your desire a level of significance of 0.05
3. A national perfume manufacturer, claims that the cost of
processing a sales order is $12.50.
The company accountant suspects that the average cost of
processing is more than $12.50.
In order to test her beliefs she obtains a random sample of 40
orders and finds the mean
processing cost is $12.75 with a population standard deviation
of $0.50. Test her suspicions
at the 0.01 level of significance.
Application
Tests of the Mean of a Normal
Population (σ Unknown)
σ Known σ Unknown
Hypothesis
The decision rule is:
α
0
c0
t
n
s
μx
Consider the test
(Assume the population is normal)
00
01
1. Ralphs supermarket claims its low fat ground beef is 15% fat
by weight. The
market’s quality assurance department would like to know if the
15% goal is
being met. Too much fat could bring on complaints from
consumer groups
while not enough fat could effect the flavor of the product.
What can the quality
control people conclude at the 0.05 level of significance if a
random sample
has values of 16.1, 16.7, 14.5, 16.4, 17.3, and 15.8%
2. The salary data for college graduates in entry level
management positions in
CA are normally distributed with a population mean of 63,200.
In San Diego a
random sample of 10 women who were college graduates in
these positions
had a mean salary of 59,870. The sample standard deviation was
2,590. Do
the sample results suggest the women earn less than 63,200
annually. Use a
level of significance of 0.05
Application
Tests of the Population Proportion
denoted by
– P) > 5, can be approximated by
a normal distribution with mean and standard
deviation:
normal, so the test statistic is a z value:
sizesample
sampleinsuccessesofnumber
n
x
Pμ p̂
n
P)P(1
σ
p̂
p̂
p̂
n
)P(1P
Pp̂
z
00
0
c
p̂
00
01
1. A marketing company claims that it receives 8% responses
from its mailing. To test this
claim, a random sample of 500 were surveyed with 25
significance level.
2. The Scenic Blight sign company claims that Alliance Medical
Group gets at least 25% of
its new patients from the sign company’s bill board on highway
101. To test this claim
the medical group randomly selects 120 patient information
forms to determine the
answers given to the question “Where did you hear about our
service?” If 26 of the
forms had the response: “The billboard on highway 101” what
should the medical group
conclude at α = 0.05?
Application
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Project #4 Due 121015 by 500pm Upload one PowerPoint and .docx

  • 1. Project #4 Due 12/10/15 by 5:00pm Upload one PowerPoint and one Excel file into Blackboard by the deadline above (e-mailed copies will not be accepted). For this project you select a Fortune 500 company of your choice and assess performance data via the SEC website EDGAR; here are the minimal requirements: 1. (PowerPoint ) Describe the company including their ticker symbol, fiscal year, major line(s) of business, history, financial projections for the future, and any affiliations currently existing. Good sources include Yahoo finance and EDGAR but you are free to use others as well. 2. (Excel and PowerPoint) Using the 10-K reports create an Excel spreadsheet (label 2 to indicate the sheet corresponds to question 2) with a column for Fiscal Years (i.e. 2014, 2013, etc.), Fiscal Year Sales. 3. (Excel and PowerPoint) Using the 10-Q reports create an Excel spreadsheet (label 3 to indicate the sheet corresponds to question 3) with a column for Quarters, (i.e. Q1- 2015, Q2-2015, etc.) and Quarterly Sales.
