Question

A local mattress manufacturer wants to know if its manufacturing process is in or out of control and has hired you, a statistics expert in the field, to analyze its process. Specifically, the business has run 20 random samples of size 5 over the past month and has determined the mean of each sample.

Sample Mean of Sample
1 95.72
2 95.44
3 95.4
4 95.5
5 95.56
6 95.72
7 95.6
8 95.24
9 95.46
10 95.44
11 95.8
12 95.2
13 94.82
14 95.78
15 95.18
16 95.32
17 95.08
18 95.22
19 95.04
20 95.48


Required:
a. Determine the estimate of the mean when the process is in control.
b. Assuming the process standard deviation is 0.50 and the mean of the process is the estimate calculated in part a , determine the Upper Control Limit (UCL) and the Lower Control Limit (LCL) for the manufacturing process.
c. Explain the results to the vice-president of the mattress manufacturer focusing on whether, based on the results, the process is in or out of control.

Answer 1

a. 95.4

b. UCL = 96.07

LCL = 94.73

c. Process is in control

Explanation:

a. The computation of estimate mean is shown below:-

`Estimate\ mean = (\Sigma X)/(N) \\\\ = (1908.000)/(20)`

= 95.4

b. The computation of Upper Control Limit (UCL) and the Lower Control Limit (LCL) for the manufacturing process is shown below:-

`UCL = Mean + (3* \sigma )/(√(Sample\ size) )`

`UCL = 95.4 + (3* (0.5))/(√(0.5) )`

= 95.4 + 0.67082

= 96.07

`UCL = Mean - (3* \sigma )/(√(Sample\ size) )`

`UCL = 95.4 - (3* (0.5))/(√(0.5) )`

= 95.4 - 0.67082

= 94.73

c. The explanation is shown below:-

From the above calculation we can see that the sample lies between LCL AND UCL that is (94.73 ,96.07) ,

The Process is in control