Case Study Focsa Cristina

8/3/2019 Case Study Focsa Cristina

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Student: FOCSA CRISTINA

Master program: International Project Management

Course: MANAGERIAL DATA ANALYSIS

1

CASE STUDY

Record 4060 data for two statistical variables (X and Y) at your choice:

Admision - sesion July 2011

"Alexandru Ioan Cuza" Police Academy

List of admitted candidates

Police - Law

X -

independent

variable

Y -

dependen

t variable

Selection

unit

(county)

NUME PRENUME GenderBaccalaureate

Mark

Admision

Mark

AG GHINESCU ANDREEA-FLORENTINA F 6,50 8,98

AG MORARU ANCA-NICOLETA F 6,50 8,90

AG ALEXE SIDONIA-CRESCENZIA F 6,50 8,75

AR RADU LUIZA-CLAUDIA F 6,80 8,75

AG ARSENE STEFAN-ALEXANDRU M 7,00 8,80

AG FLORESCU IONUT-ANDREI M 7,00 8,80

DJ MARCULESCU ROXANA F 7,50 8,85

AG TRACHE CAMELIA-ELENA F 7,50 8,85

AG CALUGAROIU LIVIU-MARIAN M 7,80 8,93

CS TEODORESCU SILVIU-PETRU M 7,80 8,93

BR STOILESCU FLORIAN M 7,90 8,95

AG ANITA ALEXANDRU M 7,90 8,95

SV DUMITRAS CORINA-LAVINIA F 8,00 8,90BT ISTRATE DANIELA-ANDREEA F 8,00 8,90

AG CIOBANU DANIEL-VALENTIN M 8,00 8,90

BT GABOR ANDREEA-CATALINA F 8,30 8,98

BZ OPREA ALEXANDRU-ION M 8,40 9,00

AG IORDACHESCU MIHAI-CIPRIAN M 8,40 9,00

GJ DRAGOESCU LAVINIA F 8,70 9,00

PH MOCANU RAZVAN-DANIEL M 8,70 9,00

VL TURCU GEORGE-IONUT M 8,70 9,00


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2

AG VASILE ROXANA-MARIA F 8,80 9,03

BC TIRON RADU-MARIAN M 8,90 9,05

AG MINCA IOANA-CATALINA F 8,90 9,05

OT CIOBANU MARIANA F 8,90 8,98

DJ BUTOI FLORIAN-COSMIN M 8,90 8,98

PH BEZNEA MIHAI-FLORIN M 9,00 9,00

OT ANDOR VLAD-CRISTIAN M 9,00 9,00

PH CONSTANTIN MADALIN M 9,00 9,00

BC SPANU CONSTANTIN M 9,00 9,00

BZ RACOREANU COSMIN-LAURENTIU M 9,00 9,00

AG ANGHELOIU LOREDANA-ELENA F 9,00 9,00

CT PETCULESCU ELENA-ADELINA F 9,00 9,00

GJ DRAGOTA LARISA-PETRUTA F 9,00 8,93

MH POPA IOANA F 9,00 8,93

AG PATRU ANDREEA-GEORGIANA F 9,00 8,93

MM DRAGOMIR RADU-STEFAN M 9,00 8,93

BZ DIMOFTE CRISTIAN-DANIEL M 9,10 8,95

DJ VOICULESCU ROBERT-CRISTIAN M 9,10 8,95

SV VAMANU IONELA-DANIELA F 9,20 8,98

GJ OGORANU IONUT-ADRIAN M 9,20 8,90

NT PETRESCU TEDY-FLORIN M 9,30 8,93

DJ RADUT ROXANA-FLORENTINA F 9,70 9,03

BZ BESNEA CATALIN-GEORGE M 9,80 9,05

VN TANASE CATALIN-TITEL F 9,80 9,05

BZ MATEI RAZVAN-COSMIN M 9,80 9,05

GJ GRECU FLAVIUS-ANDREI M 10,00 9,03

SV IVANUTA LAURA-VASILICA F 10,00 9,03

IF SIMA MARIAN-IONUT M 10,00 9,03

DB ATANASIE ALIN-IONUT M 10,00 9,03

Source:

http://www.academiadepolitie.ro/old/Facdepol/admitere/2011/rezultate/politie_drept%20-%20admisi.pdf
http://www.academiadepolitie.ro/old/Facdepol/admitere/2011/rezultate/politie_drept%20-%20admisi.pdfhttp://www.academiadepolitie.ro/old/Facdepol/admitere/2011/rezultate/politie_drept%20-%20admisi.pdfhttp://www.academiadepolitie.ro/old/Facdepol/admitere/2011/rezultate/politie_drept%20-%20admisi.pdf


