The Study Of Efficiency And Management In A Manufacturing Company

Submitted By holderle4
Words: 1098
Pages: 5

Executive Overview, Part I
As a manufacturing company, a big part of keeping up with other companies in the field is efficiency. To improve efficiency, this company in particular is looking at their staffing and researching the amount of time needed for tasks, as to not over allocate time that could be used elsewhere. To do this, they are taking a random sample of 250 times that represent the time it took to complete a task at two stages, A and B. For Times A, we can conclude that the data is in a Continuous Uniform model; this means that the intervals have an approximately equal width between the minimum (10.030) and maximum (21.980) possible value. This shows us that each of these values is equally likely in the first stage, Time A. The histogram labeled “Histogram of Times A/Times A” in Appendix A shows a pretty symmetric shape of the data. There is no real skewness as our one-variable summary for Times A shows very slight skewness to left of -0.0962, which is very small. The “Q-Q Normal Plot of Times A/Times A” in Appendix A also supports the histogram for Times A, as there is more of an S-shape to the data, instead of following the line for the normal distribution. This means there is more variability for the first stage (Times A) than for stage 2 (Times B) as Times B is normal.
For Times B, we can conclude that the data has a normal distribution. As shown in the histogram labeled “Histogram of Times B/Times B” in Appendix A the data is approximately symmetric in shape, there is no apparent skewness (as in our one-variable summary for Times B the skewness=-0.0480 which is a very minimal skewness to the right) The “Q-Q Normal Plot of Times B/Times B” in Appendix A confirms the same findings as the histogram. In the Q-Q for Times B, the plots are all on the line for the normal distribution, which also confirms that Times B is normal.
Based on our analysis, the maximum amount of time that should be allowed to complete each task 95% of the time for Time A, as it is not a normal distribution but continuous uniform, is 21.39 minutes. We used the equation f(x)=1/(b-a) and plugged in the maximum point of Times A as “b” and the minimum point of Times A as “a”. To be sure 95% of the time, we need to set .95=.0847(x) and solve for X, which gives us 11.21 which is ___?__. We then add this to the minimum value of 10.030 which gives us a total of 21.39 minutes for the maximum amount of time that should be allowed to complete each task 95% of the time for Times A.
For Times B, as it is a normal distribution, we used the equation of X-bar=Z*(St.Dev of X-bar) = 1.645(3.536)=5.8167. We got the Z* from the 95% confidence or at an alpha of 0.05 and we got the Standard Deviation for X-bar from our one-variable summary for Times B. The result 5.8167 is __?__. To find the maximum of time and be sure for 95% of the time, we need to add the 5.8167 to x-bar. Doing this, we came up with 16.031+5.8167=21.8297. This means that 21.83 minutes is the maximum amount of time that should be allowed to complete each task 95% of the time for Times B.

Executive Overview, Part II
Years ago, there was a law passed that drivers had to have insurance to be on the roads. This law was not well accepted, as there were many drivers who felt discriminated against because they could not afford the insurance. This law was challenged on these exact grounds, and in supporting this, the challengers took a random sample of 249 license plates to investigate whether or not they were insured. Each plate would go into one of three categories: 1 (insured), 2 (not insured), or 3 (missing). It was necessary to estimate with 95% confidence the proportion of passenger vehicles in the entire state that were not insured. As there are a few data points that were missing the information, we had to calculate the Interval once with the