Elementary Statistics: A Step-by-Step Approach with Formula Card 9th Edition

Published by McGraw-Hill Education
ISBN 10: 0078136334
ISBN 13: 978-0-07813-633-7

Chapter 11 - Other Chi-Square Tests - 11-2 Testing Using Contingency Tables - Exercises 11-2 - Page 635: 17

Answer

Step 1: $H_0$ : The study groups and the professors are independent from each other. $H_1$: The study groups and the professors are dependent upon each other. Step 2: Since α=0.05, the critical value using Table G with (3-1)(3-1) = (2)(2) =4 degrees of freedom is 9.488. Step 3: Expected Value: $E_1,1$ = $\frac{(62)(58)}{(174)}$ = 20.67 $E_1,2$ = $\frac{(62)(55)}{(174)}$ = 19.60 $E_1,3$ = $\frac{(62)(61)}{(174)}$= 21.74 $E_2,1$ = $\frac{(55)(58)}{(174)}$ = 18.33 $E_2,2$ = $\frac{(55)(55)}{(174)}$ = 17.39 $E_2,3$ = $\frac{(55)(61)}{(174)}$ = 19.28 $E_3,1$ = $\frac{(57)(58)}{(174)}$ = 19 $E_3,2$ = $\frac{(57)(55)}{(174)}$ = 18.02 $E_3,3$ = $\frac{(57)(61)}{(174)}$ = 19.98 Test Value : χ2 = Σ $\frac{(O-E)^{2}}{E}$ = $\frac{(25-20.67)^{2}}{20.67}$ + $\frac{(22-19.60)^{2}}{19.60}$ + $\frac{(15-21.74)^{2}}{21.74}$ + $\frac{(16-18.33)^{2}}{18.33}$ + $\frac{(16-18.33)^{2}}{18.33}$ + $\frac{(15-17.39)^{2}}{17.39}$ + $\frac{(24-19.28)^{2}}{19.28}$ + $\frac{(17-19)^{2}}{19}$ + $\frac{(18-18.02)^{2}}{18.02}$ + $\frac{(22-19.98)^{2}}{19.98}$ =0.909+0.294+2.087+0.297+0.327+1.155+0.211+0+0.204 =5.483 Step 4: Since 5.483 < 9.488, the decision is not to reject the null hypothesis. Step 5: There is not enough evidence to claim that the study groups and the professors are dependent upon each other.
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