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3

I For each of the two variables:

a) Calculate and interpret the average, standard deviation and the coefficient

of variation for row data. Interpret the results. Is the data series homogenous?

a.1) Average (or mean) is the arithmetic average of the scores (Baccalaureate

Mark and Admission Mark), also the average is a measure of central tendency

(a parameter enabling the researcher to determine the average score of a group

of scores).

In order to give an answer to this question and to offer an interpretation I have to

understand the relationship between the average, the median and the mode and

then to interpret the skewness parameter for the both cases (X and Y).

Firstly, by choosing the Descriptive Statistics tool from the Data Analysis

toolpack provided by MS EXCEL, I will have an overview picture of the

measures of central tendency, dispersion of data, skew of data and the kurtosis

of data.

Descriptive statistics will help us to examine:

1. central tendency (location) of data, i.e. where data tend to fall, as

measured by the mean, median, and mode.

2. dispersion (variability) of data, i.e. how spread out data are, as measured

by the variance and its square root, the standard deviation.

3. skew (symmetry) of data, i.e. how concentrated data are at the low or high

end of the scale, as measured by the skew index.

4. kurtosis (peakedness) of data, i.e. how concentrated data are around a

single value, as measured by the kurtosis index.

Baccalaureate Mark Admision Mark

Mean 8,6060 Mean 8,9570

Standard Error 0,1352 Standard Error 0,0106

Median 8,9000 Median 8,9750

Mode 9,0000 Mode 9,0000

Standard Deviation 0,9561 Standard Deviation 0,0751

Sample Variance 0,9140 Sample Variance 0,0056

Kurtosis -0,1215 Kurtosis 0,9812Skewness -0,6893 Skewness -1,1143


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Range 3,5000 Range 0,3000

Minimum 6,5000 Minimum 8,7500

Maximum 10,0000 Maximum 9,0500

Sum 430,3000 Sum 447,8500Count 50,0000 Count 50,0000

Largest(1) 10,0000 Largest(1) 9,0500

Smallest(1) 6,5000 Smallest(1) 8,7500

Confidence Level(95,0%) 0,2717

Confidence

Level(95,0%) 0,0213

So, I computed the average of X and Y as follows:

X Y

AVERAGE

X

AVERAGE

Y

8,61 8,96

Comparing the averages computed (X and Y) and those provided by the

Descriptive Statistics tool, I will find them identical.

In both cases the average () is different from their median and mode, meaning

that the arrays (X and Y ) arent normal distributed data.

Comparing the average with the median and mode, I can see that:

average < median < mode in the both cases (X: 8,60 < 8,90 < 9,00) and

(Y: 8,95 < 8,97 < 9,00),

so this 3 parameters show us that the distributions of data in our datas are non-

normal distributions (skewed distributions) and, in this case, I have for X and Y

a non-bell-shaped distribution of scores.

Looking at the Skewness parameter provided by the Descriptive Statistics tool, I

see that the Skewness is negative in both cases, so both of the arrays (X and Y )

are negatively skewed or skewed left, meaning that the left tail is longer.


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As the skewness in the Y array is between 1 and ( -0,6893), the distribution

is moderately skewed.

As the skewness in the X array is less than 1 (-1,1143), the distribution ishighly skewed.

If the data is very skewed, then the arithmetic mean might become misleading,

so we can conclude that in the Y data the average is not a good parameter to

measure the central tendency, while in the X data, the average might be taken

into consideration when talking about the central tendency.

a.2) Standard deviation is the square root of variance providing an index ofvariability in the distribution of scores, also the standard deviation is a measure

of variability (a parameter enabling the researcher to indicate how spread out a

group of scores are).

