Question

Problem:

A plant manager is interested in developing a quality-control program for an assembly line that produces light bulbs. To do so, the manager considers the quality of the products that come from the line. The light bulbs are packed in boxes of 12, and the line produces several thousand boxes of bulbs per day. To develop baseline data, workers test all the bulbs in 100 boxes. They obtain the following results:

No. of Defective Bulbs/Box	No. of Boxes
0	68
1	27
2	3
3	2

Run @RISK’s distribution fitting procedure on the preceding data choosing Discrete Sample Data (Counted Format) for the type of data.

a. The Fit Results windows shows that the Poisson is the best fitting theoretical distribution. Is the Poisson a good choice? Why or why not? What is the interpretation of the parameter of the Poisson in this setting?

b. Noticing that there are only two boxes with three defective bulbs, you combine the last two categories in the preceding data. Rerunning @RISK’s fitting procedure, we see that the binomial now fits best according to the Chi-Square measure, with the Poisson coming in a close second. How much has the parameter (m) for the Poisson changed from the case with four categories? Is there a reasonable interpretation of the parameters n and p for the binomial in this setting?

c. Which distribution would you use if you were the plan manager? Why?

Answer 1

a]

Using MINITAB, we test the goodness of fit test for Poisson distribution

Use Goodness-of-Fit Test for Poisson to test the hypotheses:

H0: Data follow a Poisson distribution

H1: Data do not follow a Poisson distribution

Stat > Basic Statistics > Goodness-of-Fit Test for Poisson

In Variable, enter

Click OK.

Session window output

MTB > PGoodness 'No. of defective';
SUBC>   Frequencies 'No. of Boxes';
SUBC>   GBar;
SUBC>   GChiSQ;
SUBC>     Pareto;
SUBC>   RTable.

Goodness-of-Fit Test for Poisson Distribution

Data column: No. of defective
Frequency column: No. of Boxes

Poisson mean for No. of defective = 0.38

No. of                     Poisson              Contribution
defective Observed Probability    Expected     to Chi-Sq
0                68        0.677057 67.7057        0.001279
1                27        0.264052      26.4052        0.013398
>=2             5       0.058891       5.8891     0.134229

N DF    Chi-Sq    P-Value
100     1    0.148906    0.700

Poisson parameter is 0.38.

Here P-value = 0.7 > . Hence we failed to reject null hypothesis

Conclusion: At , Data follows the Poisson distribution. Poisson is good fit for this data.

b]

If we test Goodness of fit for Binomial test

mean = np = 0.37

Variance = np(1 - p) = 0.33 Calculated from given data after merging last two entries

Parameters of Binomial distribution

p = 1 - Variance/mean = 0.1

n = 0.37/0.1 = 3.7 but n is always integer hence it will be 4

x	P(X =x)	Observed	Expected	(Obs - Exp)^2	(O - E)^2/E
0	0.6561	68	65.61	5.7121	0.087061424
1	0.2916	27	29.16	4.6656	0.16
2	0.0486	5	4.86	0.0196	0.004032922
				Chi_Sq -Test	0.25
				P-value =	0.12

Parameters of Binomial distribution

p = 0.1 and n = 4

Here P-value = 0.12 > . Hence we failed to reject null hypothesis

Conclusion: At , Data follows the Binomial distribution. Binomial is good fit for this data.

c]

If I am plant Manager, I will preferred Poisson distribution because in goodness of fit test of Binomial distribution the sum of observed frequency and sum of Expected frequency is not same. That is Expected frequencies not good.