  • 2. Complete a total of 40 quarters (10 years of data) with the last quarter of 2014 representing quarter 40. Since the 10-Q reports are not released for the 4 th quarter of each year you will need to use the information from question 2 to derive 4 th quarter data (e.g. if you find that for fiscal year 2012 ABC Inc. made 1.5 billion in sales and you calculate that the sum of quarters 1, 2, & 3 net income equals 1.2 billion, then it stands to reason that the 4 th quarter net income must be 300 million). 4. (Excel and PowerPoint) Using Excel’s scatterplot function, estimate a linear trend equations [Y = α + β t] for Sales. Plot the time series data on the X-axis (Quarters from lowest to highest) and the sales on the Y- Axis (in millions of dollars). In the graph make sure to show the trend line and regression metrics (equation & R-squared). Using the equation forecast Sales for Q= 41 (Q1-2015). 5. (Excel and PowerPoint) Using Excel’s Data Analysis Regression function create a spreadsheet that provides the computer-generated statistical details (label 5 to indicate the sheet corresponds to question
  • 3. 5). The resulting coefficients should mirror those calculated in question 4 with the scatterplot. If they don’t then you may have reversed your dependent and independent variables so you will need to correct this before proceeding. 6. (PowerPoint) Summarize and explain your forecasted and statistical results (i.e. R Square, Coefficients, T-stat, P-value, and the 95% confidence interval). WHAT ARE STANDARDS?1 Standards are published documents that establish specifications and procedures designed to maximize the reliability of the materials, products, methods, and/or services people use every day. Standards address a range of issues, including but not limited to various protocols to help maximize product functionality and compatibility, facilitate interoperability and support consumer safety and public health. Standards form the fundamental building blocks for product development by establishing consistent protocols that can be universally understood and adopted. This helps fuel compatibility and interoperability and simplifies product development, and speeds time-to-market. Standards also make it easier to understand and compare competing products. As standards are globally adopted and applied in many markets, they also fuel international trade. It is only through the use of standards that the requirements of interconnectivity and interoperability can be assured. It is only through the application of standards that the credibility of new products and new markets can be verified. In summary standards fuel the development and implementation of technologies that influence and transform the way we live, work
  • 4. and communicate. Role of IEEE Standards IEEE-SA develops and publishes standards that include but are not limited to definitions and terminology; methods of measurement and test; systems; products; technology; ratings structures; temperature limits and application guides; recommended practices; and safety. Rating and dimensional information included in an IEEE standard result from technical considerations. Rating information developed by other competent organizations may be included provided it is consistent with good engineering practice. The approval and publication of an IEEE standard implies that the document represents a consensus of the parties who have participated in its development and review. Since every attempt is made to involve all interests in the activity, it can be presumed that the document represents a consensus of interests concerned with the scope of the standard. Consensus is established when, in the judgment of the IEEE-SA Standards Board, substantial agreement has been reached by directly and materially affected interest categories. Substantial agreement means much more than a simple majority, but not necessarily unanimity. Consensus requires that all views and objections be considered, and that a concerted effort be made toward their resolution. Purpose of IEEE standardization IEEE standards provide a common ground for communication in some specific area of electrotechnology. They also provide criteria for measuring the acceptable performance of equipment or materials pertinent to the field of electrotechnology. The purpose of the review by the IEEE-SA Standards Board is to ensure that IEEE standards represent a consensus of interests from those that are materially affected by these standards and that proper procedures have been followed during the development of these standards. An active IEEE standard gives an authoritative reference that is kept up to date through review at least every ten years by the Sponsor responsible for its preparation.
  • 5. 1. http://standards.ieee.org/ - intervals is: Point Estimate ± Margin of Error Point & Confidence Interval Estimates Population Mean Estimation (σ Known) e:
  • 7. Application 1. A random sample of 75 students taking the quantitative analysis test at SDSU showed an average score of 18.3 with a population standard deviation of 5.0. What would be a 95% confidence interval for all scores on the quant? 2. A sample of 120 randomly selected cars showed highway speeds in a certain location that averaged 66.2 MPH with a population standard deviation of 13.8 MPH. Find a 99% confidence interval for the average highway speed of the cars at this location. 3. A recent study of 38 college students found they spent an average $275 per semester for books with a population standard deviation of $28. Find a 90% confidence interval for the average amount spent per semester by college students.
