Question

We have m = 5 preliminary samples of size n = 3 (some numbers have unfortunately been erased by accident by a clumsy co-op student): i 113 ri 1.9 Ti, 1 T.2 1 27.1 29.4 2 30.6 32.5 3 25.7 35.5 4 31.1 23.2 5 24.1 34.2 Si 1.3 1.07 4.91 32.4 30 25 27.4 total: 27.9 31.83 30.4 26.43 28.57 145.13 7.9 10.1 32 5.15 16.57 What is the control chart (give the interval) for X from R?

Answer 1

Answer:-

Given That:-

We have m = 5 preliminary samples of size n = 3 (some numbers have unfortunately been erased by accident by a clumsy co-op student):

Given,

i	$x_{i1}$	$x_{i2}$	$x_{i3}$	C	$r_i$	$s_i$
1	27.1	29.4	27.2	27.9	2.2	1.3
2	30.6	32.5	32.4	31.83	1.9	1.07
3	25.7	35.5	30	30.4	5.5	4.91
4	31.1	23.2	25	26.43	7.9	4.14
5	24.1	34.2	27.4	28.57	10.1	5.15
			Total	145.13	32	16.57

$x_{13}$

$\bar{x}_1=\frac{\Sigma x_i}{3}$

= 27.9 * 3 = 27.1 + 29.4 + $x_{13}$

= 83.7 = 56.5 + $x_{13}$

$x_{13}$ = 83.7 - 56.5

$x_{13}$= 27.2

$r_1=29.4-27.2=max-min=2.2$

$r_3=35.5-30=5.5$

$s_4=16.57-s_1+s_2+s_3+s_5$

= 16.57 - 12.43

= 4.14

standard deviation chart formula

X

$CL=\bar{s}=\frac{\Sigma s_1}{k}$

= 16.57/5

= 3.314

$UCL = B_4\bar{s}$ where $B_4$ = constant = 2.089

= 2.089 * 3.314

= 6.922946

$LCL=B_3\bar{s}$ where $B_3=0$

LCL = 0

For   chart

X

UCL = $\bar{\bar{X}}+A_3\bar{s}$ where $A_3$ is constant

$\bar{\bar{x}}=\frac{\Sigma \bar{x_i}}{5}$

= 145.13/5

= 29.026

and  $A_3=1.427$

UCL = 29.026 + 1.427 * 3.314

= 29.026 + 4.729078

= 33.755078

LCL =  $\bar{\bar{x}}-A_3 \bar{s}$

= 29.026 - 4.729078

= 24.296922

Control chart for
X
from  $\bar{s}$ is (24.296922, 33.755078)

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