Question

A food supplement manufacturer just released a whey protein supplement. The product is expected to have 24 grams of protein per serving. The population standard deviation is 1.3553 During a period of 10 days, The workers collected half serving of the product and measured the amount of protein 5 day during a period of 10 days. The results are shown in Table 1 Table 1: Grams of protein per half serving of product 1 2 3 45 67 8 9 10 12 10 9 10 10 10 11 10 8 14 10 12 10 11 10 10 10 8 12 9 9 11 10 10 11 12 12 10 12 6 12 10 9 10 9 9 98 9 12 1.1 Plot the histogram 1.2 Range for the samples, R 1.3 Average of the ranges, R .4 Average for all observations, .5 Average for samples, f 6 Average of the averages of the samples, i .7 Determine the control charts limits .8 What is the probability of the process be in control? .9 Plot the control chart .10 Is the process under control or not? Why (name the intervals)? 11 Determine whether the process was capable of conforming to the required lim- 10 5 5 5 5 5 10 10 20 20 its.

Answer 1

Raw Data
Sample number	Item 1	Item 2	Item 3	Item 4	Item 5	Sample Average (mu)	Sample Range (R)
Sample 1	10	12	10	9	12	10.6	3
Sample 2	10	10	12	11	10	10.6	2
Sample 3	12	9	10	10	9	10	3
Sample 4	10	10	11	10	10	10.2	1
Sample 5	11	10	10	11	9	10.2	2
Sample 6	11	10	10	12	9	10.4	3
Sample 7	10	11	10	12	9	10.4	3
Sample 8	13	10	8	10	8	9.8	5
Sample 9	8	8	12	12	9	9.8	4
Sample 10	10	14	9	6	12	10.2	8
					Total	1022	34
					Overall Average	10.22	3.4

1.2:

Range of sample = maximum value of sample – minimum value of sample

1.3.

Average of ranges = mean of ranges of 10 samples = 34/10 = 3.4

1.4

Average for all observation = total of observations / (no. of sample x observation per sample) = 511 / (10 x 4) = 10.22

μ = 10.22

1.5

Average for samples = mean of four observation for a sample

1.6

Average of averages of samples = overall average = mean of sample averages = 10.22

x = 10.22

1.7

Chart Constants: for sample size n = 5, A₂ = 0.577, D₃ = 0, D₄ = 2.115

X-bar Chart Central Limit (CL) Upper control Limit (UCL) R-chart CL = R = 3.4 UCL = D 4R CL = X = 10.22 UCL= X + A,R -10.22(0.577(3.4) =12.18 (2.115)3.4) 7.191 Lower Control Limit (LCL) -L-X-AR LCL = D 10.22-(0.577)(34) = 8.258

X-chart

	X-chart	R-chart
Upper control limit	12.1818	7.191
Center line	10.22	3.4
Lower control limit	8.2582	0

1.9

The charts are as follows:

x-bar chart 14 12 Data LCL -CL UCL 10 range chart ◆ Range LCL UCL 10 Sample

1.10

All the sample means are within the limits, but the range value of sample 10 is beyond upper limit, thus the process is not under statistical control.