Question

A process sampled 20 times with a sample of size 8 resulted in

= 26.5 and R = 1.6. Compute the upper and lower control limits for the x chart for this process. (Round your answers to two decimal places.)

UCL = ____

LCL = ____

Compute the upper and lower control limits for the R chart for this process. (Round your answers to two decimal places.)

UCL =____

LCL = ____

Answer 1

Answer:

Given that,

$\rightarrow$ A process sampled 20 times with a sample of size 8 resulted in   = 26.5 and R = 1.6.

(a).

Compute the upper and lower control limits for the x chart for this process:

$\rightarrow$ From the given information,

i.e,

n=8

For x chart:

$\rightarrow$ The upper limit (UCL $\bar{x}$ ):

= 1 + A2R

Where A2 constant is a function of the sample size n.

$A_2=\frac{3}{2.847\times \sqrt{n}}$

For n=8,

$A_2=\frac{3}{2.847\times \sqrt{8}}$

=3/2.847(2.828)

=3/8.05132

=0.372609

A2=0.373(Approximately)

Therefore,

Then

= 1 + A2R

=26.5+0.373(1.6)

=26.5+0.5968

=27.0968

=27.09 (Approximately)

$\rightarrow$ The lower limit (LCL $\bar{x}$ ):

= - AUR

=26.5-0.373(1.6)

=26.5-0.5968

=25.9032

=25.90(Approximately)

Therefore,

UCL=27.09

LCL=25.90

(b).

Compute the upper and lower control limits for the R chart for this process:

For the R chart:

$\rightarrow$$\text{The UCL}=D_4\bar{R}$

From R-chart D4 is constant is 1.864.

UCL=1.864 $\times$ 1.6

=2.9824

=2.98 (Approximately)

$\rightarrow$$\text{The LCL}=D_3\bar{R}$

From R-chart D3 is constant is 0.136.

LCL=0.136 $\times$ 1.6

=0.2176

=0.22 (Approximately)

Therefore,

UCL=2.98

LCL=0.22