  • 8. freedom: en σ is unknown and n<30 d.f. = n - 1 Population Mean Estimation (σ Unknown) sample to sample so we
  • 9. use the t-distribution instead of the normal distribution. n s tx α/21,-n n σ txUCL α/21,-n n σ txLCL α/21,-n Student’s t-Table Application 1. A medical researcher would like to estimate the average resting heart rate of
  • 10. subjects that are being treated with a new medication. A sample of size n=18 gives an average resting rate of 87.2 with a sample standard deviation of 11.31. Find a 90% confidence interval for the true resting rate. 2. The average deposit of 20 customers selected at random from the depositors of a local bank is $83.60 with a sample standard deviation of $12.41. What would be a 95% confidence interval for the mean deposit of all bank depositors? 3. A researcher would like to estimate the average weight loss of people using a new high protein diet. A random sample of 12 people using the diet showed an average weight loss of 12.3 pounds with s = 2.5 pounds. Find a 90% confidence interval for the true mean weight loss. Application wing identify the appropriate distribution
  • 11. to use, t or Z, in inferring sample statistics to population parameters. 1. The average time it took a sample of 45 selected mice to complete a certain maze was 3.2 minutes. The population standard deviation is unknown but the sample standard deviation is 0.4 minutes. 2. A sample of 20 sport fishing boats working out of San Diego ports showed an average catch of 80 tuna on three day trips during tuna season. The population standard deviation is known to be 12.8 fish. 3. A random sample of 18 families found that they sent an average of 44.7 greeting cards during the holiday season with a sample standard deviation of 8.1. Population Proportion Estimation le proportion is approximately normal if the sample size is large.
  • 12. and n(1−P) > 5 val are: n )p̂ (1p̂ n P)P(1 σ P n )p̂ (1p̂
  • 13. zp̂ α/2 n )p̂ (1p̂ z-p̂ LCL α/2 n )p̂ (1p̂ zp̂ UCL α/2 → → Ep̂ M Application 1. A Cardiologist is doing a study of damage to the heart muscle due to heart attack. As part of the study, the doctor would like to know the proportion of
  • 14. patients who suffer a second heart attack within one year of their first attack. A random sample of 300 patients finds that 64 suffer a second heart attack within one year. Find a 98% confidence interval for the true proportion of patients who suffer a second heart attack within one year of their first. 2. To estimate the number of accidents that involve alcohol, 80 past accidents were chosen at random. Sixty three were found to involve alcohol. Find a 90% confidence interval for the true proportion of accidents that involve alcohol. 3. A survey asked college students the following question. Do you consider yourself a social liberal? The survey questioned 220 students and 168 answered yes to the question. Find a 94% confidence interval for the true proportion of students who consider themselves social liberals.
  • 15. Sampling and the Distribution of Sample Means Sampling ting, presenting, and describing data. population based only on sample data. sufficiently precise. Sample: each element has the same probability of being selected and is independent of other selections. Methods of Sampling • Simple Random
  • 16. • Systematic • Cluster (Area) • Stratified ion of all of the possible values of a statistic for a given size sample randomly selected from a population Sampling Distribution 1 st 2 nd Observation Obs 18 20 22 24 18 18,18 18,20 18,22 18,24
  • 17. 20 20,18 20,20 20,22 20,24 22 22,18 22,20 22,22 22,24 24 24,18 24,20 24,22 24,24 Now consider all possible samples of size n = 2 16 possible samples (sampling with replacement) 1st 2nd Observation Obs 18 20 22 24 18 18 19 20 21 20 19 20 21 22 22 20 21 22 23 24 21 22 23 24 Sampling Distribution 16 Sample Means
  • 18. Sample Mean Different samples of the same size from the same population will yield different sample means. sample is given by the Standard Error of the Mean: n decreases as the sample size increases n 1i i X
