In this paper, we will describe and discuss the application of the statistics and probability theory to a real world problem. We were hired as a manager of a bottling company and our task is to control the quality of production. According to the standards of production, a bottle of soda should contain sixteen ounces of soda. The customers of our soda claimed that the bottles of our brand of the soda contain less soda, than advertised. Due to the fact of the multiple complains on the quantity of soda in the bottles, we should perform a statistical study to examine the issue and determine its significance. Employees have measured the amount of soda in a random sample of 30 bottles and provided us with the following data:
We start our investigation with descriptive statistics. Calculate mean, median and the standard deviation of the ounces of soda in the bottles.
The mean value is calculated as follows:
x=1nx1++xn=13014.23+14.32++16.96=15.854
The sample median is the middle element of the ranged data (in this case, this is the average value between 15th and 16th element in a sorted data set). The median is equal to 15.99.
The sample standard deviation is calculated by using the formula below:
s=1n-1i=1n(xi-x)≈0.661
Using the information calculated above, we construct a 95% confidence interval for the population mean of ounces of soda in the bottles. Assuming that the distribution of the variable is normal, we got (Stattrek.com, 2016):
x±1.96*sn15.854±1.96*0.66130
The 95% CI is (15.617, 16.091)
It seems that the value of 16 ounces is within the given interval. We are 95% confident that the population mean value of ounces in the bottles is between 15.617 and 16.091 ounces.
Now we conduct a hypothesis test to examine the customers’ claim. Null hypothesis: the average amount of ounces in the bottles is not significantly different from 16 ounces. Alternative hypothesis: the average amount of ounces in the bottles is significantly less than 16 ounces.
H0: μ=16Ha: μ<16
Set the level of significance alpha at 5%.
Since the population standard deviation is unknown, we can use one-sample Student’s t-test to compare the observed and the hypothesized mean values (Statistics Solutions, 2016).
t=x-μsn=15.854-160.66130=-1.2091
The critical value of the test is t(29, 0.05)=1.6991 (Easycalculation.com, 2016). Since 1.2091<1.6991, we failed to reject the null hypothesis. We have no enough evidence to support the claim that the average amount of ounces in the bottles is significantly less than 16 ounces (at the 5% level of significance).
The reasons of the customers’ complaint are most likely due to the fact of the high value of variance. The amount of ounces of soda in bottles varies significantly. Some bottles contain only 14 ounces of soda, while others contain almost 17. We assume that those who bought the bottles of the first kind, complained most often. We would recommend to perform quality analysis of the equipment that is filling bottles with soda. The filling accuracy needs to be improved.
References
Confidence Interval. (2016). Stattrek.com. Retrieved 13 March 2016, from http://stattrek.com/estimation/confidence-interval.aspx
One Sample T-Test - Statistics Solutions. (2016). Statistics Solutions. Retrieved 13 March 2016, from http://www.statisticssolutions.com/manova-analysis-one-sample-t-test/
T Distribution Critical Values Table. (2016). Easycalculation.com. Retrieved 13 March 2016, from https://www.easycalculation.com/statistics/t-distribution-critical-value-table.php
