7.
Product and Process Comparisons
7.4.
Comparisons based on data from more than two processes
7.4.6.
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Do all the processes have the same proportion of defectives?
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The contingency
table approach
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Testing for homogeneity of proportions using the chi-square
distribution via contingency tables
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When we have samples from \(n\)
populations (i.e., lots, vendors,
production runs, etc.), we can test whether there are significant
differences in the proportion defectives for these populations using
a contingency table approach. The contingency table we construct
has two rows and \(n\)
columns.
To test the null hypothesis of no difference in the proportions
among the \(n\)
populations
\( \mbox{H}_0: \,\,\, p_1 = p_2 = \cdots = p_n \)
against the alternative that not all \(n\)
population proportions are equal
\( \mbox{H}_a: \,\,\, \mbox{Not all } p_i \mbox{ are equal } (i = 1, \, 2, \, \ldots, \, n) \, , \)
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The chi-square test statistic
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we use the following test statistic:
$$ \chi^2 = \sum_{\mbox{all cells}} \frac{(f_o - f_c)^2}{f_c} \, , $$
where \(f_o\)
is the observed frequency in a given cell of a \(2 \times n\)
contingency table, and \(f_c\)
is the theoretical count or expected frequency in a given cell
if the null hypothesis were true.
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The critical value
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The critical value is obtained from the \(\chi^2\)
distribution table with degrees of freedom \((2-1)(n-1) = n-1\),
at a given level of significance.
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An illustrative example
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Data for the example
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Diodes used on a printed circuit board are produced in lots of size
4000. To study the homogeneity of lots with respect to a demanding
specification, we take random samples of size 300 from 5 consecutive
lots and test the diodes. The results are:
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Lot
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Results
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1
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2
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3
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4
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5
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Totals
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Nonconforming
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36
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46
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42
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63
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38
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225
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Conforming
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264
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254
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258
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237
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262
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1275
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Totals
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300
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300
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300
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300
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300
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1500
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Computation of the overall proportion of nonconforming units
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Assuming the null hypothesis is true, we can estimate the single
overall proportion of nonconforming diodes by pooling the results
of all the samples as
$$ \bar{p} = \frac{36 + 46 + 42 + 63 + 38}{5(300)} = \frac{225}{1500} = 0.15 \, . $$
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Computation of the overall proportion of conforming units
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We estimate the proportion of conforming ("good") diodes by the
complement 1 - 0.15 = 0.85. Multiplying these two proportions by
the sample sizes used for each lot results in the expected
frequencies of nonconforming and conforming diodes. These are
presented below:
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Table of expected frequencies
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Lot
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Results
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1
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2
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3
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4
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5
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Totals
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Nonconforming
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45
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45
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45
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45
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45
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225
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Conforming
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255
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255
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255
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255
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255
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1275
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Totals
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300
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300
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300
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300
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300
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1500
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Null and alternate hypotheses
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To test the null hypothesis of homogeneity or equality of
proportions
\( \mbox{H}_0: \,\,\, p_1 = p_2 = \cdots = p_5 \)
against the alternative that not all 5 population proportions
are equal
\( \mbox{H}_a: \,\,\, \mbox{Not all } p_i \mbox{ are equal } (i = 1, \, 2, \, \ldots, \, 5) \, , \)
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Table for computing the test statistic
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we use the observed and expected values from the tables above to
compute the \(\chi^2\)
test statistic. The calculations are presented below:
| \(f_o\)
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\(f_c\)
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\((f_o - f_c)\)
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\((f_o - f_c)^2\)
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\((f_o - f_c)^2 / f_c\)
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36
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45
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-9
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81
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1.800
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46
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45
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1
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1
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0.022
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42
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45
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-3
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9
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0.200
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63
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45
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18
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324
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7.200
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38
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45
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-7
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49
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1.089
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264
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225
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9
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81
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0.318
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254
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255
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-1
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1
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0.004
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258
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255
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3
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9
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0.035
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237
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255
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-18
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324
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1.271
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262
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255
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7
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49
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0.192
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12.131
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Conclusions
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If we choose a 0.05 level of significance, the critical value of \(\chi^2\)
with 4 degrees of freedom is 9.488 (see the
chi square distribution table
in Chapter 1). Since the test statistic
(12.131) exceeds this critical value, we reject the null hypothesis.
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