DBM380 v14Create a DatabaseDBM380 v14Page 2 of 2Create a D.docx

DBM/380 v14
Create a Database
DBM/380 v14
Page 2 of 2Create a Database
The following assignment is based on the business scenario for
which you created both an entity-relationship diagram and a
normalized database design in Week 2.
For this assignment, you will create multiple related tables that
match your normalized database design. In other words, you
will implement a physical design (an actual, usable database)
based on a logical design.
Refer to the linked W3Schools.com articles “SQL CREATE
TABLE Statement,” “SQL PRIMARY KEY Constraint,” “SQL
FOREIGN KEY Constraint,” and “SQL INSERT INTO
Statement” for help in completing this assignment.
Note: In the industry, even the most carefully thought out
database designs can contain mistakes. Feel free to correct in
your tables any mistakes you notice in your normalized database
design. Also, note that in Microsoft® Access®, you follow the
steps below to launch the SQL editor:
Figure 1. To create a SQL query in Microsoft® Access®, begin
by clicking the CREATE tab.
To Complete This Assignment:
1. Use the CREATE TABLE statement to create each table in
your design. Note that a table in a RDMS corresponds to an
entity in an entity-relationship diagram. Recommended tables
for this assignment are CUSTOMER, ORDER,
ORDER_DETAIL, PRODUCT, EMPLOYEE, and STORE.
2. As part of each CREATE TABLE statement, define all of the
columns, or fields, that you want each particular table to
contain. Give them short, meaningful names and include
constraints; that is, describe what type of data each column
(field) is allowed to hold and any other constraints, such as

size, range, or uniqueness.
3. Note that any field you marked as a unique identifier in your
normalized database design is a key field. Key fields must be
described as both UNIQUE and NOT NULL, which means a
value must exist for each record and that value must be unique
across all records.
4. After you have created all six tables, including relationships
between the tables as appropriate (matching the primary key in
one table to a foreign key in another table), use the INSERT
INTO statement to insert 10 records into each of your tables.
You will need to make up the data you insert into your tables.
For example, to insert one record into the CUSTOMER table,
you will need to invent a customer number, a customer name,
and so on—one value for each of the fields you defined for the
CUSTOMER table—to insert into the table.
5. To ensure that your INSERT INTO statements succeeded in
populating your tables, use the SELECT statement described in
Ch. 7, “Introduction to Structured Query Language,” in
Database Systems: Design, Implementation, and Management.to
retrieve the records you inserted. For example, to see all 10
records you inserted into the CUSTOMER table, you might
apply the following SQL statement: SELECT * FROM
CUSTOMER;
After you have created all six tables and populated ten records
in each table, submit to the Assignment Files tab the database
containing all of the tables you created, or a Microsoft® Word
document listing all of the SQL statements you used.

Copyright© 2018 by University of Phoenix. All rights reserved.
Copyright© 2018 by University of Phoenix. All rights reserved.
Chapter 4
Regression Analysis
Data analysts use regressions to determine if relationships
exists between variables…...
In a simple linear regression we test whether measured values
of the dependent variable (on the Y axis) vary with provided
independent variable (on the X axis)
Does an increase in advertising (X) coincide to an increase in
sales (Y)….?
Can more training hours (X) leads to decreased scrap (Y)….?
Will more spending on HR benefits (X) prompt an increase in
employee retention (Y)…?
We use sample data (x,y) and sample y intercept

X is the independent variable (we are given)
Y is the dependent variable (we measure)
is the slope (change in Y for a change in X)
is the value of y when x=0
is the random error
X
Y
X
Y
X
Y
When performing regressions there are 3 rules we must follow:
(Rule 1) Do not predict values far beyond the data we are
working with
In the example below we see a linear relationship between X
and Y.
What is the predicted Y value at X=12?

In this case the relationship changed (from linear to curvilinear)
when x exceeded 6.
Conclusion: we can only apply extrapolate values near the test
range
(Rule 2) Data deviations from the predicted line are assumed to
be random
Sales
The data points ( ) are randomly scattered around the
regression line. Meaning there is not an underlying influence
on Y values other than the X values we are considering
(Rule 3) Variables X and Y are normally distributed
Y
X
Regression line
How do we determine if our data is normally distributed?

To test data for skewsness we use the formula =SKEW(). If
SKEW value is between -1 (negative skew) and +1 (positive
skew) we can say the data is normal in X
To test data for kurtosis we use the formula =KURT(). If the
KURT values are between -1 (flat) and +1 (peaked) we can say
the data is normal in Y
In this example the data X and Y are normally distributed
because SKEW and KURT values are all between +1 and -
1.XY820022307220321072406200421092306216SKEW-
0.415760.268996KURTOSIS-0.86776-0.99992
Now that we know the rules of regression lets try one…
We start by enabling Excel Add-ins
In Excel 2010 and later go to File > Options
22

1. Click this
2. Click this
23
3. Check these
4. Click this.
5. Click “Data”. Now you should be able to see these.
24
1. On Data tab
2. Select Data Analysis
3. Select Regression
4. Click OK
5. Click to select D3:D10
6. Click to select C3:C10
7. Click as 1st row of X & Y are labels
8. Click to make plot
What does all this mean???

Start by looking at Significance F. If F is < .05, there is < 5%
chance of incorrectly accepting a regression exists. In other
words, there is >95% chance of a regression existing. At F <
.05 we accept the regression.
Next we look at R square (i.e. r2)
The coefficient of determination () tell us the % variation in y
(“in our example electrical demand”) explained by x (“time
period”)
How does r2 do this?
r2 is a ratio of variation explained by the model to total
variation.

In our example = 0.8, so 80% of variation in electrical demand
can be explained by variation in time period.
= 56.70 + 10.54x
The F <5% means a regression exists and r2 = 0.8 that it is
strong; we can now look to coefficients to find x slope and y
intercept of the regression line
Are the regression coefficients significant?
The P values of y intercept (.0029) and slope (.006) are less
than .05. So….