  • 19. n 1 X n σ σ X Summary Measures for the Population Distribution: Summary Measures of the Sampling Distribution: Population vs. Sampling Distribution ormal with mean μ and standard deviation σ, the sampling distribution is also normally distributed with: replacement use a
  • 20. Finite Population Correction Factor to adjust the standard error to μμ X n σ σ X 1N nN n σ σ X If it is unknown whether the population is
  • 21. distributed normal, assume such when n ≥ 30. N - n N -1 Finite Population Correction Factor Approaches 1 as N move toward infinity. Population vs. Sampling Distribution Unbiased Estimator: mean of the sampling distribution = parameter being estimated. Consistent Estimator: when the difference between the unbiased estimator and the parameter becomes smaller as the sample size becomes larger. Application For each of the following find the mean of the sample means and the standard error of the mean. Determine what the shape of the sample distribution is. 1. All samples of size 12 are taken, without replacement, from a
  • 22. population of size 90 with μ=127 and σ=28. 2. Calculate the above but this time with replacement. 3. All samples of size 21 are taken, without replacement, from a normal population of size 125 with μ=22.4 and σ=4.52. 4. All samples of size 53 are taken, with replacement, from a population of size 805 with μ=36.2 and σ=6.20. Standard Normal Distribution for the Sample Means -value for the sampling distribution of: where: = sample mean = population mean = standard error of the mean Z is a standardized normal random variable with mean of 0 and a variance of 1
  • 24. Distribution Normal Sampling Distribution x x x μ x μ (both distributions have the same mean) n σ σ x (the distribution of has a reduced standard deviation) x Central Limit Theorem mple means from the
  • 25. population will be approximately normal as long as the sample size is large enough. variables having identical distributions with mean µ, variance σ2, and X as the mean of these random variables. distribution of approaches the standard normal distribution X x σ μX Z
  • 26. If the Population is Not Normal Population Distribution Sampling Distribution (becomes normal as n increases) Central Tendency Variation x x Larger sample size Smaller sample size Sampling distribution properties: μμ x n
  • 27. σ σ x x μ μ ng distribution that is nearly normal. of the mean is always normally distributed. Example deviation σ = 3. Suppose a random sample of size n = 36 is selected. and 8.2? 0.5 36 3
  • 28. n σ σ x x μ = 8 n ≥ 30 justifying central limit theorem. Application In the following exercises assume that the sample is taken from a large population. 1. The mean weight of 10-year-old girls is 80 pounds, and the standard deviation is 6 pounds. If a sample of 31 girls is selected, what is the probability that the mean of the sample is between 78.5 and 80.5 pounds. 2. The average purchase by a customer in a large department store the week before Christmas is $120 with a standard deviation of $15.20. If 50 customers
  • 29. are selected at random, what is the probability that their average purchase will be more than $124? 3. The average mileage between oil changes of cars taken to a local quick oil change store is 4200 miles with a standard deviation of 680 miles. What is the probability that a randomly selected sample of 45 cars have a mean time between oil changes of more than 4000 miles? Copyright © 2013 Pearson Education, Inc. Publishing as Prentice Hall Regression Analysis Population Linear Regression Model
  • 30. in Y are assumed to be influenced by changes in X. House Price in $1000s (Y) Square Feet (X) 245 1400 312 1600 279 1700 308 1875
  • 31. 199 1100 219 1550 405 2350 324 2450 319 1425 255 1700 0 50 100 150 200 250 300 350 400 450 0 500 1000 1500 2000 2500 3000 Square Feet
  • 33. Sum of Squares Total Sum of Squares Regression Sum of Squares Error (residual) SST = (y i - y) 2 ii 2 i )yy(SSR ˆ Sample Linear Regression Model their mean, y.
