Case Study - Popescu Laura

8/3/2019 Case Study - Popescu Laura

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MANAGERIALDATAANALYSIS

Case Study : Catering Business

Student: Popescu Laura Elena


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CASE STUDY:CATERING BUSINESS

A trader has a network of catering business units with 40 vendors employed.For the 40 vendors, considered as a general community, has made a survey done inalphabetical order, which is a criterion and were randomly given a code (no CRT.).

We considered 2 variables:- Variable x : the number of worked hours- Variable y : the amount of the monthly net salary

Table 1: Collected Data

Crt. No Age Hours Worked Net Monthly Salary ( lei )

1 19 150 1313

2 20 170 1488

3 22 190 1663

4 21 184 1472

5 26 180 1575

6 24 188 1504

7 23 174 1392

8 25 178 1558

9 27 168 1470

102

1

196

1715

11 23 176 1540

12 23 183 1601

13 20 157 1374

14 24 175 1531

15 21 191 1671

16 25 184 1610

17 22 147 1286

18 26 135 1181

19 23 187 1636

20 20 166 1453

21

19

146

1

278

22 24 158 1383

23 20 157 1374

24 25 173 1514

25 21 170 1488

26 26 180 1575

27 22 169 1479

28 27 168 1470


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29 22 211 1846

30 23 176 1540

31 19 160 1400

32 21 150 1313

33 20 165 1444

342

1

140

122

5

35 21 151 1321

36 24 193 1689

37 25 178 1558

38 26 135 1181

39 23 187 1636

40 24 179 1566

The average, standard deviation and the coefficient ofvariation:

1. a. The average worked hours of the vendors:

15.16640

68251 !!!!

n

x

x

n

i

i

b. The average monthly net salary:

lei73,72540

593131 !!!

!

n

y

y

n

i

i

2. a. The standard deviation for the number of worked hours

51,17=

2

!

n

xxiW

This result tells us that the between the real and estimated number of worked hours and

the amount of salary there is a difference of , plus/minus, 17,51 hours

b. The standard deviation for the amount of the vendors monthly net salary

52,150=

2

!

n

yyiW


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This result tells us that the between the estimated and the real amount of monthly salary and

the number of worked hours there is a difference of , plus/minus, 150,52 hours

3. a. The coefficient of variance of the worked hours%68,101100 !!

xCv

W

b. The coefficient of variance of the monthly net salary

%35,100100 !!y

CvW

Because the level of the coefficient of variance is over the level of 35% we can conclude

that the average worked hours and the average monthly net salary is not representative and the

data is not heterogeneous.

Table 2: The Average, Standard Error and Standard Deviation

Frequency distributionThe 40 vendors will be grouped in 5 intervals with the calculated range of15.2 for the

variable x and of133 for the variable y.

LCL UCL

135 150.2 7 142.6

150.2 165.4 6 157.8

165.4 180.6 16 173

180.6 195.8 9 188.2

195.8 211 2 203.4

Worked Hours Salary

Minimum 135 1181

Maximum 211 1846

Max-Min 76 665

No. of Classes 5 5Range 15.2 133


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LCL UCL

1181 1314 7 1247.5

1314 1447 7 1380.51447 1580 17 1513.5

1580 1713 7 1646.5

1713 1846 2 1779.5

1. a. The arithmetic mean of worked hours of the n=40 vendors:34,170!

!

i

ii

n

fxx

b. The arithmetic mean of the amount of the monthly net salary of the n=40 vendors:

44,3289!

!

i

ii

n

fyy

2. a. The standard deviation for variable x:5,13

2!! xx WW

This result tells us that between the real and estimated number of worked hours and the

amount of salary there is a difference of, plus/minus, 13, 05 hours.

b. The standard deviation for variable y:

35,572 !! yy WW

This result tells us that the between the estimated and the real amount of monthly salary and

the number of worked hours there is a difference of , plus/minus, 57,35 hours

3. a. For variable x, the number of worked hours the coefficient of variance is :%66,7100 !!

xCv

W


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b. For the y variable, for the amount of the monthly net salary the coefficient of variance

is:

%74,1100 !!y

CvW

Because the coefficient of variance of the two variables is below 35% it results that the

grouping of x and y is eloquent.

Table 3: The frequency distributions

This charts give us useful information about the shape of the distribution and as we can see

above for both of the variables the highest number of data is found the third interval.