There is < 5% chance of incorrectly accepting these
coefficients. In other words, there is >95% chance of a
regression existing with these coefficients.
Let’s try another…
Determine if a relationship exists between how much Triple A
Construction Co. sells and how much it pays in payroll.
The null hypothesis () at 95% confidence (is no relationship
between sales and payroll
The X and Y data are normally distributed so we can test for a
regression
1. On Data tab
3. Select regression
4. Click OK

5. Click to select D8:D14
6. Click to select E8:E14
7. Click as 1st row of X & Y are labels
We look at the Significance F
From our Significance F, there is only a 3.9% chance of
incorrectly rejecting the null hypothesis () that no relationship
exists between sales and payroll
Since our null hypothesis () is tested at 95% confidence (a 3.9%
chance of error is acceptable. We reject no relationship
between sales and payroll
With a correlation coefficient (r) of .69, the regression is
moderate.
With an intercept of 2 and slope coefficient of 1.25 our
estimated linear regression equation

With an intercept p value of .3, we cannot accept this value at
95% confidence. We need to consider standard error.
What does standard error mean?
The Standard Errors are errors associated with regression
coefficients. Think of it standard deviation of coefficients.
At a 95% confidence interval (i.e. 2 standard deviations) payroll
and y intercept coefficients could vary from:
Coefficient
Lowest value of predicted sales () using payroll (x) is:

Highest value of predicted sales () using payroll (x) is:
Is it possible when we collected sales and payroll numbers,
there were external factors we didn’t control that affected
results (such as years service, or employee performance ratings,
or economy strength)?
From the “residual plots” we can see
Residual error is on the vertical axis. The independent variable
on the horizontal axis.
Since the points in this example are randomly scattered around
the horizontal axis (sum approximately to 0), we can reject
external factors and accept a single variable linear regression.
Payroll (X) Residual Plot
3464250.251-0.5-201.25
Payroll (X)
Residuals
A multiple regression model allows us to predict an output
value Y using multiple independent variables X1, X2 ….
Lets look at an example….

Can square footage of a house () or age () or both be used to
predict the selling price (Y) of a house?
Y
The null hypothesis () at 95% confidence (is no relationship
between sales price and square footage or age
1. On Data tab
3. Select regression
4. Click OK
5. Click to select B4:B18
6. Click to select C4:D18
7. Click as 1st row of X1, X2 & Y are labels
We look at Significance F

From our Significance F (.0021), there is only a 0.22% chance
of incorrectly rejecting the null hypothesis () that no
relationship exists between Y, X1 and X2
Since our null hypothesis () is tested at 95% confidence (a
0.22% chance of error is acceptable. We reject that no
relationship exists.
The r2=0.67 tells us the linear regression explains 67% of the
variance in the dependent variable (i.e. house selling price).
So, we have a moderately strong model.
Since the p-values for square feet (.0013) and age (.0039) are
both below .05, square feet and age can both be used to predict
price
A non-significant P value (>.05) would have told us the variable
does not have predictive capability in the presence of the other;
so we would have removed it and refit the model without it.
P values shouldn’t be used to eliminate more than one variable
at a time
Why? Because a variable that doesn’t have predictive capability
in the presence of other variables may have predictive
capability when some of those variables are removed from the

model.
With an intercept of 146,630 and slope coefficients of 43.8 & -
2,898 our estimated linear regression equation is
At higher values of square feet () and lower values of age ()
home sale prices are larger
Lowest value of predicted home sales price () using square feet
() and age () is:
Highest value of predicted home sales price () using square feet
() and age () is:
What do t values tell us?

In multiple linear regression, the absolute size of the coefficient
for each independent variable gives you the size of the effect
that variable is having on your dependent variable, and the sign
on the coefficient (positive or negative) gives you the direction
of the effect.
In our case square feet (t=4.26) has a bigger effect on house
price than age (t=3.64)
What is the adjusted R2
As additional variables are added to a multiple regression
equation, R² increases even when the new variables have no real
predictive capability.
When variables are added and adjusted R² doesn't increase the
new variables do not improve predictive capability.
Is it possible when we collected house price, house age and
square footage, there were external factors we didn’t control
that affected price (such as school district, builder, or taxes)?
From the “residual plots” we can see
The points are randomly dispersed around the horizontal axis

for both square feet and age; we can reject external factors are
impacting our age and square feet multiple regression with
house price
Square feet Residual Plot
19262069172013961706184719502323228537522300252538001
740-49066.406014619977-2345.7129556209984-
10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413
Square feet
Residuals
Age Residual Plot
3040301532382730263518174012-49066.406014619977-
2345.7129556209984-10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413
Age
Residuals
When we do linear regressions, there are certain assumptions we
make….
Sample sizes are large enough (>30) the t distributions
approximates normal distributions
Correlation does not equal causality
An action or occurrence can cause another (such as smoking
causes lung cancer), or it can correlate with another (such as

smoking is correlated with high alcohol consumption). If one
action causes another, then they are most certainly correlated.
But just because two things occur together does not mean that
one caused the other, even if it seems to make sense.
SUMMARY OUTPUT
Regression Statistics
Multiple R0.894909611
R Square0.800863211
Adjusted R Square0.761035854
Standard Error12.43238858
Observations7
ANOVA
dfSSMSFSignificance F
Regression13108.0357143108.03571420.108370.006493257
Residual5772.8214286154.5642857
Total63880.857143
CoefficientsStandard Errort StatP-valueLower 95%Upper
95%Lower 95.0%Upper 95.0%
Intercept56.7142857110.50728615.3976150610.00294829.7044
469283.7241245129.7044469283.72412451
Time
Period10.535714292.3495005984.4842356260.0064934.496130
72516.575297854.49613072516.57529785
Sheet1Time PeriodElectrical
Demand2001174200227920033802004490200551052005614220
077122SUMMARY OUTPUTRegression StatisticsMultiple
R0.8949096107R Square0.8008632114Adjusted R
Square0.7610358536Standard
Error12.4323885764Observations7ANOVAdfSSMSFSignificanc
e
FRegression13108.03571428573108.035714285720.1083691483
0.0064932569Residual5772.8214285714154.5642857143Total6

