Question 6.9: Use the data in Table 6.2 to test the null hypothesis that t......

We begin by using Equation (6.5) to compute the expected values E_{ij}. We show the calculations of E_{11} and E_{23} in detail:

                               E_{ij} = \frac{O_{i.} O_{.j}}{O_{..}} (6.5)

E_{11} = \frac{(120)(66)}{500} = 15.84

E_{23} = \frac{(200)(32)}{500} = 12.80

      The complete table of expected values is as follows:

      We note that all the expected values are greater than 5. Therefore, the chi-square test is appropriate. We use Equation (6.6) to compute the value of the chi-square statistic:

\chi^{2} = \sum\limits_{i=1}^{I}{} \sum\limits_{j=1}^{J}{} \frac{(O_{ij}\ −\ E_{ij})^{2}}{E_{ij}} (6.6)

            \chi^{2} = \frac{(10\ −\ 15.84)^{2}}{15.84} + ⋯ + \frac{(10\ −\ 5.12)^{2}}{5.12}

                                = \frac{34.1056}{15.84} + ⋯ + \frac{23.8144}{5.12}

                                = 15.5844

Since there are four rows and three columns, the number of degrees of freedom is (4 − 1)(3 − 1) = 6. To obtain the P-value, we consult the chi-square table (Table A.5). Looking under six degrees of freedom, we find that the upper 2.5% point is 14.449, and the upper 1% point is 16.812. Therefore 0.01 < P < 0.025. (Software yields P = 0.016.) It is reasonable to conclude that the machines differ in the proportions of pins that are too thin, OK, or too thick.