Bottling Company Case Study

Title: Bottling Company Case Study

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Bottling Company Case Study

Complaints have been received that the bottles of soda produced in our company contain less than the advertised sixteen (16) ounces of product. The purpose of ensuing study is to verify the claim of the complainers. A random sample of 30 bottles was drawn from among all the shifts of the plant.

Bottle Number | Ounces | Bottle Number | Ounces | Bottle Number | Ounces |

1 | 14.5 | 11 | 15 | 21 | 14.1 |

2 | 14.6 | 12 | 15.1 | 22 | 14.2 |

3 | 14.7 | 13 | 15 | 23 | 14 |

4 | 14.8 | 14 | 14.4 | 24 | 14.9 |

5 | 14.9 | 15 | 15.8 | 25 | 14.7 |

6 | 15.3 | 16 | 14 | 26 | 14.5 |

7 | 14.9 | 17 | 16 | 27 | 14.6 |

8 | 15.5 | 18 | 16.1 | 28 | 14.8 |

9 | 14.8 | 19 | 15.8 | 29 | 14.8 |

10 | 15.2 | 20 | 14.5 | 30 | 14.6 |

Summary Statistics

Sample mean, x = (Sum of product volumes of all bottles)/ (No. of bottles in sample)

= 446.1/30 = 14.87 ounces

After arranging the data in ascending order, the median was calculated. Since there are even number of data values,

Median = Average of the (n/2)th data and (n/2 + 1)th data [after sorting]

= (14.8+14.8)/2 = 14.8 ounces

Sample Variance = [Sum of all (data values – mean) 2] / (n-1)

= 8.873/29 = 0.303

Sample Standard Deviation, s = √ (Variance)

= 0.550329 ounces

Sample mean = 14.87 ounces, Sample median = 14.8 ounces, Sample sd = 0.550329 ounces

Confidence Interval for ounces in the bottles

x = 14.87, s = 0.550329

Since, n ≥ 30, we can approximate the sample as large sample and use σ = s

For 95% confidence interval, α = 0.05, so Zα/2 = 1.96

Margin of Error E = Zα/2 * s/√n

= 1.96*0.550329/√30 = 0.197 ounces

Lower limit of confidence interval = x – E

= 14.87 – 0.20 = 14.67 ounces

Upper limit of confidence interval = x + E

= 14.87 + 0.20 = 15.07 ounces

We can say with 95% confidence that the population...