Computation in excel:

Total No. of

observations

= N

Standard

Deviation

X

(X -

X)

variance

of X

=

standard

dev of X

Standard

Deviation

Y

(Y -

Y)

variance

of Y

=

standard

dev of Y

50 0,9561 4,4352 0,9140 0,9561 0,0751 0,0003 0,0056 0,0751

4,4352 0,0032

4,4352 0,0428

3,2616 0,0428

2,5792 0,0246

2,5792 0,0246

1,2232 0,0114

1,2232 0,0114

0,6496 0,0010

0,6496 0,0010

0,4984 0,0000

0,4984 0,0000

0,3672 0,0032

0,3672 0,0032

0,3672 0,0032

0,0936 0,0003


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0,0424 0,0018

0,0424 0,0018

0,0088 0,0018

0,0088 0,0018

0,0088 0,0018

0,0376 0,0046

0,0864 0,0086

0,0864 0,0086

0,0864 0,0003

0,0864 0,0003

0,1552 0,0018

0,1552 0,0018

0,1552 0,0018

0,1552 0,0018

0,1552 0,0018

0,1552 0,0018

0,1552 0,0018

0,1552 0,0010

0,1552 0,0010

0,1552 0,0010

0,1552 0,0010

0,2440 0,0000

0,2440 0,0000

0,3528 0,0003

0,3528 0,0032

0,4816 0,0010

1,1968 0,0046

1,4256 0,0086

1,4256 0,0086

1,4256 0,0086

1,9432 0,0046

1,9432 0,0046

1,9432 0,0046

1,9432 0,0046


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Comparing standard deviation computed ( X and Y) and those provided by

the Descriptive Statistics tool, I will find them identical.

I have X= 0,9561 and Y= 0,0751.

The values computed are not so relevant measure of dispersion in their

relationship with the average because in both cases (X and Y) I have non-bell-

shaped distributions, as I have negatively skewed or skewed left distributions.

Even though, by comparing the two values of the standard deviation, I can see

that X is bigger than Y even if the Y average is bigger than the X average.

With this information I conclude by saying that the X array has a great variety ofvariables while the Y array has every variable proximate to the Y average.

a.3) Coefficient of variation: measures relative dispersion.

I have chosen to express the coefficient of variation in percentage and the values

computed are:

Coefficient of variation

CVx= x

/ X

CVy= y /

Y

11% 1%

The coefficient of variation values certify what I concluded in the interpretation

of the standard deviation values. Once again, I can say that Y has a lower

relative variability than X.

a.4) Is the data series homogenous?

Homogeneity measures the differences or similarities between the several

variables.


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Having a 11% coefficient of variation in the first array, I can say that

homogeneity is low while the 1% coefficient of variation show us that the

second array has a high homogeneity.

b) Summarize the data in an appropriate number of classes. Construct the

frequency distribution.

b.1) To solve this point, I have to identify the lowest and highest values in the

list for X and Y, so I compute the min and max:

Min X Max X Min Y Max Y

6,50 10,00 8,75 9,05

Secondly, I have to compute the Range (Maximum Value Minimum Value)

for each variable (X, Y):

Range X Range Y

3,50 0,30

Thirdly, I compute the number of classes by using the grouping rule:

> N.

In our case, I have N (no. of observations) =50 and Ill have k=6 classes.

For identifying the exact classes I have to divide each ranges by k (Lx= 0, 5 and

Ly: 0, 05) to establish the length of the interval:

For X: For Y:

LL UL

6,500 7,0007,000 7,500


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b.2) Frequency distribution for X:

LL ULMidpointmarks =

mxi

Nostudents=

xfi

6,500 7,000 6,750 6

7,000 7,500 7,250 2

7,500 8,000 7,750 7

8,000 8,500 8,250 3

8,500 9,000 8,750 19

9,500 10,000 9,750 13

Total 50

Frequency distribution for Y:


myi

Nostudents=

yfi

8,750 8,800 8,775 4

8,800 8,850 8,825 1

8,850 8,900 8,875 6

8,900 8,950 8,925 9

8,950 9,000 8,975 19

9,000 9,050 9,025 11

total 50

c)Calculate and interpret for the frequency distribution the average, standard

deviation and coefficient of variance. Compare with the results from point a).

