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Assignment Determine the mean proportion defective, the UCL, and the LCL?

Category: Engineering Paper Type: Assignment Writing Reference: APA Words: 850

Sample

The proportion of defective item

Total Defective Proportions

1

0.01

0.21

2

0.03

3

0

4

0.04

5

0.01

6

0.01

7

0

8

0.01

9

0.02

10

0.02

11

0.03

12

0.03


Particulars

Formulas

Calculations

Sample Taken

99

Mean of Proportionate defective

m

Standard Deviation

Sigma

Mean of Proportionate defective

Total defective Proportions/number of samples

0.0175

Standard Deviation (Sigma)

STDEV.S function in excel

0.0129

Upper Control Limit

m+3*Sigma

0.056

Lower Control Limit

m-3*upper limit

-0.151


The mean of Proportionate defective is calculated by using the arithmetic mean formula, which is mentioned as follows:

The mean in the above table is denoted with m. In the first step, the sum of all the proportion of defective items is calculated that is 0.21. After calculating total defective proportions the mean of the proportionate defective is evaluated by dividing the total with the number of samples. The mean of proportionate defective is 0.0175.

For evaluating the UCL & LCL first standard deviation is calculated. The formula of standard deviation is:

When the standard deviation is calculated, the UCL and LCL can be calculated. In the above table, it can be seen that the UCL is 0.056 and LCL is -0.151 (Anderson, Sweeney, & Williams, 2010).

a.  Draw a control chart and plot each of the sample measurements on it?

 

                The sample measures, which include upper limit, lower limit and mean of the proportionate defective are done in order to know the statistical control. In the above control chart the proportionate defective is plotted, which can be seen in the chart as blue line. The upper control limit in the chart can be seen with the orange dot above the blue line with a value of 0.056. The lower control limit can be seen with the grey dot below the blue line having a value of -0.151 (Brown & Bessant, 2013).

b.      Does it appear that the process for making tees is in statistical control?

                From the control chart above, it can be seen that the proportion of defective items is within the control limits. In the process of making tees, no control limit has been breached, which is evident from the chart above. Therefore it can be said that the process of making tees is in statistical control. If the blue line which is indicating the proportion of defective items was above or below the UCL & LCL than it can be said that the process is not in statistical control

 (Wegner, 2010).

Question No. 2

a.      Forecast the demand for week 7 using a five-period moving average?

Week

Demand

5 year moving Average

1

649

 

2

524

 

3

561

 

4

738

 

5

511

 

6

600

596.6

7

 

586.8

 

                The moving average is calculated with the formula:

                By using a 5-year moving average the demand for week seven is forecasted. In week 7 it is forecasted that the demand will be 586.8 (Mahadevan, 2009).

b.      Forecast the demand for week 7 using a three-period weighted moving average.

Week

Demand

3 year weighted moving Average

1

649

 

2

524

 

3

561

 

4

738

567.50

5

511

642.10

6

600

589.10

7

 

600.90

The weighted moving average is calculated with the formula, which is mentioned as follows:

     By using 3 years weighted moving average the demand for week seven is forecasted. In week 7 it is forecasted that the demand will be 600.90 (Anderson, Sweeney, & Williams, 2010).

c.       Forecast the demand for week 7 using exponential smoothing. Use an α value of .1 and assume the forecast for week 6 was 602 units?

Week

Demand

3 year moving Average

1

649

 

2

524

649

3

561

637

4

738

629

5

511

640

6

600

602

7

 

601.8

 The demand for week 7 is calculated by using exponential smoothing. The formula of exponential smoothing is mentioned as follows:

                It is forecasted that the demand for week 6 will be 602. Therefore with the exponential smoothing method, the demand in week 7 will be 601.8.

d.      What assumptions are made in each of the above forecasts?

                The main assumption in the moving average model and the exponential smoothing model is the time series, which remains stationary and have slow-changing mean. Therefore the moving average is used for evaluating the current mean which is then utilized for forecasting the future. The exponential smoothing technique can be described as the weighted average approach. This technique is usually used for short term forecasting, and that data which is used in this technique is historical (Anderson, Sweeney, & Williams, 2010).

References of Assignment Determine the mean proportion defective, the UCL, and the LCL?

Anderson, Sweeney, D. J., & Williams, T. A. (2010). Fundamentals of Business Statistics (illustrated ed.). Cengage Learning.

Brown, S., & Bessant, J. (2013). Strategic Operations Management. Routledge.

Mahadevan, B. (2009). Operation Management: Theory and Practice. Pearson Education, India.

Wegner, T. (2010). Applied Business Statistics: Methods and Excel-based Applications (illustrated ed.). Juta and Company Ltd.

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