3880.8571428571CoefficientsStandard Errort StatP-valueLower
95%Upper 95%Lower 95.0%Upper
95.0%Intercept56.714285714310.50728610185.39761506110.00
2947951729.704446919283.724124509329.704446919283.7241
245093Time
Period10.53571428572.34950059834.48423562590.0064932569
4.49613072516.57529784644.49613072516.5752978464RESID
UAL OUTPUTObservationPredicted Electrical
DemandResiduals167.256.75277.78571428571.2142857143388.
3214285714-8.3214285714498.8571428571-
8.85714285715109.3928571429-
4.39285714296119.928571428622.07142857147130.464285714
3-8.4642857143
Time Period Line Fit Plot
Electrical Demand123456774798090105142122Predicted
Electrical
Demand123456767.2577.78571428571429288.32142857142858
498.857142857142861109.39285714285714119.9285714285714
3130.46428571428572
Time Period
Electrical Demand
Sheet2
Sheet3
Demand2001174200227920033802004490200551052005614220
e
FRegression13108.03571428573108.035714285720.1083691483
0.0064932569Residual5772.8214285714154.5642857143Total6
95.0%Intercept56.714285714310.50728610185.39761506110.00
2947951729.704446919283.724124509329.704446919283.7241

245093Time
Period10.53571428572.34950059834.48423562590.0064932569
4.49613072516.57529784644.49613072516.5752978464RESID
3214285714-8.3214285714498.8571428571-
8.85714285715109.3928571429-
4.39285714296119.928571428622.07142857147130.464285714
3-8.4642857143
Electrical
Demand123456767.2577.78571428571429288.32142857142858
498.857142857142861109.39285714285714119.9285714285714
3130.46428571428572
Time Period
Electrical Demand
Sheet2
Sheet3
Demand2001174200227920033802004490200551052005614220
e
FRegression13108.03571428573108.035714285720.1083691483
0.0064932569Residual5772.8214285714154.5642857143Total6
95.0%Intercept56.714285714310.50728610185.39761506110.00
2947951729.704446919283.724124509329.704446919283.7241
245093Time
Period10.53571428572.34950059834.48423562590.0064932569
4.49613072516.57529784644.49613072516.5752978464RESID

3214285714-8.3214285714498.8571428571-
8.85714285715109.3928571429-
4.39285714296119.928571428622.07142857147130.464285714
3-8.4642857143
Electrical
Demand123456767.2577.78571428571429288.32142857142858
498.857142857142861109.39285714285714119.9285714285714
3130.46428571428572
Time Period
Electrical Demand
Sheet2
Sheet3
SUMMARY OUTPUT
R Square0.694444444
Observations6
ANOVA
Regression115.62515.6259.0909090.039351852
Residual46.8751.71875
Total522.5
Intercept21.7425436391.1477470.31505-
2.8380767576.838076757-2.8380767576.838076757
Payroll
(X)1.250.4145780993.0151130.0393520.0989466672.40105333
30.0989466672.401053333
Sheet1Triple A Construction Co.Sales (Y)Payroll
(X)638496544.529.55SUMMARY OUTPUTRegression

StatisticsMultiple R0.8333333333R
Square0.6944444444Adjusted R Square0.6180555556Standard
Error1.3110110602Observations6ANOVAdfSSMSFSignificance
FRegression115.62515.6259.09090909090.0393518519Residual
46.8751.71875Total522.5CoefficientsStandard Errort StatP-
valueLower 95%Upper 95%Lower 95.0%Upper
95.0%Intercept21.74254363891.14774743970.3150499206-
2.83807675676.8380767567-2.83807675676.8380767567Payroll
(X)1.250.41457809883.01511344580.03935185190.0989466669
2.40105333310.09894666692.4010533331RESIDUAL
OUTPUTObservationPredicted Sales
(Y)Residuals15.750.2527139.5-0.547-254.5068.251.25
3464250.251-0.5-201.25
Payroll (X)
Residuals
Payroll (X) Line Fit Plot
Sales (Y)34642568954.59.5Predicted Sales
(Y)3464255.7579.574.58.25
Payroll (X)
Sales (Y)
Sheet2
Sheet3
95.0%Intercept21.74254363891.14774743970.3150499206-
2.83807675676.8380767567-2.83807675676.8380767567Payroll
(X)1.250.41457809883.01511344580.03935185190.0989466669
2.40105333310.09894666692.4010533331RESIDUAL

(Y)Residuals15.750.2527139.5-0.547-254.5068.251.25
3464250.251-0.5-201.25
Payroll (X)
Residuals
(Y)3464255.7579.574.58.25
Payroll (X)
Sales (Y)
Sheet2
Sheet3
95.0%Intercept21.74254363891.14774743970.3150499206-
2.83807675676.8380767567-2.83807675676.8380767567Payroll
(X)1.250.41457809883.01511344580.03935185190.0989466669
2.40105333310.09894666692.4010533331RESIDUAL
(Y)Residuals15.750.2527139.5-0.547-254.5068.251.25
3464250.251-0.5-201.25
Payroll (X)
Residuals
(Y)3464255.7579.574.58.25
Payroll (X)
Sales (Y)
Sheet2

Sheet3
95.0%Intercept21.74254363891.14774743970.3150499206-
2.83807675676.8380767567-2.83807675676.8380767567Payroll
(X)1.250.41457809883.01511344580.03935185190.0989466669
2.40105333310.09894666692.4010533331RESIDUAL
(Y)Residuals15.750.2527139.5-0.547-254.5068.251.25
3464250.251-0.5-201.25
Payroll (X)
Residuals
(Y)3464255.7579.574.58.25
Payroll (X)
Sales (Y)
Sheet2
Sheet3
95.0%Intercept21.74254363891.14774743970.3150499206-
2.83807675676.8380767567-2.83807675676.8380767567Payroll

(X)1.250.41457809883.01511344580.03935185190.0989466669
2.40105333310.09894666692.4010533331RESIDUAL
(Y)Residuals15.750.2527139.5-0.547-254.5068.251.25
3464250.251-0.5-201.25
Payroll (X)
Residuals
(Y)3464255.7579.574.58.25
Payroll (X)
Sales (Y)
Sheet2
Sheet3
95.0%Intercept21.74254363891.14774743970.3150499206-
2.83807675676.8380767567-2.83807675676.8380767567Payroll
(X)1.250.41457809883.01511344580.03935185190.0989466669
2.40105333310.09894666692.4010533331RESIDUAL
(Y)Residuals15.750.2527139.5-0.547-254.5068.251.25
3464250.251-0.5-201.25
Payroll (X)
Residuals
(Y)3464255.7579.574.58.25