Explain the differences.

c.1) Average of the frequency distribution ():


No students=xfi

mxi*xfi X

7,500 8,000

8,000 8,500

8,500 9,000

9,500 10,000

LL UL

8,750 8,800

8,800 8,850

8,850 8,900

8,900 8,950

8,950 9,000

9,000 9,050


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10

mxi

6,500 7,000 6,750 6 40,500 8,540

7,000 7,500 7,250 2 14,500

7,500 8,000 7,750 7 54,250

8,000 8,500 8,250 3 24,750

8,500 9,000 8,750 19 166,250

9,500 10,000 9,750 13 126,750

Total 50 427,000


myi

Nostudents=

yfi

myi*yfi Y

8,750 8,800 8,775 4 35,100 8,946

8,800 8,850 8,825 1 8,825

8,850 8,900 8,875 6 53,250

8,900 8,950 8,925 9 80,325

8,950 9,000 8,975 19 170,525

9,000 9,050 9,025 11 99,275

total 50 447,300

c.2) Standard deviation of the frequency distribution ():


mxi

Nostudents=

xfimxi*xfi X

xfi(mxi-X)

Standard

Deviation

X

6,500 7,000 6,750 6 40,500 8,540 19,225 0,970

7,000 7,500 7,250 2 14,500 3,328

7,500 8,000 7,750 7 54,250 4,369

8,000 8,500 8,250 3 24,750 0,252

8,500 9,000 8,750 19 166,250 0,838

9,500 10,000 9,750 13 126,750 19,033

Total 50 427,000 47,045


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myi

Nostudents=

yfimyi*yfi Y

yfi(myi-

Y)

Standard

Deviation

Y

8,750 8,800 8,775 4 35,100 8,946 0,117 0,071

8,800 8,850 8,825 1 8,825 0,015

8,850 8,900 8,875 6 53,250 0,030

8,900 8,950 8,925 9 80,325 0,004

8,950 9,000 8,975 19 170,525 0,016

9,000 9,050 9,025 11 99,275 0,069

total 50 447,300 0,250

c.3) Coefficient of variance of the frequency distribution (CV):


mxi

Nostudents=

xfimxi*xfi X

xfi(mxi-

X)

Standard

Deviation

X

CVx=

x / X

6,500 7,000 6,750 6 40,500 8,540 19,225 0,970 11%

7,000 7,500 7,250 2 14,500 3,328

7,500 8,000 7,750 7 54,250 4,369

8,000 8,500 8,250 3 24,750 0,252

8,500 9,000 8,750 19 166,250 0,838

9,500 10,000 9,750 13 126,750 19,033

Total 50 427,000 47,045


myi

Nostudents=

yfimyi*yfi Y

yfi(myi-

Y)

Standard

Deviation

Y

CVy=

y /

Y

8,750 8,800 8,775 4 35,100 8,946 0,117 0,071 1%

8,800 8,850 8,825 1 8,825 0,015

8,850 8,900 8,875 6 53,250 0,030

8,900 8,950 8,925 9 80,325 0,004

8,950 9,000 8,975 19 170,525 0,016

9,000 9,050 9,025 11 99,275 0,069


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total 50 447,300 0,250

c.4) Compare with the results from point a). Explain the differences.

Values computed for the row data: Values computed for the grouped data:

Average_X 8,61

Average_Y 8,96

Standard_deviation_X 0,9561

Standard_deviation_Y: 0,0751

Coefficient_of_variation_X 11%

Coefficient_of_variation_Y 1%

Average_X 8,64

Average_Y 8,94

Standard_deviation_X 0,970

Standard_deviation_Y: 0,071

Coefficient_of_variation_X 11%

Coefficient_of_variation_Y 1%

There are small differences between the average and the standard deviationparameters computed for the row data and those computed for the grouped data

and the values resulted in the second place are more accurate because I interpret

them by grouping the arrays and narrow the errors that might occur.

d) Construct a histogram and describe the shape of the distribution based on

the histogram.

d.1) Construct a histogram:

Firstly, Ive created the intervals (bins) by using the CONCATENATE function,

to merge the low limit with the upper limit and then Ive copied the frequencies

as follows (I have chosen to do this way because its more elegant than using the

Histogram Tool from the Data Analysis Tool pack):

HYSTOGRAM X HYSTOGRAM Y

Intervals Xfi Intervals Yfi

6,5-7 6 8,75 - 8,8 4

7-7,5 2 8,8 - 8,85 1

7,5-8 7 8,85 - 8,9 6


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8-8,5 3 8,9 - 8,95 9

8,5-9 19 8,95 - 9 19

9,5-10 13 9 - 9,05 11

The next step was to copy and paste special values in the sheet called

HISTOGRAM and to select the frequencies, click Insert - Column Chart, delete

the Series legend, right click on the edge of the graph and choose Select data,

and enter the Intervals for the Horizontal-Axis.

