Difference Between Mean Volume Of Machines In Study Assessment Answer

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Question :

Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The filling processes can be assumed to be normal, with standard deviations of al = 0.015 and a2 = 0.018. The quality engineering department suspects that both machines fill to the same net volume, whether or not this volume is 16.0 ounces. An experiment is per-formed by taking a random sample from the output of each machine.

c) See the "Ch2 Prob26" worksheet in the "Module 3 Homework Assignment Data" for this data. Copy the data into JMP. It is best to put all data in one column, and put a label for machine (M1 or M2) in a second column. The label should be text versus a continuous number.

d) Plot the data in a way that will help determine if there is a difference in the means. Include at least one plot in the homework that you turn in. (Refer to the note on the next page as a reminder of how to do this.) What does this plot indicate to you about the averages? Why? Note: To copy and paste graphs from JMP, produce the graph, then look for the symbol at the top of the graph that looks like a cross or plus symbol (+). Click on this, and then select the graph you wish to copy and paste. The graph should now be highlighted. Click on copy, and then you can paste it into Word, PowerPoint, or other documents. This usually works best with "paste special", using either pdf or "picture". These enable you to properly size the graph in Word.

e) Suppose the data produced are totally inconsistent with your null hypothesis in Question a. What would be the conclusion of a hypothesis test? How would you explain your conclusion to an engineer who has no background in statistics? That is, how would you explain it in plain English, without using any statistical terms?

f) Repeat question e, but assume that the data are fairly consistent with the null hypothesis in Question a - approximately what you would have expected. What would you conclude, and how would you explain this, again using no statistical terms?

g) Now perform at two-sample t test comparing the population means in JMP. Use the default option in JMP, which does NOT assume equal variances. Then answer the following questions: • What is the p value, assuming we use a two-sided alternative? • What is the p value if we use the alternative: Hi: pi > p2? • Explain why the answers to the above two questions are different • Assuming we use a two-sided alternative, what can we conclude about the null hypothesis? Explain this not only in statistical terms but also in plain English.

h) Assume that we have been told that the Commerce Department is auditing the plant tomorrow, to verify that our 16oz soft drinks are actually 16oz on average. By federal law, if the population average volume for a soft drink labeled 16oz were less than 16oz, this would constitute criminal fraud. Assume further that the Commerce Department is not concerned about machine 1 versus machine 2; they only care about the population of soft drinks coming out of the plant. What would be appropriate hypotheses for the Commerce Department to use when statistically evaluating the plant, to determine if there is evidence of fraud or not? Explain your answer

Research question: Is there difference between mean volume of machines in study?

Hypotheses:

H0: μM1 = μM2

HA: μM1 ≠ μM2

c) I created both files. First MachineIS as required by task. The “Volume” column contains values (continuous). The “Machine” column contains text data about machine – M1 and M2.

d) Distribution for Volume

Next chart helps to see the difference between means.

This is a bar plot of means. From first view, this plot indicates that there is not difference between machines. I did this plot Graph->Graph builder. Then I selected bar chart. Drug and drop volume on target zone, and the machine variable on X scale. Then I pressed “done” to see the chart in chart window.

To see exact statistics I made descriptive stats using Analyze->Distribution.  “Y column” = Volume, “By” = Machine. Then I pressed “Ok” to see the output.

Despite a fact that plot does not indicate the visual difference we can see in descriptive stats that actual data mean difference by groups is very small. As task in the beginning tells that standard deviations are small too, I prefer to build better chart. I add customization to current chart by the customizing of axis Y and an adding a confidence interval for mean.

It can be seen that M1 has higher mean value compare to M2. But the 95% confidence intervals for means are intersected. I think that we may not expect the difference between M1 and M2, and the machines are equal.

e) The conclusion of hypothesis test will be that we have to reject H0 and accept Ha with 95% probability if I’ll use α=0.05. This means that mean values differ by machines (One more potential outcome is that sample size is not adequate - if observations count is less than 3 in each group). In addition to this I have to confirm the equality of variances (assumption of t-test), or use a correction if variances are not equal. In reporting of differences I have to show the value of difference, and I have to confirm the statistical significance of difference. If t value is higher than tcritical for 95% probability then there is difference, if it is not significant p≥0.05 then there is not difference between volumes. In this case we have to expect that t>tcritical  and the difference exists. I can post the p-value of t-test also. If p<α=0.05 then H0 is not consistent.

f) Similar to the e the data is consistent; we must have more than 3 answers in each group (assumption of t-test). Then we have to confirm the equality of variances or use correction for t-values. The equality of variances can be confirmed by Levene’s test. If test result is not significant at α=0.05 than “equal variances assumed”. If it is significant then “equal variances not assumed”. With consistent data we have to expect non-significant pvalue≥α=0.05.  This means that with 95% probability we can guarantee the equality of means.

g) Analyze -> fit Y by X

I put Volume in Y and X in Factor. And then “Ok” button pressed.

• The probability in a two-sided alternative: p=0.435 (H0 confirmed)
• If we use Ha the probability should be less than 0.05 (Should be H0 rejected, Ha confirmed)
• Because the actual machines are equal. The answers on two questions above are different because they state different hypotheses. The actual data for machines are equal but the second question asks about hypothesis of difference confirmation.
• We have to confirm the H0 and reject Ha. The machines mean volumes are equal with 95 confidence. The mean difference between machines = -0.01 is not significant because t=-0.79894<tcritical. p=0.435>0.05=α. The upper and Lower CL confirms that difference is not significant for mean difference between machines: “-0.3635” – “0.01252”

h) As we plan to confirm 16oz. We can run one sample t=test to confirm the mean value of volume. This should be done for the whole volume (both machines). I can recommend also the 99% or 99.9% confidence interval to guarantee that there is no evidence of fraud.  This will guarantee the 99% or 99.9% for this particular sample.

In addition to this I can recommend to increase the sample size. This will help to reduce a sample error of investigation.

As we see the t-test may confirm the equality of bottles to 16oz for particular sample. I think that federal law should give the recommendations about precision and allowed gap. Then we have to design correct sample to confirm that volume is in particular gap and does not exceeds the bounds with 99.9% probability.

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