Payroll (X)
Sales (Y)
Sheet2
Sheet3
95.0%Intercept21.74254363891.14774743970.3150499206-
2.83807675676.8380767567-2.83807675676.8380767567Payroll
(X)1.250.41457809883.01511344580.03935185190.0989466669
2.40105333310.09894666692.4010533331RESIDUAL
(Y)Residuals15.750.2527139.5-0.547-254.5068.251.25
3464250.251-0.5-201.25
Payroll (X)
Residuals
(Y)3464255.7579.574.58.25
Payroll (X)
Sales (Y)
Sheet2
Sheet3

95.0%Intercept21.74254363891.14774743970.3150499206-
2.83807675676.8380767567-2.83807675676.8380767567Payroll
(X)1.250.41457809883.01511344580.03935185190.0989466669
2.40105333310.09894666692.4010533331RESIDUAL
(Y)Residuals15.750.2527139.5-0.547-254.5068.251.25
3464250.251-0.5-201.25
Payroll (X)
Residuals
(Y)3464255.7579.574.58.25
Payroll (X)
Sales (Y)
Sheet2
Sheet3
SUMMARY OUTPUT
R Square0.671875802
Observations14
ANOVA
Regression2133139369686.66E+0911.261950.002178765
Residual1165021316035.91E+08
Total1319816068571
Intercept146630.893625482.082875.7542740.00012890545.207
31202716.579890545.20731202716.5798
Square
feet43.8193664910.280965074.2621840.00133821.1911149466.

4476180421.1911149466.44761804
Age-2898.686247796.5649421-3.638980.003895-4651.913863-
1145.45863-4651.913863-1145.45863
Sheet1Selling PriceSquare
feetAge9500019263011900020694012480017203013500013961
51428001706321450001847381590001950271650002323301820
00228526183000375235200000230018211000252517215000380
040219000174012SUMMARY OUTPUTRegression
Error24312.6072850603Observations14ANOVAdfSSMSFSignifi
cance
FRegression213313936968.45536656968484.2276611.26194574
30.0021787652Residual116502131602.97325591102872.997569
Total1319816068571.4286CoefficientsStandard Errort StatP-
95.0%Intercept146630.89355597425482.08286875785.7542742
6050.000127566490545.2073136126202716.57979833590545.2
073136126202716.579798335Square
feet43.819366490110.28096507024.26218416180.00133809482
1.191114939166.44761804121.191114939166.447618041Age-
2898.686246708796.5649420672-3.6389829550.0038949963-
4651.9138632471-1145.4586301689-4651.9138632471-
1145.4586301689RESIDUAL OUTPUTObservationPredicted
Selling PriceResiduals1144066.40601462-
49066.406014622121345.712955621-
2345.7129556213135039.616517664-
10239.61651766444164322.4354755-
29322.43547549965128628.77289338714171.227106612661174
15.1860882427584.81391175987153814.1295505065185.87044
949438161462.694511183537.305488829171392.303571389106
07.69642861110209587.137991958-
26587.137991958111195239.0840424044760.915957596122079
97.1277493793002.87225062113197197.03634994217802.9636
50058114188092.3562882130907.6437117904

19262069172013961706184719502323228537522300252538001
740-49066.406014619977-2345.7129556209984-
10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413
Square feet
Residuals
Age Residual Plot
3040301532382730263518174012-49066.406014619977-
2345.7129556209984-10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413
Age
Residuals
Square feet Line Fit Plot
Selling
Price192620691720139617061847195023232285375223002525
38001740950001190001248001350001428001450001590001650
00182000183000200000211000215000219000Predicted Selling
Price192620691720139617061847195023232285375223002525
38001740144066.40601461998121345.712955621135039.61651
76644164322.4354754996128628.77289338737117415.1860882
4018153814.12955050575161462.69451117999171392.3035713
8903209587.13799195812195239.08404240402207997.1277493
7903197197.03634994186188092.35628820959
Square feet
Selling Price
Age Line Fit Plot
Selling
Price304030153238273026351817401295000119000124800135

00014280014500015900016500018200018300020000021100021
Price3040301532382730263518174012144066.40601461998121
345.712955621135039.6165176644164322.4354754996128628.
77289338737117415.18608824018153814.12955050575161462.
69451117999171392.30357138903209587.13799195812195239.
08404240402207997.12774937903197197.03634994186188092.
35628820959
Age
Selling Price
Sheet2
Sheet3
feetAge9500019263011900020694012480017203013500013961
51428001706321450001847381590001950271650002323301820
00228526183000375235200000230018211000252517215000380
cance
FRegression213313936968.45536656968484.2276611.26194574
30.0021787652Residual116502131602.97325591102872.997569
95.0%Intercept146630.89355597425482.08286875785.7542742
6050.000127566490545.2073136126202716.57979833590545.2
073136126202716.579798335Square
feet43.819366490110.28096507024.26218416180.00133809482
1.191114939166.44761804121.191114939166.447618041Age-
2898.686246708796.5649420672-3.6389829550.0038949963-
4651.9138632471-1145.4586301689-4651.9138632471-
49066.406014622121345.712955621-
2345.7129556213135039.616517664-

10239.61651766444164322.4354755-
29322.43547549965128628.77289338714171.227106612661174
15.1860882427584.81391175987153814.1295505065185.87044
949438161462.694511183537.305488829171392.303571389106
07.69642861110209587.137991958-
26587.137991958111195239.0840424044760.915957596122079
97.1277493793002.87225062113197197.03634994217802.9636
50058114188092.3562882130907.6437117904
19262069172013961706184719502323228537522300252538001
740-49066.406014619977-2345.7129556209984-
10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413
Square feet
Residuals
Age Residual Plot
3040301532382730263518174012-49066.406014619977-
2345.7129556209984-10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413
Age
Residuals
Selling
Price192620691720139617061847195023232285375223002525
38001740950001190001248001350001428001450001590001650
Price192620691720139617061847195023232285375223002525
38001740144066.40601461998121345.712955621135039.61651

