Question

Sample

1	2	3	4
12.3	11.9	12.0	12.1
12.2	12.2	12.2	11.8
12.1	12.2	11.8	11.8

Calculate UCL and LCL for a mean chart. Assume the standard deviation is unknown. Round to 1 decimal.

a.

UCL=11.7, LCL=12.4

b.

UCL=11.7, LCL=0

c.

UCL=12.4, LCL=0

d.

UCL=12.4, LCL=11.7

e.

cannot be determined

Using the data provided in Question 38 above, how many up/down runs are present?

a.

0

b.

1

c.

2

d.

3

e.

4

Using the data provided in Question 38 above, calculate the UCL and LCL for a range chart. Assume the standard deviation is unknown.

a.

UCL=0, LCL = 1.2

b.

UCL=0.8, LCL=0

c.

UCL=1.2, LCL=0

d.

UCL=1.5, LCL=0

e.

none of the choices

Using the data provided in Question 38 above, if the median is 12.05 how many above/below runs are present?

	a.	0
	b.	1
	c.	2
	d.	3

Answer 1

= 12.05

= 0.3

Mean

UCL = + A₂*

LCL = - A₂*

For n=3 A₂ = 1.023

UCL = 12.05+1.023*0.3 = 12.4

LCL =   12.05+1.023*0.3 = 11.7

d)UCL=12.4, LCL=11.7

there are 0 up/down runs are present

a) 0

Range:

UCL = D₄ = 2.574*0.3 = 0.8

LCL = D₃ = 0*0.3 = 0

b) UCL=0.8, LCL=0