Calculation of Mean and Median and Standard Deviation x ̅ = (∑▒x)/n= 446.1/30=14.87 Mean
14.8 + Median, based on the middle score of all measured bottles. σ=√(1/(N-1) ∑_(i=1)^N▒〖(x_i-μ)〗^2 …show more content…
As such hypothesis needs to reflect the average bottle in production, it should be as follows: If a sample of x bottles taken at random throughout all production shifts during a day does not equal 16 ounces of fluid in them, then the claim that content per bottle is less than the advertised 16 ounces is …show more content…
Each of these possible reasons will be examined separately. The error stemming from miscalibration in volumetric weight assumes that the bottling system utilized by the bottling company verifies bottle content through the means of taking the weight of the filled bottle. If such a system is not calibrated to account for the weight of the bottle itself, the verification or checking system of such a bottling machine would assume that the finished weight (including the weight of the bottle) is meant to represent 16 ounces. If this is the case, the system would need to be recalibrated to expect a finished weight, which is that of the 16 ounces of fluid in addition to the weight of the bottle. This would represent only a minor repair that would need to be undertaken. If the error is stemming from a miscalibration of the dispensing system, meaning that the amount of fluid dispersed is wrongly set, the amount dispensed in non-bottle scenarios should be determined. If the amount dispensed proves to be constant, an adjustment to bring the filling volume back to par is required. If the amount dispensed proves to be non-constant, the filling mechanism would need to be completely new calibrated to dispense