# 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.