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 < -1.96, or > 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.