  • 34. linear relationship between x and y. factors other than the linear relationship between x and y. Coefficient of Determination (R2) variable that is explained by variation in the independent variable. 1R0 2 squares of sum regression SST SSR R 2 Excel Output
  • 35. Regression Statistics Multiple R 0.76211 R Square 0.58082 Adjusted R Square 0.52842 Standard Error 41.33032 Observations 10 ANOVA df SS MS F Significance F Regression 1 18934.9348 18934.9348 11.0848 0.01039 Residual 8 13665.5652 1708.1957 Total 9 32600.5000 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720 232.07386 Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580 58.08% of the variation in house prices is explained by variation in square feet
  • 36. 0.58082 32600.5000 18934.9348 SST SSR R 2 Standard Error Y Y X X e s small e s large se is a measure of the variation of observed y values from the regression line The magnitude of se should always be judged relative to the size of the y values in the sample data
  • 37. i.e., se = $41.33K is large relative to house prices in the $200 - $300K range Excel Output Regression Statistics Multiple R 0.76211 R Square 0.58082 Adjusted R Square 0.52842 Standard Error 41.33032 Observations 10 ANOVA df SS MS F Significance F Regression 1 18934.9348 18934.9348 11.0848 0.01039 Residual 8 13665.5652 1708.1957 Total 9 32600.5000 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720 232.07386 Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580
  • 38. 41.33032s e Standard Error Y X Y X 1b S small 1b S large is a measure of the variation in the slope of regression lines from different possible samples 1b S Regression Statistics Multiple R 0.76211
  • 39. R Square 0.58082 Adjusted R Square 0.52842 Standard Error 41.33032 Observations 10 ANOVA df SS MS F Significance F Regression 1 18934.9348 18934.9348 11.0848 0.01039 Residual 8 13665.5652 1708.1957 Total 9 32600.5000 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720 232.07386 Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580 0.03297s 1b Excel Output
  • 40. H0: β1 = 0 (no linear relationship) Inference about the Slope: t Test 1b 11 s βb t where: b1 = regression slope coefficient β1 = hypothesized slope sb1 = standard
  • 41. error of the slope H0: β1 = 0 Test Statistic: t = 3.329 From Excel output: Coefficients Standard Error t Stat P-value Intercept 98.24833 58.03348 1.69296 0.12892 Square Feet 0.10977 0.03297 3.32938 0.01039 1b s t b1 Decision: Reject H0 Conclusion: There is sufficient evidence that square footage affects a homes price. Reject H0 Reject H0 a/2=.025 -tn-2,α/2 Do not reject H0
  • 42. 0 a/2=.025 -2.3060 2.3060 3.329 d.f. = 10-2 = 8 t8,.025 = 2.3060 tn-2,α/2 Inference about the Slope: t Test H0: β1 = 0 P-value = 0.01039 From Excel output: Coefficients Standard Error t Stat P-value Intercept 98.24833 58.03348 1.69296 0.12892 Square Feet 0.10977 0.03297 3.32938 0.01039 P-value Decision: P-value < α so Reject H0 Conclusion: There is sufficient evidence
  • 43. that square footage affects house price Inference about the Slope: P value Confidence Interval Estimate 11 bα/22,n11bα/22,n1 stbβstb Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 98.24833 58.03348 1.69296 0.12892 -35.57720 232.07386 Square Feet 0.10977 0.03297 3.32938 0.01039 0.03374 0.18580 d.f. = n - 2 This 95% confidence interval does not include 0. Conclusion: There is a significant relationship between house price and square feet at the .05 level. we are 95% confident that the average impact on sales price is between $33.70 and $185.80 per square foot of house size
  • 44. Forecasting with Regression Analysis The predicted price for a house with 2000 square feet is 98.24833+2000(.10977)=317,788 Multiple Regression εXβXβXββY KK22110 Population Multiple Regression Equation with K Independent Variables: Y-intercept Population slopes Random Error KiK2i21i10i Estimated (or predicted) value of y Estimated slope coefficients Sample Multiple Regression Equation with K Independent Variables: Estimated intercept
  • 45. Example: Julian Pie Company Week Pie Sales Price ($) Advertising ($100s) 1 350 5.50 3.3 2 460 7.50 3.3 3 350 8.00 3.0 4 430 8.00 4.5 5 350 6.80 3.0 6 380 7.50 4.0 7 430 4.50 3.0 8 470 6.40 3.7 9 450 7.00 3.5
  • 46. 10 490 5.00 4.0 11 340 7.20 3.5 12 300 7.90 3.2 13 440 5.90 4.0 14 450 5.00 3.5 15 300 7.00 2.7 Sales = b0 + b1(Price) + b2(Advertising) Excel Output Regression Statistics Multiple R 0.72213 R Square 0.52148 Adjusted R Square 0.44172 Standard Error 47.46341 Observations 15 ANOVA df SS MS F Significance F Regression 2 29460.027 14730.013 6.53861 0.01201 Residual 12 27033.306 2252.776 Total 14 56493.333