Then I modified some Layout and Design elements, I added a trendline, and

now looks like this:

0

2

4

6

8

10

12

14

16

18

20

6,5-7 7-7,5 7,5-8 8-8,5 8,5-9 9,5-10

F

r

e

qu

e

n

c

y

Baccalaureate Marks grouped in classes

Frequency distribution of the baccalaureate marks


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d.2) Describe the shape of the distribution based on the histogram.

Our both histograms are one right heaped with a longer tail on the left this is

way this histograms are negatively skewed (and moderately skewed).

e)In which interval is expected that about 95% of the data will fall? Is this

assumption true for this data?

X Y

LL UP freq Percentageof data

LL UP freq Percentageof data

7,65 9,56 42,00 84% 8,88 9,03 45 90%

6,69 10,52 8,00 16% 8,81 9,10718356 5 10%

5,74 11,47 0,00 0% 8,73172466 9,18227534 0 0%

0

2

4

6

8

10

12

14

16

18

20

8,75 - 8,8 8,8 - 8,85 8,85 - 8,9 8,9 - 8,95 8,95 - 9 9 - 9,05

F

r

e

q

u

e

n

c

y

Admision Marks grouped in classes

Frequency distribution of the admision marks


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total obs 50,00 total obs 50

II Using the Pivot Table Wizard in EXCEL, build a pivot table on your

spreadsheet (using also the second variable). You may have to change the

order of the rows (You should define the intervals first using VLookup

function).

I start by creating a new work sheet called VLookUp which contains the

following columns:

Unitatea

selectoareGender Baccalaureate

Mark

Baccalaureate

Categories

Admision

Mark

Admision

Categories

The next step will be to create 2 table arrays with the intervals and categories foreach mark (baccalaureate and admission):

Using the VLookUp formula I will fill the Categories columns with the values

presented .above:

table array

LL Baccalaureate

6,5 lucky

,25 normal

10 smart

table array

LL Admision

8,75 extreme low

8,98 normal

9,05 extreme high

Unitatea

selectoare

Gender Baccalaureate

Mark

Baccalaureate

Categories

Admision

Mark

Admision

Categories

AG F 6,50 lucky 8,75 extreme low



AR F 6,80 lucky 8,75 extreme low

AG M 7,00 lucky 8,80 extreme low


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Using the data above, I clicked on the Pivot Table button from the Insert Fieldand I have created the pivot table below:


DJ F 7,50 lucky 8,85 extreme low



CS M 7,80 lucky 8,93 extreme low

BR M 7,90 lucky 8,95 extreme low


SV F 8,00 lucky 8,90 extreme low

BT F 8,00 lucky 8,90 extreme low


BT F 8,30 normal 8,98 extreme low

BZ M 8,40 normal 9,00 normal

AG M 8,40 normal 9,00 normal

GJ F 8,70 normal 9,00 normal

PH M 8,70 normal 9,00 normalVL M 8,70 normal 9,00 normal

AG F 8,80 normal 9,03 normal

BC F 8,90 normal 8,98 extreme low

AG M 8,90 normal 8,98 extreme low

OT M 8,90 normal 9,05 normal

DJ F 8,90 normal 9,05 extreme high

PH F 9,00 normal 8,93 extreme low

OT F 9,00 normal 8,93 extreme low

PH F 9,00 normal 8,93 extreme low

BC M 9,00 normal 8,93 extreme low

BZ M 9,00 normal 9,00 normal

AG M 9,00 normal 9,00 normal

CT M 9,00 normal 9,00 normal

GJ M 9,00 normal 9,00 normal

MH M 9,00 normal 9,00 normal

AG F 9,00 normal 9,00 normal

MM F 9,00 normal 9,00 normal

BZ M 9,10 normal 8,95 extreme low

DJ M 9,10 normal 8,95 extreme low

SV M 9,20 normal 8,90 extreme low

GJ F 9,20 normal 8,98 extreme lowNT M 9,30 normal 8,93 extreme low

DJ F 9,70 normal 9,03 normal

BZ M 9,80 normal 9,05 extreme high

VN F 9,80 normal 9,05 extreme high

BZ M 9,80 normal 9,05 extreme high

GJ M 10,00 smart 9,03 normal

SV M 10,00 smart 9,03 normal

IF F 10,00 smart 9,03 normal

DB M 10,00 smart 9,03 normal


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Admision

Categories(All)

