Question

A machine is used to fill containers with a liquid product. It can be assumed that the filling volume has a normal distribution. A random sample of 32 containers is selected and the net contents are those shown below. The manufacturer wants to make sure that the average net content exceeds 12 oz and does not exceed 12.5 oz., What is the process capability to meet these specifications? Explain your results. 12.09 12.35 12.06 11.61 12.53 11.85 11.65 12.07 12.29 12.18 12.15 12.07 12.18 12.07 12.35 12.19 11.76 11.80 12.00 12.54 12.09 12.17 11.81 12.11 12.44 11.85 12.40 11.98 12.08 12.28 12.10 11.76

Answer 1

Let X be the amount of liquid in container.

n = 32 = number of containers to be selected .

Assume that X follows normal distribution.

: average net contents.

From the information

the product meet specfication when the average contents is greater than 12.0 oz but not exceed 12.5 oz.

The average net contents lies between 12 and 12.5 i.e. 12 < xbar < 12.5.

Upper specification limit = 12.5

Lower specification limit = 12.00

The process capability to meet these specificatiions is given by

The sampling distribution of mean is

  and estimate of population variance is S² .

The standard error of sample mean is S/sqrt(n)

X	(X-Xbar)^2
12.09	3.9E-07
12.35	6.8E-02
12.06	8.6E-04
11.61	2.3E-01
12.53	1.9E-01
11.85	5.7E-02
11.65	1.9E-01
12.07	3.8E-04
12.29	4.0E-02
12.18	8.2E-03
12.15	3.7E-03
12.07	3.8E-04
12.18	8.2E-03
12.07	3.8E-04
12.35	6.8E-02
12.19	1.0E-02
11.76	1.1E-01
11.8	8.4E-02
12	8.0E-03
12.54	2.0E-01
12.09	3.9E-07
12.17	6.5E-03
11.81	7.8E-02
12.11	4.3E-04
12.44	1.2E-01
11.85	5.7E-02
12.4	9.6E-02
11.98	1.2E-02
12.08	8.8E-05
12.28	3.6E-02
12.1	1.1E-04
11.76	1.1E-01
386.86	1.8E+00

S =sqrt(0.0581) = 0.2410

standard error of sample mean = s/sqrt(n) = 0.2410/sqrt(32) = 0.0426

Hence process capability is given by

C_p = 1.9560

Since value of Cp > 1.33. The process is highly capabale.