76644164322.4354754996128628.77289338737117415.1860882
4018153814.12955050575161462.69451117999171392.3035713
8903209587.13799195812195239.08404240402207997.1277493
7903197197.03634994186188092.35628820959
Square feet
Selling Price
Age Line Fit Plot
Selling
Price304030153238273026351817401295000119000124800135
00014280014500015900016500018200018300020000021100021
Price3040301532382730263518174012144066.40601461998121
345.712955621135039.6165176644164322.4354754996128628.
77289338737117415.18608824018153814.12955050575161462.
69451117999171392.30357138903209587.13799195812195239.
08404240402207997.12774937903197197.03634994186188092.
35628820959
Age
Selling Price
Sheet2
Sheet3
feetAge9500019263011900020694012480017203013500013961
51428001706321450001847381590001950271650002323301820
00228526183000375235200000230018211000252517215000380
cance
FRegression213313936968.45536656968484.2276611.26194574
30.0021787652Residual116502131602.97325591102872.997569
95.0%Intercept146630.89355597425482.08286875785.7542742
6050.000127566490545.2073136126202716.57979833590545.2

073136126202716.579798335Square
feet43.819366490110.28096507024.26218416180.00133809482
1.191114939166.44761804121.191114939166.447618041Age-
2898.686246708796.5649420672-3.6389829550.0038949963-
4651.9138632471-1145.4586301689-4651.9138632471-
49066.406014622121345.712955621-
2345.7129556213135039.616517664-
10239.61651766444164322.4354755-
29322.43547549965128628.77289338714171.227106612661174
15.1860882427584.81391175987153814.1295505065185.87044
949438161462.694511183537.305488829171392.303571389106
07.69642861110209587.137991958-
26587.137991958111195239.0840424044760.915957596122079
97.1277493793002.87225062113197197.03634994217802.9636
50058114188092.3562882130907.6437117904
19262069172013961706184719502323228537522300252538001
740-49066.406014619977-2345.7129556209984-
10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413
Square feet
Residuals
Age Residual Plot
3040301532382730263518174012-49066.406014619977-
2345.7129556209984-10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413

Age
Residuals
Selling
Price192620691720139617061847195023232285375223002525
38001740950001190001248001350001428001450001590001650
Price192620691720139617061847195023232285375223002525
38001740144066.40601461998121345.712955621135039.61651
76644164322.4354754996128628.77289338737117415.1860882
4018153814.12955050575161462.69451117999171392.3035713
8903209587.13799195812195239.08404240402207997.1277493
7903197197.03634994186188092.35628820959
Square feet
Selling Price
Age Line Fit Plot
Selling
Price304030153238273026351817401295000119000124800135
00014280014500015900016500018200018300020000021100021
Price3040301532382730263518174012144066.40601461998121
345.712955621135039.6165176644164322.4354754996128628.
77289338737117415.18608824018153814.12955050575161462.
69451117999171392.30357138903209587.13799195812195239.
08404240402207997.12774937903197197.03634994186188092.
35628820959
Age
Selling Price
Sheet2
Sheet3
feetAge9500019263011900020694012480017203013500013961
51428001706321450001847381590001950271650002323301820
00228526183000375235200000230018211000252517215000380

cance
FRegression213313936968.45536656968484.2276611.26194574
30.0021787652Residual116502131602.97325591102872.997569
95.0%Intercept146630.89355597425482.08286875785.7542742
6050.000127566490545.2073136126202716.57979833590545.2
073136126202716.579798335Square
feet43.819366490110.28096507024.26218416180.00133809482
1.191114939166.44761804121.191114939166.447618041Age-
2898.686246708796.5649420672-3.6389829550.0038949963-
4651.9138632471-1145.4586301689-4651.9138632471-
49066.406014622121345.712955621-
2345.7129556213135039.616517664-
10239.61651766444164322.4354755-
29322.43547549965128628.77289338714171.227106612661174
15.1860882427584.81391175987153814.1295505065185.87044
949438161462.694511183537.305488829171392.303571389106
07.69642861110209587.137991958-
26587.137991958111195239.0840424044760.915957596122079
97.1277493793002.87225062113197197.03634994217802.9636
50058114188092.3562882130907.6437117904
19262069172013961706184719502323228537522300252538001
740-49066.406014619977-2345.7129556209984-
10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413
Square feet

Residuals
Age Residual Plot
3040301532382730263518174012-49066.406014619977-
2345.7129556209984-10239.616517664399-
29322.43547549960214171.22710661262727584.813911759818
5185.87044949425043537.305488820013110607.696428610972
-
26587.1379919581234760.91595759597843002.8722506209742
17802.9636500581430907.643711790413
Age
Residuals
Selling
Price192620691720139617061847195023232285375223002525
38001740950001190001248001350001428001450001590001650
Price192620691720139617061847195023232285375223002525
38001740144066.40601461998121345.712955621135039.61651
76644164322.4354754996128628.77289338737117415.1860882
4018153814.12955050575161462.69451117999171392.3035713
8903209587.13799195812195239.08404240402207997.1277493
7903197197.03634994186188092.35628820959
Square feet
Selling Price
Age Line Fit Plot
Selling
Price304030153238273026351817401295000119000124800135
00014280014500015900016500018200018300020000021100021
Price3040301532382730263518174012144066.40601461998121
345.712955621135039.6165176644164322.4354754996128628.
77289338737117415.18608824018153814.12955050575161462.
69451117999171392.30357138903209587.13799195812195239.
08404240402207997.12774937903197197.03634994186188092.
35628820959
Age

Selling Price
Sheet2
Sheet3
feetAge9500019263011900020694012480017203013500013961
51428001706321450001847381590001950271650002323301820
00228526183000375235200000230018211000252517215000380
cance
FRegression213313936968.45536656968484.2276611.26194574
30.0021787652Residual116502131602.97325591102872.997569
95.0%Intercept146630.89355597425482.08286875785.7542742
6050.000127566490545.2073136126202716.57979833590545.2
073136126202716.579798335Square
feet43.819366490110.28096507024.26218416180.00133809482
1.191114939166.44761804121.191114939166.447618041Age-
2898.686246708796.5649420672-3.6389829550.0038949963-
4651.9138632471-1145.4586301689-4651.9138632471-
49066.406014622121345.712955621-
2345.7129556213135039.616517664-
10239.61651766444164322.4354755-
29322.43547549965128628.77289338714171.227106612661174
15.1860882427584.81391175987153814.1295505065185.87044
949438161462.694511183537.305488829171392.303571389106
07.69642861110209587.137991958-
26587.137991958111195239.0840424044760.915957596122079
97.1277493793002.87225062113197197.03634994217802.9636
50058114188092.3562882130907.6437117904

