Question
In: Statistics and Probability
A new technology for the production of liquid-crystal displays (LCD) is being mastered. 8,000,000 pixels are...
A new technology for the production of liquid-crystal displays (LCD) is being mastered. 8,000,000 pixels are located on the LCD. The LCD is defective if there are more than 5 malfunctioning pixels. Let denote the number of defective pixels observed on the LCD. Of 3,000 LCD inspected, the following data were observed for the values of :
| Values | 1 or less | 2 | 3 | 4 | 5 | 6 | 7 | 8 or more | |
|---|---|---|---|---|---|---|---|---|---|
| Observed frequency |
146 | 255 | 401 | 552 | 545 | 422 | 306 | 373 |
Does the assumption of the Poisson distribution seem appropriate as a probability model for these data? Use α=0.9.
Solutions
Expert Solution
| Values | f: Observed Frequency |
| 1 or less | 146 |
| 2 | 255 |
| 3 | 401 |
| 4 | 552 |
| 5 | 545 |
| 6 | 422 |
| 7 | 306 |
| 8 or more | 373 |
Poisson distribution Probability mass function: 
| x | f | fx |
| 1 | 146 | 146 |
| 2 | 255 | 510 |
| 3 | 401 | 1203 |
| 4 | 552 | 2208 |
| 5 | 545 | 2725 |
| 6 | 422 | 2532 |
| 7 | 306 | 2142 |
| 8 | 373 | 2984 |
| Total | 3000 | 14450 |
Theoretical Distribution is then : poisson with mean : 
=
4.82; And the Probability mass function
And the Expected Frequency
E= N x p(x) = 3000p(x)
| x | p(x) | E: 3000*p(x) |
| 0 | 0.008093 | 24.28035 |
| 1 | 0.038984 | 116.95118 |
| 2 | 0.093886 | 281.65938 |
| 3 | 0.150741 | 452.22292 |
| 4 | 0.181519 | 544.55553 |
| 5 | 0.174864 | 524.59213 |
| 6 | 0.140378 | 421.13381 |
| 7 | 0.096594 | 289.78218 |
| 8 | 0.058158 | 174.47423 |
| Total | 2829.65171 |
Total Frequency = 3000
Therefore Frequency for X> 8 = 3000 - 2829.65171 = 170.34829
Frequency for For X 
8 =
174.47423+170.34829 = 344.82252
Frequency for X : 1 or Less = Frequency for X=0 + Frequency X=1
= 24.28035+ 116.95118=141.23153
| O | E |
| 146 | 141.23153 |
| 255 | 281.6593831 |
| 401 | 452.2229169 |
| 552 | 544.555531 |
| 545 | 524.5921252 |
| 422 | 421.1338149 |
| 306 | 289.782178 |
| 373 | 344.82252 |
| O | E | O-E | (O-E)2 | (O-E)2/E |
| 146 | 141.23153 | 4.76847 | 22.73831 | 0.161 |
| 255 | 281.6593831 | -26.6594 | 710.7227 | 2.523341 |
| 401 | 452.2229169 | -51.2229 | 2623.787 | 5.801978 |
| 552 | 544.555531 | 7.444469 | 55.42012 | 0.101771 |
| 545 | 524.5921252 | 20.40787 | 416.4814 | 0.793915 |
| 422 | 421.1338149 | 0.866185 | 0.750277 | 0.001782 |
| 306 | 289.782178 | 16.21782 | 263.0177 | 0.907639 |
| 373 | 344.82252 | 28.17748 | 793.9704 | 2.302548 |
| Total | 12.59397 |
Degree of freedom = 8-2 = 6
:
0.9
As p-value : 0.050 < :0.9
Reject null hypothesis.
Poisson distribution does not seem to fit the data.
The assumption of the Poisson distribution does not seem appropriate as a probability model for these data
p-value is computed using excel function,
CHISQ.DIST.RT(12.59397,6) = 0.050
CHISQ.DIST.RT function
Returns the right-tailed probability of the chi-squared distribution.
The χ2 distribution is associated with a χ2 test. Use the χ2 test to compare observed and expected values. For example, a genetic experiment might hypothesize that the next generation of plants will exhibit a certain set of colors. By comparing the observed results with the expected ones, you can decide whether your original hypothesis is valid.
Syntax
CHISQ.DIST.RT(x,deg_freedom)
The CHISQ.DIST.RT function syntax has the following arguments:
-
X Required. The value at which you want to evaluate the distribution.
-
Deg_freedom Required. The number of degrees of freedom.
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units of flat panel displays are estimated as follows:
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Direct labor
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