Question

14) The operations manager of a company that manufactures tiles wants to determine whether the quality of work is independent of the daily shifts. He randomly selects 384 tiles and carefully inspects them. Each tile is either classified as perfect, satisfactory, or defective, and the shift that produced it is also recorded. The data are summarised in the two-way table below and a ? ଶ test of independence is performed.

Perfect Satisfactory Sempurna) Memuaskan 106 124 Defective (Rosak/ Total Jumlah/ 231 67 85 153 Shift 1 (Syif 11 Shift 2 (Syif 2] Total [Jumlah 173 209 2 384

A) Find the critical value for the test at ? = 0.01.

A. 6.635 C. 10.597 B. 4.605 D. 9.210

Calculate the test statistics, ? ^2 to determine whether the level of quality is independent of the shifts.

A. 0.2354 C. 1.6452 B. 2.9513 D. 0.8662

Answer 1

Solution:

Column 1 Column 2 Column 3 Total Expected Values Row 1 173x231 384 = 104.07 209x231 384 = 125.727 2.231 = 1.203 231 384 Row 2 173x153 384 = 68.93 209% 153 = 83.273 2x153 = 0.797 153 354 384 Total 173 209 384 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E-0)°/E. The table with squared distances is shown below:

Squared Distances Column 1 Column 2 Column 3 Row 1 (106-104.07 104.07 = 0.036 (124-125.727) 125.727 0.024 (1-1.203)2 1.203 = 0.034 Row 2 (67-68.93) 68.93 = 0.054 (85-83.273) 83.273 = 0.036 (1-0.7972 0.79 = 0.052 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: The two variables are independent He: The two variables are dependent This corresponds to a Chi-Square test of independence. Rejection Region Based on the information provided, the significance level is a = 0.01, the number of degrees of freedom is df = (2-1) (3 - 1) = 2, so then the rejection region for this test is R = {x?: x> 9.21) Test Statistics The Chi-Squared statistic is computed as follows: x (Oj - E;) = 0.036 +0.054 +0.024 +0.036 +0.034 +0.052 = 0.235 Decision about the null hypothesis Since it is observed that y? = 0.235 < x = 9.21, it is then concluded that the null hypothesis is not rejected. Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at 0.01 significance level. The corresponding p-value for the test is p = Pr(x3 > 0.235) = 0.889.