Age
Residuals
Selling
Price192620691720139617061847195023232285375223002525
38001740950001190001248001350001428001450001590001650
Price192620691720139617061847195023232285375223002525
38001740144066.40601461998121345.712955621135039.61651
76644164322.4354754996128628.77289338737117415.1860882
4018153814.12955050575161462.69451117999171392.3035713
8903209587.13799195812195239.08404240402207997.1277493
7903197197.03634994186188092.35628820959
Square feet
Selling Price
Age Line Fit Plot
Selling
Price304030153238273026351817401295000119000124800135
00014280014500015900016500018200018300020000021100021
Price3040301532382730263518174012144066.40601461998121
345.712955621135039.6165176644164322.4354754996128628.
77289338737117415.18608824018153814.12955050575161462.
69451117999171392.30357138903209587.13799195812195239.
08404240402207997.12774937903197197.03634994186188092.
35628820959
Age
Selling Price
Sheet2
Sheet3
Ch. 4 Homework (SCM 386)
1. What is a simple linear regression?
_____________________________________________________
_____________________________________________________
_____________________________________________________

_____________________________________________________
_____________________________________________________
_______________________
2. In this equation define what each of the variables are:
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_______________________
3. In terms of slope, what is the differences between positive,
negative and no linear relationship
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_______________________
4. What is coefficient of determination (r2)?
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_______________________
5. What is the meaning of a 0.8 coefficient of correlation?
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________

_______________________
6. In a multiple regression, P values shouldn’t be used to
eliminate more than one variable at a time. Why?
_____________________________________________________
___________________
7. What is the use of an adjusted r2 value in a multiple
regression?
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_______________________
Using the Excel Data Analysis add-on solve:
8. As HR manager you wonder how effective training is at
reducing scrap. For the last 5 years you track training hrs. vs
scrap level.
Year
Training
Scrap
2012
200
5000
2013
300
4900
2014
400
4300
2015

500
4200
2016
600
4000
What is the F value and at 95% significance does it support a
linear relationship? If so, what is the equation? What % of the
variation is explained by the model? What are the upper and
lower values for the regression coefficients?
9. You are tracking production output and years’ experience.
Years’ experience
output
1
2000
2
2012
3
1900
4
2020
At 95% significance, does the F value support a linear
relationship between years’ experience and output?
10. As the production manager, you have been putting more
time in preventative maintenance & operator training. You
want to quantify what this has meant for output.

Year
PM hrs
Training
Ouput
2011
300
200
20000
2012
330
210
20200
2013
500
260
21000
2014
510
280
21050
2015
600
330
22000
At 95% significance, what is the F value?
What is the equation?
What is the correlation coefficient? Is this high or low?
What % of the variability in output is explained by the linear
regression model?
Which variable (PM hrs or training) has a higher impact on
output?
What is the sum of residual values? What does this sum tell
you about the model?
11. Using the VizDataEffectivelyPractice File complete Dotplot

(Ch 3 Effective Data Visualization) and SmallMultiples (Ch 3
Effective Data Visualization)
Chapter 2
Random variables & distributions
As analysts, we work with discrete random variables or
continuous random variables.
The probability p(x) associated with each discrete random
variable (x) taken together forms a discrete probability
distribution.
An example would be calculating the expected # of radios sold
per week
p(x)≥0
The mean () value is the expected # of radios sold

=0(0.03)+1(0.2)+2(0.5)+3(0.2)+4(.05)+5(.02)
=2.1 radios expected to be sold each week (over a large number
of weeks)
We can calculate the expected # of radios sold in Excel ()
X values are 0,1,2,3,4,and 5 radios sold per week (column B.)
The probability P(x) of radios sold are in column C (i.e. 0
radios sold is 3%; 1 radio sold is 20%, 2 radios sold is 50%...)
In column D (x*p(x))
we multiply each value
in column B by each
value in column C
We sum x*p(x) values in column D by selecting “ “ giving
us the () expected value of 2.1.
We can calculate the variance in # of radios sold in Excel ()

For each X value in column B we subtract 2.1 and square the
difference
We multiply this value by the p(x) value in column C
=(B6-$D$12)^2*C6
We copy and paste the equation in E6 to cells E7 –E11.
……Then sum E6-E11 values
To put variance (into meaningful units (e.g. radios2 has no
meaning) we take the square root
= radios
In Excel the formula to take a square root of .89 is =Sqrt(.89)
Knowing the standard deviation allows us to find the probability
the average # of radios sold will fall in a range:
=[2.1-1(0.94) up to 2.1+1(0.94)]=[1.16,3.04]
[2.1-(2*0.94)…2.1+(2*0.94)]=[0.21,3.99]
Discrete values of radios sold
in the +/- 2 range are
1, 2, 3

2.1
+2
-2
The probability radios sold are 2
standard deviations around the mean
are 0.2 + 0.5 + 0.2 = 0.9 or 90%
As an analyst you will be faced with 2 special types of discrete
probability distributions.
Binominal
Poisson
Binomial distributions occur when you are faced with a number
of successes in a sequence of n independent binary (yes/no)
outcomes, each of which yields success with probability p.

Examples of Binominal Distributions in Business
A company is making transistors. Every hour a supervisor
takes a random sample of n=5. The probability p(x) a transistor
is bad is 0.15.
What is the probability of finding r= 3,4 or 5 bad transistors?
P (X=r) in Excel is binomdist(…)
Number is 3; trials is 5; probability is 0.15, cumulative is False
since we want to know the probability of finding exactly r=3
bad transistors
P(X=r) for 3 (e.g. .024), 4 (.002) and 5 (.0001). We add up all
3 values to find probability of 3 or more defects .0266 or
2.66%.
We could have done this another way using the TRUE
cumulative function in Excel…
Since n=5, the probability of 3 or more transistor defects is the
same as 100% minus the probability of 2 or fewer P(x≤2).
=binomdist(..)
B11=r=2
B9=n=5

B10=p=.15
Cumulative=true
since we are looking at x
of 0,1 and 2
P(X≥3)=1-P(X≤2)
The expected (µ = mean) value of a binominal distribution is
n*p (in our example µ= 5*.15=.75 defective transistors)
The spread (σ = standard deviation) in a binominal distribution
is or σ = = 0.8 transistors in our example
But, how do we know if the binomial distribution is normal?
It depends……on the number of trials (n) and the probability of
success (p)
If np(1-p)≥10 the binomial
distribution is a bell shaped

normal distribution
In our example, the probability (p) of finding a bad transistor is
0.15….so our binominal distribution is skewed right. We
cannot use σ and µ to define a defect range.
Poisson distributions are discrete probability distributions that
express the probability of a number of events occurring in a
fixed period of time if these events occur (1) with a known
average rate µ and (2) independently of the time since the last
event.
Examples of Poisson Distributions in Business
Let’s say defects in a factory occur randomly at an average rate
(µ) of 1.8 defects per hour.
What is the probability p(x) of observing x=4 defects in a given
hour at the factory?

