Soda Volumes Troubleshooting Bottling Errors

Due to customer complaints of low product volume an investigation was conducted to check whether these complaints had any merit. Bottles (n = 30) of soda were randomly taken off the production line and the volumes measured. The total amount of soda measured was ?X = 446.1 oz, so the mean (MX) amount of soda per bottle was ?X/n = 446.1/30 = 14.87 oz. The median value is [(n + 1)/2] = 31/2 = 15.5, so the two middle values were averaged to obtain the median. The two middle values are 14.8 and 14.8, so the median is 14.8. Since the mean and median have similar values, the distribution of soda volumes is not skewed. The standard deviation is SDX = ?{[?(X -- MX) 2]/n-1} = = ?{[(14.5 -- 14.87)2 + (14.6 -- 14.87)2 + . . . + (14.8 -- 14.87)2 + (14.6 -- 14.87)2]/29} = 0.550329.

Calculating the 95% confidence interval (CI) for the data requires determining tcrit, which based on the t table for a two-tailed test with an alpha = .05 would be 2.045. The 95% CI is equal to t (s/?n) = 2.045(0.550329/?30)...

Based on these calculations, the upper and lower limits for the 95% CI would be 15.07 and 14.67, respectively.
The experimental hypothesis (H1) is that the volumes in the soda bottles are on average significantly lower than the 16 ounces advertised, which would support customer allegations. By comparison, the null hypothesis (H0) is that there is not a significant difference between the advertised volume of 16 ounces and the average amount of soda found in the sample. Stated more formally: H1 ? 16 oz and H0 = 16 oz. A two-tailed test is appropriate because it is unknown whether the volumes are greater than, less than, or equal to 16 oz.

If the population mean is assumed to be 16 oz, then t = (MX -- 16 oz)/(s/?n) = (14.87 -- 16.00)/(0.550329/?30) = -1.13/0.1005 = -11.24. For a two-tailed test with df = 29 and an alpha of .05, tcrit = 2.045. Since -11.24 > -2.045, the null hypothesis must be rejected. Based on this analysis, the chance that a randomly chosen bottle…