Baccalaure

ate

Categories

(All)

Count of

Gender

Admisi

on

Mark

Unitatea

selectoare9,05

9,

0

5

9,

0

3

9,

0

3

9,

0

0

8,

9

8

8,

9

8

8,

9

5

8,

9

5

8,

9

3

8,

9

3

8,

9

0

8,

9

0

8,

8

5

8,

8

5

8,

8

0

8,

7

5

Gran

d

Total

AG 1 3 2 1 1 1 1 1 2 1 14

AR 1 1

BC 1 1 2BR 1 1

BT 1 1 2

BZ 2 2 1 5

CS 1 1

CT 1 1

DB 1 1

DJ 1 1 1 1 4

GJ 1 2 1 4

IF 1 1

MH 1 1MM 1 1

NT 1 1

OT 1 1 2

PH 1 2 3

SV 1 1 1 3

VL 1 1

VN 1 1

Grand Total 4 1 3 31

21 4 2 2 5 2 3 2 1 1 2 2 50

The pivot tables shows us:

- the most students who were enrolled this year came from ARGES county

(14 students from AG);

- the most students who were enrolled this year passed the admission exam

with 9,00 (12 students). Notice that we can make the same observations


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for the baccalaureate marks too if we switch Admission Mark with

Baccalaureate Mark in the Column Labels section of the pivot table.

- By choosing the from report filter the extreme high admission

category we can find which mark was the extreme high mark at the

admission exam, how many students took it and from which counties:

Admision Categories extreme high

Baccalaureate Categories (All)

Count of Gender Admision Mark

Unitatea selectoare 9,05Grand

Total

BZ 2 2

DJ 1 1

VN 1 1

Grand Total 4 4

III Calculate the regression line and interpret and test the regression

coefficients, coefficient of determination and coefficient of correlation.

Interpret the results.

a) Calculate the regression line

The regression line is described as: y = a+ bx, can be computed like this:

430.3 =50 a+ 447.85 b

3856.865 = 447.85 a + 3747.95b

a= (430.3-447.85 b)/50

3856.865=447.85(430.3-447.85 b)/50 +3747.95

b=0,059


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a= 8,444

Coefficients

Intercept 8,44436615

Baccalaureate

Mark 0,059567029

The a parameter is 8,444 and it represents the intercept of the regression

function and it does not have any economic significance. Geometrically, it is

the point where the regression line intersects OX axis.

The b parameter 0,059 is the slope of the regression line and it is called

regression coefficient. Because it is positive we can say that the relationship

between the two marks is a positive relationship. This parameter shows the fact

that when the Baccalaureate makrs increase by 0,5 points, the Admision Marks

increases by 0,059 points.

b) Test the regression coefficients, coefficient of determination and

coefficient of correlation.

Regression Statistics

y = 0,059x + 8,444

R = 0,575

r=0,758

8.7

8.8

8.9

9

9.1

6.5 7 7.5 8 8.5 9 9.5 10

A

d

m

i

s

i

o

n

M

a

r

k

s

BAC grades

Bac grades vs

Admision Mark


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Multiple R 0,758398211

R Square 0,575167847

AdjustedR

Square 0,566317177

Standard Error 0,049451391

Observations 50

The coefficient of correlation

r =

[ [ = 0,758

R = 0,575

We can say by interpreting the coefficient of correlation (Multiple R =

0,758398211) that we have a relationship between the two variable and the fact

that the coefficient of correlation is very close to 1 leads to the conclusion that

between the baccalaureat marks and the admision one is a strong relationship(75%).

The coefficient of determination

R square is 0,575167847, meaning that 57.51 % of the variation of the

Admission Marks can be explained by the variation of the Baccalaureate Marks

and rest of percentage by the variation of other factors.

The Adjusted R Square value is 0,566317177, meaning that 56% of the

evolution of the Admission Marks can be explained by the regression model

y = 0,059x + 8,444

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Case Study Focsa Cristina