In Excel, the formula is =Poisson(...)
D7=x=4
D4=µ=1.8
Cumulative =False since we are concerned only with x=4
defects
The probability of 4 defects happening in a given hour is 7.2%
What is the probability of observing 2 or less P(x≤2) defects in
a given hour at the factory?
=Poisson(...)
x=2
E5=µ=1.8
Cumulative= True since we are concerned with 0,1 & 2 defects
in an hour
The probability of observing 2 or less P(x≤2) defects in a given
hour at the factory is 73%

Can we find a range of values for the # of defects?
It depends on the average rate of occurrence µ per unit time
As the average rate of occurrence in a Poisson distribution
increases so does the spread (i.e. standard deviation σ)
µ
µ
µ
In a Poisson distribution σ=
In our example, defects in a factory occur randomly at an
average rate (µ) of 1.8 defects per hour so the standard
deviation in defects is =1.34
Up till now we have been talking about discrete distributions.
With continuous probability distributions we figure out the
probability a random variable (x) will fall in an interval (a-b)
In a continuous probability distribution:
f(x) is ≥0 for all values of x
The total area under the curve f(x)=1

There are different types of continuous probability
distributions. We will look at 4:
Normal
F
t
Exponential
A normal distribution is one
where the data is evenly distributed around the mean, which
when plotted results in a bell curve
Why are normal distributions important?
Because of the Empirical Rule….
Empirical Rules for normal curves
Probability x is within +/- 1 standard deviation of the mean is
68%
95%
99.7%

So, if we knew on average ( µ ) company sales were
$10,000/day and the standard deviation ( σ ) in sales was
$2,000, there is a:
68% probability sales are from $8,000 to $12,000
95% probability sales are from $6,000 to $14,000
99.7% probability sales are from $4,000 to $16,000
We can “standardize” normal distributions to tell us how many
standard deviations (�) the value (x) is from the mean (µ)
Suppose travel time to work (in minutes)
Ok, so z scores tell us how many standard deviations we are
from the mean. Why is that important?
Remember, the Empirical Rule relates standard deviations to
probability so…

Based on Z scores we can calculate probabilities of occurrence.
Suppose build time for a product is normally distributed with an
average of 100 days & standard deviation of 20 days. The sales
team promises the customer no more than a 125 day built time.
What is the probability the factory can produce on-time?
In Excel, we can calculate the z score..
=(C5-C3)/C4
The P(Z≤1.25) in Excel =normdist(..)
C5=125; C3=100; C4=20; cumulative is True
since we are looking at all values through 125
For a product which takes on average 100 days to build with a
20 day standard deviation there is an 89.4% chance it can be
built in 125 days.
Now, suppose you were a Quality manager in a factory. A
process change was made. You want to know if the change
reduced process variation.

From samples collected before and after the process change,
suppose the average (µ = 52) is the same for both group but the
standard deviations (σ = 6 vs 12) are quite different.
Is the different large enough to say population variance is
different?
The F distribution tests for differences in variance at a specified
level of confidence (α).
Let’s look at an example.
You take two independent random samples of sizes n1 = 9 and
n2 = 7 from two normally distributed populations. Measured
sample variances are = 100 and = 20.
Test the null hypothesis H0: = at a confidence level of α = .05.
What’s a null hypothesis?
What does a confidence α = .05 mean?

A hypothesis is a theory which requires testing. The Quality
manager hypothesized after the process change variation
changed.
A null hypothesis ( ) is a hypothesis that says there is no
statistical significance between the test variable and the
outcome. It’s the hypothesis that you are trying to disprove. In
our example, the null hypothesis is no statistically significant
change in variation before and after the process change.
Ok. But what’s Alpha ( α )?
An α of 5% means there’s a 5% chance we say variance changed
when
in reality it didn’t.
So, there’s a 5% chance we’re wrong and 95% chance we’re
right..
The F statistic is s12/ s22
=5
Is the 5x difference in sample variance large enough to say
population variance is different?
If our Calculated F (s12/ s22) is greater than the Critical F we
can reject and conclude the variance before and after the
process change is different.

To find the critical F we need to know the shape of the F
distribution.
The shape depends on how many degrees of freedom our sample
numbers have
At different degrees of freedom (df) the shape of the F
distribution changes.
What are degrees of freedom?
Degrees of freedom are the number of values that have the
freedom to vary.
For example, a student needs to take nine courses to graduate,
and there are only nine courses offered the student can take.
There are eight degrees of freedom. Why? The student is able
to choose classes one through eight in any order; but after
taking these 8 classes we know what the ninth class must
be…it’s is the only class left.
The degrees of freedom (df) for group 1 (v1) is n1 - 1 = 9 – 1
=8
The degrees of freedom (df) for group 2 (v2) is n2 - 1 = 7 – 1 =
6

Since we are testing H0: = our null hypothesis is that the
variances are equal.
A two-tailed test will test if and if at a confidence level of α =
.05 ( on each side).
In Excel we can calculate the critical F for our & degrees of
freedom df1 & df2.
=finv(…)
=.025
deg. freedom 1=8
deg. freedom 2=6
Since our calculated F value (5) is less than the critical F value
(5.59) we cannot reject the null hypothesis that the variances in
the 2 groups is equal
What if our manager said to test if process variance was less
after the process change.
In this case we do not have a 2 tailed test since we are only
interested in less than. This is now a one directional test.

