Question

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?

Answer 1

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
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