0

2

4

6

8

10

12

14

16

18

135-135 150.2-150.2 165.4-165.4 180.6-180.6 195.8-195.8

Frequency

Range Intervals

Frequency Distribution Of Worked Hours

0

5

10

15

20

1181-1181 1314-1314 1447-1447 1580-1580 1713-1713

Frequency

Range Intervals

Frequency Distribution Of The Monthly Net

Salary


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The Empirical RuleThe empirical rule Worked Hours

erval

Worked Hours

(lower) Worked Hours (upper)

Work Hours

frequency %

s 1W 153.1179501 188.1320499 35

s 2 W 135.6109003 205.6390997 39

The empirical rule Monthly Net Salary

erval Salary (lower) Salary (upper) Salary frequency %

s 1W 1332.300977 1633.349023 33

s 2 W 1181.776955 1783.873045 39

.

The empirical rule states that for a normal distribution:

y 68% of the data will fall within 1 standard deviation of the meany 95% of the data will fall within 2 standard deviations of the meany Almost all (99.7%) of the data will fall within 3 standard deviations of the meanRegression

From this scatter diagram we can see that the model is almost a perfect line, a linear

model. This means that the two variables are interdependent and for example the amount of the

monthly net salary depends on the number of worked hours in the company.

y = 8.3514x + 57.868

R = 0.9435

0

500

1000

1500

2000

0 50 100 150 200

Salary

Worked Hours

Relationship Betweem Worked Hours And The Monthly Net Salary


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The Analysis of variance ( ANOVA )Regression Statistics

Multiple R 0.971328786

R Square 0.94347961

Adjusted R Square 0.941992231

Standard Error 36.25341277

Observations 40

Because Multiple R = 0,97 we can draw the conclusion that the link between the amount

of the vendors monthly salary and the number of hours they work are linked together.

The Standard Error reflects the difference between the estimations made about that there

is an average difference of 36,25 between the two chosen variables.

Coefficients

Standard

Error t Stat P-value

Intercept 57.868 56.86746539 1.017593443 0.315307Hours

Worked 8.3514 0.331591679 25.18578452 2.59E-25

Lower 95% Upper 95% Lower 95.0% Upper 95.0%

-57.25420516 172.9901249 -57.25420516 172.9901249

7.680124313 9.022668832 7.680124313 9.022668832

From the table above, we can see that the intercept variable is5

7,868

and it mean thatwhen the explanatory level is 0, when the number of worked hours is 0, the amount of the

monthly net salary will be of57,868 lei.

Because the value of P-value is high 0,31 the coefficient is insignificant.

The coefficient b = 8,35 and it means that if the level of worked hours will increase also

the salary will increase with 8,35 lei.

ANOVA

df SS MS F Significance F

Regression 1 833697.9974 833697.9974 634.3237 2.58866E-25

Residual 38 49943.77762 1314.309937

Total 39 883641.775


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ICoefficient of correlation test.

1. Formulation of hypotheses:H0: r = 0

H1: r { 0

2. We choose a significance level of95% ( = 0.05);3. Because the number of observations (40) is higher than 30 we will choose Student test (z)

with n-2 degrees of freedom.

4. 96,125;025,02;2/ !!! zzz ntab E

5.

6. The critical areas: | calcz |>| tabz |

7. Since calcz > tabz for a significance level = 0.05 we reject the null hypothesis andaccept the alternative hypothesis, therefore the model is valid

II. Testing the significance of the a parameter:

1. Formulation of hypotheses:H0: = 0

H1: {0

2. We choose a significance level of95% ( = 0.05);3. Because the number of observations is higher than 30 we will choose Student test (z)

with n-2 degrees of freedom ( in our case 38 d.f)

4. 38,225;025,02;2/ !!! zzz ntab E

5.Where Sa is the standard error of the parametera:

41,2406.0

3897.0

1

22

!!

!

r

nrzcalc

53,1930!

!

!

aa

calcs

a

s

az

E


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299,093,112

80,33

12754,4

25

28564,76

)(

2

2

!!!

!

xx

n

SSE

S

i

a

6. The critical areas: | calcz |>| tabz |

7. Since calcz > tabz , for a significance level of = 0.05 we reject the null hypothesis, so weaccept alternative hypothesis.

III. Testing the significance of the b parameter:

1. Formulation of hypotheses:H0: = 0

H1: {0

2. We choose a significance level of95% ( = 0.05);

3. Because the number of observations is higher than 30 we will choose Student test (z) withn-2 degrees of freedom ( in our case 38 d.f);

4. 38,225;025,02;2/ !!! zzz ntab E 5. Stat)zANOVAfromextracted(27,93!!

b

calcSbz

Where Sb is the standard error of the parameterb:

299.0

)(

2

2

!

!

xx

n

SSE

S

i

b

6. The critical areas: | calcz |>| tabz |7. Since calcz > tabz , for a significance level of = 0.05 we reject the null hypothesis, so we

accept alternative hypothesis.

Case Study - Popescu Laura

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