A quality analyst wants to construct a sample mean chart for controlling a packa

Question

A quality analyst wants to construct a sample mean chart for controlling a packaging process. He knows from past experience that the process standard deviation is two ounces. Each day last week, he randomly selected four packages and weighed each. The data from that activity appear below.

A. Calculate upper 2-sigma x-bar chart control limit that allow for natural variations

B. Calculate lower 2-sigma x-bar chart control limit that allow for natural variations

C. Based on the x-bar chart, is this process in control?

Process is in control

Process is out of control

Process is almost within control limits

We don't have enough information

Weight Day Package 1 Package 2 Package 3 Package 4 Monday 23 22 23 24 Tuesday 23 21 19 21 Wednesday 20 19 20 21 Thursday 18 19 20 19 Friday 18 20 22 20

Explanation / Answer

Mean of everyday samples is calculated below:

Monday = (23+22+23+24)/4 = 23

Tuesday = (23+21+19+21)/4 = 21

Wednesday = ( 20+19+20+21)/4 = 20

Thursday = (18+19+20+19)/4 = 19

Friday = (18+20+22+20)/4 = 20

Now, the mean of all sample means:

= (23 + 21 + 20 + 19 + 20) / 5 = 20.6

Standard deviation is given as 2

A) Upper 2-sigma x-bar chart control limit = Mean of Means + 2 x Standard deviation

= 20.6 + 2 x 2

= 24.6

B) Lower 2-sigma x-bar chart control limit = Mean of Means - 2 x Standard deviation

= 20.6 - 2 x 2

= 16.6

C)

Since all the mean values calculated above are within the UCL limits, the process is under control. So, option A in part C of the problem is the correct option.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.