Bottling Company Case Solution
Bottling Company Case Solution
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This 6 page Study Guide was uploaded by an elite notetaker on Wednesday January 6, 2016. The Study Guide belongs to a course at Zuni Comprehensive Community Health Center taught by a professor in Fall. Since its upload, it has received 41 views.
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Date Created: 01/06/16
RUNNING HEAD: BOTTLING COMPANY CASE ANALYSIS 1 Assignment Title Student Name Course Name Instructor Name Date BOTTLING COMPANY CASE ANALYSIS 2 1) Calculate the mean, median, and standard deviation for ounces in the bottles. Answer: Mean 14.87 sd 0.5503 3 Media 14.8 n Refer Appendix 1 to see the calculation. 2) Construct a 95% Confidence Interval for the ounces in the bottles. Answer: x=14.87,s=0.5503,n=30,α=0.05 Level of confidence is at 95%. Formula for determining the confidence interval: s s (x−t2(√)n( )+t2 √n ) tα=t0.025.045 2 Confidence interval after substituting the value: 0.5503 0.5503 (14.87−2.04( √ 30 ),14.872.04( √30 )) = (14.665,15.0)5 BOTTLING COMPANY CASE ANALYSIS 3 Thus, from the above calculation it is clear that the confidence interval at 95% confidence is 14.665 ~ 15.075 ounces. 3) Conduct a hypothesis test to verify if the claim that a bottle contains less than sixteen (16) ounces is supported. Clearly state the logic of your test, the calculations, and the conclusion of your test. Answer: Statistical hypothesis provides information about the distribution of data X. It will specify the possible set of distribution of the X variables with the statement set to be true. The main purpose of making use of hypothesis is to determine whether there is sufficient support from the statistics to reject the assumed null hypothesis in favor to the alternative hypothesis. H0 is denoted as null hypothesis. H1 is denoted as alternative hypothesis. In this case, hypothesis is simple as it considers single variable X. Null hypothesis: H0: µ ≥ 16 ounces Alternative hypothesis: H1 < 16 ounces. ´=14.87 ,s=0.5503 ,n=30 ,α=0.05 Statistic = t (df = n1 =29) BOTTLING COMPANY CASE ANALYSIS 4 ´−16 14.87−16 t= = =−11.25 s 0.5503 ( √n ( 30 ) Critical valu0.95 1.699 Area of rejection tt 1.25 From the above test with 5% level of significance it is clear that bottle contains quantity lesser than 16 ounces and it falls in the rejection area. Thus, null hypothesis H0: µ ≥ 16 ounces is rejected and mean is lesser than 16 ounces. 4) Provide the following discussion based on the conclusion of your test: a) If you conclude that there are less than sixteen (16) ounces in a bottle of soda, speculate on three (3) possible causes. Next, suggest the strategies to avoid the deficit in the future Answer: From the analysis, it is evident that customer complaint is true, and a bottle does not contain 16 ounces of soda. Following might be some of the reasons for the causes: Dispenser of soda may have some fault that results in the dispenser to dispense lesser than 16 ounces of soda in the bottle. Cover of the containers may not be appropriate that may cause the spill and reduces the quantity. No effective quality control metrics in the company to check whether the soda supplied in the appropriate quantity. Lack of calibration in the program or the software used by the company. Effective strategy must be such that it must avoid such complaints in the future completely. It is necessary for the company to look at the performance of the soda dispenser. If there is a problem with the dispenser, it is essential to replace with advanced technology machines that will dispense appropriate quantity of soda. Review BOTTLING COMPANY CASE ANALYSIS 5 of effectiveness of all machine and technology used by the company is essential. It will enable the company to know whether it is the time to upgrade to new technology or to continue with the existing technology. Software integration must be effective, and it requires an update to implement effective control system. Company must bring in effective quality control metrics to avoid more consequences and in avoiding more problems in the future. Continuous monitoring and performance update is essential to improve the performance of the company. Appendix 1: 1 14.5 2 14.6 3 14.7 4 14.8 5 14.9 6 15.3 7 14.9 8 15.5 9 14.8 10 15.2 11 15 12 15.1 13 15 14 14.4 15 15.8 16 14 17 16 18 16.1 19 15.8 20 14.5 21 14.1 22 14.2 23 14 24 14.9 25 14.7 26 14.5 27 14.6 28 14.8 BOTTLING COMPANY CASE ANALYSIS 6 29 14.8 30 14.6 Mean 14.87 =AVERAGE(C2:C31) sd 0.5503 =_xlfn.STDEV.S(C2:C3 3 1) Median 14.8 =MEDIAN(C2:C31)
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