Question

In: Operations Management

Question 1: A manager wants to investigate a bottling process by using a sample mean chart....

Question 1: A manager wants to investigate a bottling process by using a sample mean chart. He knows from his experience that the process standard deviation is 4 mL (milliliter). Each day last week, he randomly selected 9 bottles and measured each. The data (in mL) from that activity appear below.

Weight

Day

Bottle 1

Bottle 2

Bottle 3

Bottle 4

Bottle 5

Bottle 6

Bottle 7

Bottle 8

Bottle 9

Monday

323

322

323

324

323

322

323

324

322

Tuesday

323

321

319

321

323

321

319

321

323

Wednesday

320

319

320

321

320

319

320

321

320

Thursday

318

319

320

319

318

319

320

319

328

Friday

318

320

322

320

318

320

322

320

320

  1. Calculate all the sample means and the mean of all the sample means.
  2. Calculate upper and lower x-bar chart control limits that allow for natural variations with a z value of 3.

c. Based on the x-bar chart, is this process in control? Create a X-bar chart in Excel

Use the data of question 1 to (a) create Range chart in Excel. (b) Based on the R chart, is this process in control?

Solutions

Expert Solution

The information given in the question is as under:

Bottles

Days

1

2

3

4

5

6

7

8

9

Monday

323

322

323

324

323

322

323

324

322

Tuesday

323

321

319

321

323

321

319

321

323

Wednesday

320

319

320

321

320

319

320

321

320

Thursday

318

319

320

319

318

319

320

319

328

Friday

318

320

322

320

318

320

322

320

320

Answer 1. (a):

Calculation of all the sample means (X Bar) and the mean of all the sample means (X double bar).

Sample mean (X Bar) for each days 9 sample = Average of all 9 sample on a particular day.

Like,

Sample mean (X Bar) for Monday = (323+322+323+324+323+322+323+324+322)/9 = 322.889

Mean of all the sample means (X double bar) = Average of means of all 5 days = (322.889+321.222+320.000+320.000+320.000)/5 = 320.822

Bottles

Days

1

2

3

4

5

6

7

8

9

Mean (X bar)

Monday

323

322

323

324

323

322

323

324

322

322.889

Tuesday

323

321

319

321

323

321

319

321

323

321.222

Wednesday

320

319

320

321

320

319

320

321

320

320.000

Thursday

318

319

320

319

318

319

320

319

328

320.000

Friday

318

320

322

320

318

320

322

320

320

320.000

X double bar

320.822

Answer 1. (b):

For the sample size of 9, the Control chart constant A2 is as under:

Sample size

9

A2

0.337

For the calculation of UCL and LCL of X bar chart, value of R bar needs to be calculated.

Range for a particular day = Max of value – Min of value of the particular day

Like,

Range for Monday = (Max of Monday)-(Min of Monday) = 324-322 = 2.000

R bar = Average of all Range from Monday through Friday = (2+4+2+10+4)/5 = 4.400

Bottles

Days

1

2

3

4

5

6

7

8

9

Range (R )

Monday

323

322

323

324

323

322

323

324

322

2.000

Tuesday

323

321

319

321

323

321

319

321

323

4.000

Wednesday

320

319

320

321

320

319

320

321

320

2.000

Thursday

318

319

320

319

318

319

320

319

328

10.000

Friday

318

320

322

320

318

320

322

320

320

4.000

R bar

4.400

Upper x-bar chart control limits = UCLx bar = X double bar + A2*R bar = 320.822+(0.337*4.400) = 322.305

Lower x-bar chart control limits = LCLx bar = X double bar - A2*R bar = 320.822-(0.337*4.400) = 319.339

Answer 1. (c):

X bar chart:

Populate the values in the table:

Sample

1

2

3

4

5

Average (X bar)

322.889

321.222

320.000

320.000

320.000

UCL

322.305

322.305

322.305

322.305

322.305

LCL

319.339

319.339

319.339

319.339

319.339

CL

320.822

320.822

320.822

320.822

320.822

Create the X bar chart in Excel by using above data and selecting “Line Chart”:

Based on the X bar chart, the process in not in process control as the X bar value is observed beyond UCL.

R chart:

Populate the values in the table:

Sample

1

2

3

4

5

Range (R )

2.000

4.000

2.000

10.000

4.000

UCL

7.990

7.990

7.990

7.990

7.990

LCL

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