4. Sample Size

The ability of control charts to detect shifts in process quality is described by their operating characteristic curves (OHC), and from the curve it is possible to determine sample sizes.

Example 1:

Configure as shown in the figure below to Use the Sample size tool.

Then click Calculate we obtain the results. You can also generate the analyses and download them in Word format.

The results are:

OC curves for x-bar graph

	0.5	1	1.5	2	2.5	3	3.5	4
n = 2	0.989	0.944	0.81	0.568	0.296	0.107	0.026	0.004
n = 3	0.984	0.898	0.656	0.321	0.092	0.014	0.001	0
n = 4	0.977	0.841	0.5	0.159	0.023	0.001	0	0
n = 5	0.97	0.778	0.362	0.07	0.005	0	0	0
n = 6	0.962	0.709	0.25	0.029	0.001	0	0	0
n = 7	0.953	0.638	0.166	0.011	0	0	0	0
n = 8	0.944	0.568	0.107	0.004	0	0	0	0
n = 9	0.933	0.5	0.067	0.001	0	0	0	0
n =10	0.922	0.436	0.041	0	0	0	0	0

OC curves for standard deviation chart

	0.5	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6
n = 2	1	0.991	0.918	0.807	0.703	0.615	0.544	0.485	0.438	0.398	0.364	0.336
n = 3	1	0.994	0.9	0.726	0.563	0.438	0.345	0.277	0.226	0.187	0.157	0.134
n = 4	1	0.996	0.879	0.648	0.447	0.307	0.215	0.155	0.114	0.086	0.067	0.052
n = 5	1	0.996	0.856	0.574	0.35	0.212	0.132	0.085	0.056	0.039	0.027	0.02
n = 6	1	0.996	0.833	0.506	0.271	0.144	0.079	0.046	0.027	0.017	0.011	0.007
n = 7	0.999	0.997	0.808	0.442	0.208	0.097	0.047	0.024	0.013	0.007	0.004	0.003
n = 8	0.996	0.997	0.784	0.385	0.158	0.065	0.028	0.013	0.006	0.003	0.002	0.001
n = 9	0.988	0.997	0.759	0.333	0.119	0.043	0.016	0.007	0.003	0.001	0.001	0
n =10	0.973	0.997	0.734	0.287	0.089	0.028	0.009	0.003	0.001	0.001	0	0

Average run length

	0.5	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6
n = 2	90.646	17.731	5.269	2.315	1.421	1.12	1.026	1.004	1	1	1	1
n = 3	60.688	9.765	2.908	1.473	1.101	1.014	1.001	1	1	1	1	1
n = 4	43.895	6.303	2	1.189	1.023	1.001	1	1	1	1	1	1
n = 5	33.401	4.495	1.566	1.076	1.005	1	1	1	1	1	1	1
n = 6	26.358	3.437	1.333	1.03	1.001	1	1	1	1	1	1	1
n = 7	21.383	2.766	1.2	1.011	1	1	1	1	1	1	1	1
n = 8	17.731	2.315	1.12	1.004	1	1	1	1	1	1	1	1
n = 9	14.968	2	1.072	1.001	1	1	1	1	1	1	1	1
n =10	12.825	1.772	1.042	1	1	1	1	1	1	1	1	1

Expected individual units

	0.5	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6
n = 2	181.292	35.462	10.538	4.631	2.841	2.24	2.053	2.008	2.001	2	2	2
n = 3	182.064	29.294	8.724	4.42	3.303	3.043	3.003	3	3	3	3	3
n = 4	175.579	25.212	8	4.754	4.093	4.005	4	4	4	4	4	4
n = 5	167.004	22.477	7.832	5.379	5.024	5.001	5	5	5	5	5	5
n = 6	158.145	20.62	8.001	6.178	6.005	6	6	6	6	6	6	6
n = 7	149.679	19.36	8.397	7.078	7.001	7	7	7	7	7	7	7
n = 8	141.847	18.523	8.959	8.032	8	8	8	8	8	8	8	8
n = 9	134.709	18	9.644	9.012	9	9	9	9	9	9	9	9
n =10	128.251	17.716	10.424	10.004	10	10	10	10	10	10	10	10

Example 2: