Question 6.9: Use the data in Table 6.2 to test the null hypothesis that t...
Use the data in Table 6.2 to test the null hypothesis that the proportions of pins that are too thin, OK, or too thick are the same for all the machines.
| Expected values for Table 6.2 | ||||
| Too Thin |
OK | Too Thick |
Total | |
| Machine 1 | 15.84 | 96.48 | 7.68 | 120.00 |
| Machine 2 | 26.40 | 160.80 | 12.8 | 200.00 |
| Machine 3 | 13.20 | 80.40 | 6.4 | 100.00 |
| Machine 4 | 10.56 | 64.32 | 5. 12 | 80.00 |
| Total | 66.00 | 402.00 | 32.00 | 500.00 |
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_{i j}=\frac{O_{i_{.}}O_{. j}}{O_{..}} (6.5)
\begin{aligned}& E_{11}=\frac{(120)(66)}{500}=15.84 \\& E_{23}=\frac{(200)(32)}{500}=12.80\end{aligned}
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{\left(O_{i j}-E_{i j}\right)^2}{E_{i j}} (6.6)
\begin{aligned}x^2 &=\frac{(10-15.84)^2}{15.84}+\cdots+\frac{(10-5.12)^2}{5.12}\\&=\frac{34.1056}{15.84}+\cdots+\frac{23.8144}{5.12}\\&=15.5844\end{aligned}
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 < 0.025.1t is reasonable to conclude that the machines differ in the proportions of pins that are too thin, OK, or too thick.