Bottling Company, Case Study Example

Microsoft Excel was used to calculate the mean, median, and standard deviation for the ounces in the bottles. The AVERAGE command was utilized to calculate the mean, which yielded 14.87 ounces. The MEDIAN command was utilized to calculate the median, which yielded 14.8 ounces. Since the calculated mean and median, two different measures of central tendency, were approximately equal, it can be concluded that the distribution of product volume in the selected bottles is approximately normal. The STDEV.S command was utilized to calculate the standard deviation, as this command is specifically used to determine the standard deviation of a sample rather than a population. This calculation determined that the standard deviation of the sample was 0.550. Since there was not a large standard deviation, one can conclude that a majority of the bottles fall within the same range of volume.

The 95% confidence interval for these measurements can be calculated since it was previously determined that the distribution is approximately normal. Since the 95% confidence interval is of interest, a value of 1.96 will be used to represent the proportion of the normal curve that is being utilized for this approximately. In addition, the sample mean (x) of 14.87, the sample size (n) of 30, and the sample standard deviation (sigma) of 0.550 are needed to complete this calculation. Plugging these values into the confidence interval equation, we find that the 95% confidence interval is 14.66 ? x ? 15.078. Therefore, there is a 95% chance that the average volume of the entire bottle population falls between these two values.

To determine whether the bottle contains less than 16 ounces, a student’s t-test can be used. The null hypothesis states that the volume of the bottle sample would be equivalence to 16 ounces. Since knowledge of deviation from this volume in either direction would be valuable, a two-tailed will be utilized that focuses on the comparison of means. The alternative hypothesis for this question would be that the volume of the bottle sample will deviate from 16 ounces. A result value of less than 0.05 will be necessary to refute the null hypothesis. To do so, the TTEST function in Microsoft Excel was utilized. The mean of the sample population that was inserted was 14.87 ounces, while the mean that this value was compared to was 16 ounces. In both instances, the number of cases was set to 30. The standard deviation for the sample population was 0.55 ounces and equal variance was assumed. The t-test yielded a t value of -11.25 and a p value of .10. Therefore, the null hypothesis cannot be refuted, and we can assume that the bottle sample contains 16 ounces.

The statistical test indicated that there were 16 ounces of volume per bottle because many of the bottles have close to 16 ounces. The mean and the median are useful measures of tendency, but the mean is affected by data outliers, while the median fails to consider these outliers. In order to generate a better picture of central tendency in addition to an understanding of the variance within a sample, it would be ideal in the future to calculate the range and mode of the sample. Furthermore, it is important to consider that the values received in these tests may not necessarily reflect the central tendency of the bottle population. Therefore, it would be valuable to collect a larger random sample of bottles before conducting similar studies in the future.