Question

Q1: For the data given below: a) Calculate the 3o control limits for X-bar and R charts based on the first 12 samples reflecting the process before any problems were denounced. b) Plot X-bar and R charts labeling the data points, upper and lower control limits, and center lines on both charts. Sample Screw 1 Screw 2 Screw 3 0.276 0.238 0.261 2 0.249 0.263 0.234 3 0.264 0.255 0.258 4 0.255 0.279 0.269 5 0.262 0.273 0.234 6 0.268 0.267 0.270 7 0.266 0.244 0.270 8 0.232 0.261 0.278 9 0.242 0.277 0.253 10 0.246 0.253 0.236 11 0.279 0.230 0.235 12 0.238 0.243 0.237

Answer 1

Solution:

a)Let find the
X
and $\bar{R}$ values for the given data.

X-bar is the average of three samples.

Range is the difference between the maximum and minimum value.

is the average of X-bar values(12 X-bar values)

X

$\bar{R}$ is the average of range.

Sample No	Screw 1	Screw 2	Screw 3	X-bar	Range
1	0.276	0.238	0.261	0.258	0.038
2	0.249	0.263	0.234	0.249	0.029
3	0.264	0.255	0.258	0.259	0.009
4	0.255	0.279	0.269	0.268	0.024
5	0.262	0.273	0.234	0.256	0.039
6	0.268	0.267	0.27	0.268	0.003
7	0.266	0.244	0.27	0.260	0.026
8	0.232	0.261	0.278	0.257	0.046
9	0.242	0.277	0.253	0.257	0.035
10	0.246	0.253	0.236	0.245	0.017
11	0.279	0.23	0.235	0.248	0.049
12	0.238	0.243	0.237	0.239	0.006
				X =0.255	$\bar{R}$=0.027

From the control chart constant table for n=3,we have

D4	2.574
D3	0
A2	1.023

X-bar:UCL=
X
+A2*$\bar{R}$=0.255+(1.023*0.027)=0.283

CL==0.255

X

LCL=
X
- A2*$\bar{R}$=0.255-(1.023*0.027)=0.228

R-bar:UCL=D4*$\bar{R}$=2.574*0.027=0.069

CL=$\bar{R}$=0.027

LCL=D3*$\bar{R}$=0*0.027=0

b)The two charts are as follows:

X-bar chart 0.290 0.280 0.270 0.260 0.250 0.240 0.230 0.220 0.210 0.200 1 3 4 Lл 6 7 00 9 10 11 12 N LCL -CL- UCL-X-bar

Range chart 0.08 0.07 0.06 0.05 0.04 0.03 M 0.02 0.01 0.00 1 N 3 4 5 6 7 8 9 10 11 12 LCL CL UCL Range

From the above two graphs, we can see that the process is in control since all the observations fall within UCL and LCL.