Bottling Company Case Study Part 1

BOTTLING COMPANY CASE STUDY 7

BottlingCompany Case Study

Part1

Bottle No.

Ounces

Deviations

(x-15.85)

(x-15.85)^2

1

14.23

-1.62

2.6244

2

14.32

-1.53

2.3409

3

14.98

-0.87

0.0289

4

15

-0.85

0.7225

5

15.11

-0.74

0.5476

6

15.21

-0.64

0.4096

7

15.42

-0.43

0.1849

8

15.47

-0.38

0.1444

9

15.65

-0.20

0.0049

10

15.74

-0.11

0.0121

11

15.77

-0.08

0.0064

12

15.80

-0.05

0.0025

13

15.82

-0.03

0.0009

14

15.87

0.02

0.0004

15

15.98

0.13

0.0169

16

16

0.15

0.0225

17

16.02

0.17

0.0289

18

16.05

0.20

0.04

19

16.21

0.36

0.1296

20

16.21

0.36

0.1296

21

16.23

0.38

0.1444

22

16.25

0.40

0.16

23

16.31

0.46

0.2116

24

16.32

0.47

0.2209

25

16.34

0.49

0.2401

26

16.46

0.61

0.3721

27

16.47

0.62

0.3844

28

16.51

0.66

0.4356

29

16.91

1.06

1.1236

30

16.96

1.11

1.2321

Total

475.62

11.9227

Mean= sum of the ounces per bottle number / total number of bottles

Mean= 446.1 /30

=15.85

Medianwhen arranged in ascending order, the ounces for the 30bottles are as follows

14.23

14.32

14.98

15

15.11

15.21

15.42

15.47

15.65

15.74

15.77

15.8

15.82

15.87

15.98

16

16.02

16.05

16.21

16.21

16.23

16.25

16.31

16.32

16.34

16.46

16.47

16.51

16.91

16.96

Therefore,the median = (15.98 + 16) /2

=15.99

Variance= 11.9227 /(30-1)

Thereason for using (30-1) is because the variance is for a samplepopulation

=0.4111

Standarddeviation = 0.4111

=0.64

Part2

Thenormal distribution value for the 95% confidence interval from thestatistics table is 1.96.

Bottle No.

Ounces (X)

Lower Limit (X-1.96)

Upper Limit (X+1.96)

1

14.23

12.27

16.19

2

14.32

12.36

16.28

3

14.98

13.02

16.94

4

15

13.04

16.96

5

15.11

13.15

17.07

6

15.21

13.25

17.17

7

15.42

13.46

17.38

8

15.47

13.51

17.43

9

15.65

13.69

17.61

10

15.74

13.78

17.7

11

15.77

13.81

17.73

12

15.80

13.84

17.76

13

15.82

13.86

17.78

14

15.87

13.91

17.83

15

15.98

14.02

17.94

16

16

14.04

17.96

17

16.02

14.06

17.98

18

16.05

14.09

18.01

19

16.21

14.25

18.17

20

16.21

14.25

18.17

21

16.23

14.27

18.19

22

16.25

14.29

18.21

23

16.31

14.35

18.27

24

16.32

14.36

18.28

25

16.34

14.38

18.30

26

16.46

14.50

18.42

27

16.47

14.51

18.43

28

16.51

14.55

18.47

29

16.91

14.95

18.87

30

16.96

15

18.92

Part3

H0:a bottle contains less than 16 ounces

HA:a bottle contains 16 ounces

Letα = 0.05 therefore, from the z-table, z critical value is 1.96.Hence, in case z &lt -1.96, or &gt 1.96, the null hypothesis shouldbe rejected.

Calculationof z-statistic z = (15.85 – 16) / 0.64

=-0.15 / 0.64

Z= 0.2344

Conclusionz (0.2344) is less than 1.96, which implies that the null hypothesisshould be accepted. Therefore, the alternative hypothesis isrejected. Hence, it can be concluded that a bottle contains less than16 ounces.

Part4

Fromthe conclusion of the hypothesis testing, it is apparent that thereare less than 16 ounces in a bottle. The three likely reasons forhaving less than 16 ounces per bottle may be inaccurate measuringtools, inexperienced employees, and leakages during packaging. Thesefaults can be corrected through the inspection of measuring tools orreplacing old tools will new ones, training employees on the rightmeasures, and ensuring correct packaging.

References

Moore,D. S. (2008). Thebasic practice of statistics.New York: W.H. Freeman and Co.

Peck,R. (2014). Statistics:Learning from data.Australia: Brooks/Cole.

Smithson,M. (2003). Confidenceintervals.Thousand Oaks, Calif: Sage Publications.