Question

Lee Flower Company is a flower producer located in Auburn. It has a large contract with Wal-Mart to supply azaleas to the stores in the southeast region of the country. From past sales data, it has become evident that the larger, healthier plants sell better than the smaller plants. Thus, the flower producer wants to improve the growing process so that they will have larger, healthier azaleas. Wal-mart is very tough on their suppliers to provide healthy azaleas because they don’t want a lot of small, weak plants in inventory after the planting season has passed. Thus, Wal-Mart’s expectations are that the plants must have a composite score of 9 +/- 0.2 inches. They have an upper limit on the size because they want to get as many flowers on the shelves as possible. So the target score is 9 inches. The green thumbs went out and measured 3 plants at random over 33 batches of plants. This data is shared below. Each batch, or subgroup of 3 represents a short-term sample that might have been taken within a day, and they would have done this for, say, 33 days.   

Comment on the Sigma level (both long and short term). Also, how many, out of a million, would you expect to be below the Wal-mart spec and how many out of a million would you expect to see above the Wal-Mart spec. What recommendations would you make for improvement?

Data:-

Batch	Observation 1	Observation 2	Observation 3
1	8.8	8.9	8.9
2	8.8	8.8	8.8
3	9.1	8.8	8.9
4	9.0	8.9	9.1
5	8.9	8.9	8.8
6	8.8	8.8	8.9
7	8.8	8.8	8.9
8	8.9	9.0	9.0
9	8.8	8.8	8.9
10	8.9	9.0	8.9
11	8.9	8.8	8.9
12	8.6	8.9	9.1
13	8.8	8.8	9.0
14	8.9	9.0	9.0
15	8.9	9.0	8.9
16	8.8	8.7	8.7
17	9.1	8.9	9.1
18	9.1	9.0	9.1
19	8.9	8.8	8.9
20	8.8	8.7	8.7
21	8.9	9.1	9.0
22	9.1	8.9	9.0
23	9.1	9.0	9.1
24	8.7	8.8	8.9
25	8.8	8.9	8.8
26	9.1	8.8	9.1
27	9.1	9.1	9.0
28	8.9	8.9	8.8
29	8.8	8.9	8.8
30	8.9	9.0	9.0
31	8.7	8.6	8.7
32	9.0	9.0	8.8
33	9.1	9.1	9.0

Answer 1

Solution

We will find answers for the questions applying the xbar-R control chart concepts

Back-up Theory

Let X = Plant score (inches).

Specifically, let xij = the plant score for the jth observation (plant score) in the ith sub-group; j = 1, 2, 3 and i = 1 to 33

Define: xbari = (1/3)∑[j = 1, 2, 3]xij; xdoublebar = (1/33)∑[i = 1 to 33]xbari

Ri = Maxxij – Minxij and Rbar = (1/33)∑[i = 1 to 33]Ri   

Xdoublebar estimates the mean plant score and Rbar/d2 estimates the sigma, where d2 = 1.693 is a constant obtained from Control Chart Constants Table.

Number of plants below and above specification will be estimated employing Normal Distribution Theory as explained below briefly:

If a random variable X ~ N(µ, σ²), i.e., X has Normal Distribution with mean µ and variance σ², then,

Z = (X - µ)/σ ~ N(0, 1), i.e., Standard Normal Distribution ……………..………………..(1)

P(X ≤ or ≥ t) = P[{(X - µ)/σ} ≤ or ≥ {(t - µ)/σ}] = P[Z ≤ or ≥ {(t - µ)/σ}] .………………(2)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or

can be found using Excel Function NORMDIST(x,Mean,Standard_dev,Cumulative) …..(53

Now, to work out the answer,

Preparatory Work

i	xibar	Ri
1	8.867	0.1
2	8.8	0
3	8.933	0.3
4	9	0.2
5	8.867	0.1
6	8.833	0.1
7	8.833	0.1
8	8.967	0.1
9	8.833	0.1
10	8.933	0.1
11	8.867	0.1
12	8.867	0.5
13	8.867	0.2
14	8.967	0.1
15	8.933	0.1
16	8.733	0.1
17	9.003	0.2
18	9.067	0.1
19	8.867	0.1
20	8.733	0.1
21	9	0.2
22	9	0.2
23	9.067	0.1
24	8.8	0.2
25	8.833	0.1
26	9	0.3
27	9.067	0.1
28	8.867	0.1
29	8.833	0.1
30	8.867	0.1
31	8.667	0.1
32	8.933	0.2
33	9.067	0.1

k	33
n	3
Xdoublrbar	8.902152
Rbar	0.142424
d2	1.693
sigma	0.084125

Part (a)

Sigma = 0.084125 [vide last Table above] ANSWER

Part (b)

Expected number of plants to be below the Wal-mart spec (out of a million)

= 10⁶ x Probability of plants being below the Wal-mart spec

= 10⁶ x P(X < 8.8)

= 10⁶ x P[Z < {(8.8 – 8.902152)/0.084125} [vide (2)]

= 10⁶ x P(Z < - 1.2142)

= 10⁶ x 0.1123 [vide (3)]

= 112300 ANSWER

Part (c)

Expected number of plants to be above the Wal-mart spec (out of a million)

= 10⁶ x Probability of plants being above the Wal-mart spec

= 10⁶ x P(X > 9.2)

= 10⁶ x P[Z > {(9.2 – 8.902152)/0.084125} [vide (2)]

= 10⁶ x P(Z > 3.5408)

= 10⁶ x 0.0002 [vide (3)]

= 200 ANSWER

Part (d)

Out of spec number in total is very high basically because of lower plant scores. That happens because the average score well below the central value 9.0 of the spec. So, Wal-mart must go in for a larger size of the plant. ANSWER