  • 47. Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Intercept 306.52619 114.25389 2.68285 0.01993 57.58835 555.46404 Price -24.97509 10.83213 -2.30565 0.03979 -48.57626 -1.37392 Advertising 74.13096 25.96732 2.85478 0.01449 17.55303 130.70888 44.2% of the variation in pie sales is explained by the variation in price and advertising, taking into account the sample size and number of independent variables Adjusted Coefficient of Determination, added to the model, even if the new variable is not an important predictor variable e a disadvantage when comparing models variable is added
  • 48. explanatory power to offset the loss of one degree of freedom? 2 R Hypothesis Testing Concepts of Hypothesis Testing Hypothesis: a claim about a population parameter. The proportion of adults in this city with cell phones is P = 88% Null Hypothesis ue (e.g. H0: P=.88). Alternative Hypothesis
  • 49. the status quo Hypothesis Testing Process Level of Significance and Rejection Region sample statistic if the null hypothesis is true. ion region of the sampling distribution. Testing the Mean of a Normal Distribution (σ Known) ample result ( ) to a computed Z value Zc
  • 50. The decision rule is: α 0 c0 z n σ μx σ Known σ Unknown Hypothesis Consider the test 00 01 (Assume the population is normal) x
  • 51. Hypothesis Testing Steps Reject H0 Do not reject H0 zα 0 μ0 H0: μ ≤ μ0 H1: μ > μ0 Critical value n σ zμ α0 α 0 c0 z n σ
  • 52. μx 1. State the null and alternative hypotheses. 2. Specify the desired level of significance (.05, .10, .01). 3. Choose a sample size (n=?). 4. Determine statistics. 5. Determine Critical Value. 6. Calculate Computed Value. 7. Decision Rule. 1. A phone industry manager thinks that customer monthly cell phone bills have increased, and now average over $52 per month. The company wishes to test Suppose a sample is taken with the following results: n = 64, x = 53.1. Calculate the critical and computed value at a .05 level of significance and determine whether to accept or reject
  • 53. the hypothesis. 2. Test the hypothesis that the average number of T.Vs in U.S. households is 3. Your sample consists of 100 households with a mean of 2.84 T.Vs. You know the population standard deviation to be .8. Your desire a level of significance of 0.05 3. A national perfume manufacturer, claims that the cost of processing a sales order is $12.50. The company accountant suspects that the average cost of processing is more than $12.50. In order to test her beliefs she obtains a random sample of 40 orders and finds the mean processing cost is $12.75 with a population standard deviation of $0.50. Test her suspicions at the 0.01 level of significance. Application Tests of the Mean of a Normal Population (σ Unknown)
  • 54. σ Known σ Unknown Hypothesis The decision rule is: α 0 c0 t n s μx Consider the test (Assume the population is normal) 00 01
  • 55. 1. Ralphs supermarket claims its low fat ground beef is 15% fat by weight. The market’s quality assurance department would like to know if the 15% goal is being met. Too much fat could bring on complaints from consumer groups while not enough fat could effect the flavor of the product. What can the quality control people conclude at the 0.05 level of significance if a random sample has values of 16.1, 16.7, 14.5, 16.4, 17.3, and 15.8% 2. The salary data for college graduates in entry level management positions in CA are normally distributed with a population mean of 63,200. In San Diego a random sample of 10 women who were college graduates in these positions had a mean salary of 59,870. The sample standard deviation was 2,590. Do the sample results suggest the women earn less than 63,200 annually. Use a level of significance of 0.05
  • 56. Application Tests of the Population Proportion denoted by – P) > 5, can be approximated by a normal distribution with mean and standard deviation: normal, so the test statistic is a z value: sizesample sampleinsuccessesofnumber n x Pμ p̂
  • 58. 01 1. A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 significance level. 2. The Scenic Blight sign company claims that Alliance Medical Group gets at least 25% of its new patients from the sign company’s bill board on highway 101. To test this claim the medical group randomly selects 120 patient information forms to determine the answers given to the question “Where did you hear about our service?” If 26 of the forms had the response: “The billboard on highway 101” what should the medical group conclude at α = 0.05? Application