Question

A machine cuts pieces of electromagnetic interference gaskets. The length of each gasket is measured and classified as being too short, OK, or too long. Likewise, the width of each gasket is measured and classified as being too narrow, OK, or too wide. The results of a sample of 1029 gaskets are shown in the table below. Test the claim that the width classification is independent of the length classification at the 90% confidence level.

	Length
Width	Too Short	OK	Too Long	Totals
Too Narrow	10	98	7	115
OK	55	712	70	837
Too Wide	8	57	12	77
Totals	73	867	89	1029

Answer 1

Chi-Square Independence test

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0​: The two variables - Width classification and Length classification are independent Ha​: The two variables - Width classification and Length classification are dependent This corresponds to a Chi-Square test of independence. (2) Degrees of Freedom The number of degrees of freedom is df = (3 - 1) * (3 - 1) = 4 (3) Critical value and Rejection Region Based on the information provided, the significance level is α=0.1, the number of degrees of freedom is df = (3 - 1) * (3 - 1) = 4, so the critical value is 7.7794. Then the rejection region for this test becomes R={χ2:χ2>7.7794}. Chi-Square Independence test Level of significance - a 0.10 Variable 1 Width classificati Variable 2 Length classification OBSERVED VALUES Too Long Too Narrow OK Too Wide Width 10 55 Too Short 98 712 OK 7 70 Total 115 837 8 57 12 77 Total 73 867 89 1029 EXPECTED VALUES Width Too Short OK Too Long Total 96.8950 115 8.1584 59.3790 9.9466 72.3936 705.2274 837 Too Narrow OK Too Wide 0 5.4626 64.8776 6.6599 77 Total 73 867 89 1029 (OBSERVED = EXPECTED) ^ 2 / EXPECTED OK Too Long Sum Width 0.4157 Too Short 0.0126 0.8729 1.3012 0.0791 Too Narrow OK Too Wide 0 0.3229 1.1786 0.0650 0.957 0.4671 6.4171 4.2819 Sum 1.9173 1.0341 5.2339 8.1854 (4)Test Statistics The Chi-Squared statistic is computed as follows: $\chi^2 = \sum_{i=1}^n \frac{(O_{ij}-E_{ij})^2}{E_{ij}}=8.1854$ (5)P-value The corresponding p-value for the test is p=Pr(χ2​>8.1854)=0.085 (6)The decision about the null hypothesis Since it is observed that χ2=8.1854>χ2_c​rit=7.7794, it is then concluded that the null hypothesis is rejected. (7)Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables - Width classification and Length classification are dependent, at the 0.1 significance level. Conditions: a. The sampling method is simple random sampling. b. The data in the cells should be counts/frequencies c. The levels (or categories) of the variables are mutually exclusive.