Dr. Soosan Shahrokh
December 12, 2014
Bottling Company Case Study
In this project we were given the case of customer complaints that the bottles of the brand of soda produced in our company contained less than the advertised sixteen ounces of product. Our boss wants us to solve the problem at hand and has asked me to investigate. I have asked my employees to pull Thirty (30) bottles off the line at random from all the shifts at the bottling plant.
The next step in solving this problem is to calculate the mean (x bar), the median (mu), and the standard deviation (s) of the sample. All of those calculations were easily computed in excel. The mean was ...view middle of the document...
Since the claim is that the sample is less than the mean. The mean will be calculated by averaging the amount of ounces in each bottle and dividing the total by the number of bottles. The data below shows the ounces in each of the thirty bottles that were pulled. The mean among the sample bottles is 14.87. The calculation to find the mean is to add all the ounces per bottle. The total is 446.1 divided by the random sample of 30. The average ounces in the bottles are less than 16 ounces. The median for the soda bottles is 14.8. The median is imputed by dividing the random number of 30 by 2 which equals 15. Arrange the ounces from smallest to largest, and select the number that falls on 15. This will provide the median for the thirty bottles. The standard deviation for the ounces in the bottles is 0.55. The standard deviation must be known in order to compute the confidence interval. To find the...
Total: | 446.1/ 30 = | 14.87 |
Mean: (Average) | 14.87 | |
Median: | (14.87 + 14.87) / 2 = | 14.8 |
Standard Deviation: | 0.55033 | |
In this case the null hypothesis is there ARE sixteen ounces in each bottle. The alternate hypothesis is that there are NOT sixteen ounces per bottle. In order to understand the outcome of this hypothesis, we will analyze the data set and our projections in the previous questions to answer them.
Analyzing the data from above, we have a mean of 14.9 and a standard deviation of 0.55. When we add one standard deviation to the mean, we still do not get to sixteen (16) ounces, but compute it to be 15.45 ounces. It is clear from this calculation that the null hypothesis is incorrect and has been disproven, indicating that the alternate hypothesis is in fact the correct assessment.
14.5 | Mean = | 14.87 | | | | | | |
14.6 | Median = | 14.8 | | | | | | |
14.7 | Standard
Deviation = | 0.541079169 | | | | | | |
14.8 | Margin of
Error = | 0.193619136 | 30 | 5.477226 | 0.098787 | 1.96 | 0.193623 | |
14.9 | Confidence
Interval L.L.= | 14.67638086 | | | | | | |