Test H0: with α = .05.
=5
Degrees of freedom
n1=9…9-1=8
n2=7…7-1=6
Since the calculated F(5)
is in the rejection region (>4.15) we reject H0 and say process
variance after the change is less.
Often as analysts we are asked to determine if differences in
sample averages are statistically significant or not.
If we have 2 groups we conduct a t test.
Below are study hours for 6 female and 5 males. Is average
study time different by gender?
First perform an F test to see if variance between groups is
different or not. This determines the type of t test we do.

We cannot reject unequal variance; this tells us which type of t
test to use. Select Data Analysis on Data Tab
On the Pop up select 2 samples assuming equal variance. Click
OK.
On average females in our test group study more than males.
But can we reject the null hypothesis and say females study
more or less than males?
Since the t stat calculated (1.36) is less than the t critical 2 tail
(2.26) we can’t say females study more or less than males. If
we did there would be a 20.5% chance of error.

Since the t stat calculated (1.36) is less than the t critical 1 tail
(1.83) we can’t say females study more than males. If we did
there would be a 10.2% chance of error.
Sometimes, we need to test for differences in means across
more than 2 groups. We use ANOVA in Excel.
Real Estate Agent, Architect and Stockbrokers were asked to
report their degree of job-related stress. Below is the Excel file
with 3 of the groups' data:
Click on the DATA tab and select DATA ANALYSIS. In the
Pop up select "Single factor" since we are only considering one
factor (Stress)
In the Pop up "Input Range" highlight the entire range of data.
Be sure to include the labels (row 1) and click on "Labels in
First Row."
Specify critical level .05.
Real estate agents tested had the highest stress. But, the results

are not significant because the calculated F (1.19) is less than
the Critical F (3.2).
The last type of continuous distribution we will look at is an
exponential distribution.
An exponential distribution arises naturally when modeling the
time between independent events that happen at a constant
average rate.
That sounds a lot like a Poisson Distribution….. how is an
exponential distribution different?
The Poisson distribution models the average number of
occurrences in a certain fixed time (µ). It is a discrete
distribution, taking on values 0,1,2,…0,1,2,….
The exponential distribution models expected time (λ=1/µ)
between events. It is a continuous distribution.
In our factory example, the average number of defects per hour
was µ=1.8 (a Poisson distribution)
The mean time between defects is λ =1/ µ =.56 hours per defect
(Exponential distribution)
For a ride at Disney world, the mean time to wait in line is 22

min. What is the probability of waiting ≤ 15 min?
A mechanic installs 3 mufflers per hour. What is the
probability the time to install a muffler will be ½ hr. or less?
In Excel, probability install time (X) will be ≤ t (0.5 hrs) given
we can install 3 per hour is:
There is a 15.3% chance a muffler can be installed in 0.5 hrs.
Why is it called an exponential distribution?
Because the probability is following an exponential function
n5
p0.15
r2
P(X≤2)0.973388
P(X≥3)0.026612
Ch. 2 Homework (SCM 386)
1. What is the difference between discrete and continuous
random variables?
_____________________________________________________
_____________________________________________________

_____________________________________________________
_____________________________________________________
_____________________________________________________
_______________________
2. What are the meanings of: binomial, Poisson and exponential
distributions?
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
______________________________________________
3. In a continuous distribution f(x) must be >___________ and
the total area under the curve must equal ______________?
4. Explain the Empirical Rule for normal curves
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_______________________
5. Explain how a z score standardizes a distribution
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________

_____________________________________________________
_______________________
6. F distributions test for differences in what?
_____________________________________________________
___________________
7. What effect do degrees of freedom have on F distributions?
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_______________________
Use Excel to solve:
8. A company makes cars. Probability of 0 defective cars is
10%; 2 defects is 30%; 4 defects is 25%; 5 defects is 25% and 8
defects is 10%. Using x p(x) to calculate variance, what is the
expected number of defects at +/- 2 sigma.
9. A company is making soap. Every day a supervisor takes a
random sample of n=10. The probability p(x) a soap sample is
bad is 0.1. Using a binomial distribution, find what is the
probability of r= 3,4 or 5 defective soaps?
10. Machine breakdowns occur randomly at an average rate (λ)
of 2 per day. Using a Poisson distribution, what is the
probability p(x) of observing x=3 breakdowns in a given day at
the factory?
11. Suppose manufacturing time for a component is normally
distributed with an average of 5 minutes & standard deviation

of 1 min. What is the probability a part can be made in ≤ 2.5
min? What is the probability a part can be made in ≥2.5 min?
12. Your factory has 2 ways to make a product. The
engineering manager is trying to determine if the variance in
both processes is the same. 2 independent random samples of
sizes n1 = 12 and n2 = 8 are pulled from two normally
distributed populations. Measured sample variances are 10 and
18. Using F testing at 95% confidence, can your manager say
the variances are the same?
13. In your factory a repaired part lasts 15 years. Using an
exponential distribution, what is the probability the part will
last less than 6 years?
14. Using the VizDataEffectivelyPractice File complete graphs
for: StandardDeviation (Ch 2 Effective Data Visualization),
BacktoBack1 (Ch 3 Effective Data Visualization), Slopegraph
(Ch 3 Effective Data Visualization)
Chapter 1
Business Analytics
Copyright © 2018 Pearson Education, Inc.
Meaningful
Information

Quantitative
Analysis
A data driven approach to managerial decision making
What is Business Analytics?
Raw Data
“Big data” is a term that describes the large volume of data that
inundates a business on a day-to-day basis.
Raw Data
The internet of things (IoT) in US, Industrie 4.0 in Germany and
物联网 (wù lián wăng) in China are all centered on the
application of big data to business decision making
https://www.youtube.com/watch?v=QSIPNhOiMoE
But….
The more data companies have the even more complex the
problems of managing it can become. Do you buy hardware? Do
you store it in the cloud? How often will you need to access it?
The more data you have, sometimes the harder it can be to find
true value from the data.
Five of the six most damaging data thefts of all time have
happened in the last two years. At the same time, failing to
comply with data protection laws can lead to expensive
lawsuits.

Python is an important programming language for data science
and data mining because it is free, open source, easy to learn,
easy to read, and scalable.
Python is particularly well suited when data analysis tasks
involve integration with web apps or when there is a need to
incorporate statistics into databases
Implementing the Results
Analyzing the Results
Testing the
Solution

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