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 636: 31

Answer

$H_0$ : The proportions of the customers in the three stores who made a list before going shopping and the stores are independent from each other. $H_1$: The proportions of the customers in the three stores who made a list before going shopping and the stores are dependent upon each other. Step 2: Since α=0.1, the critical value using Table G with (2-1)(3-1) = (1)(2) =2 degrees of freedom is 4.605. Step 3: Expected Value: $E_1,1$ = $\frac{(219)(96)}{(288)}$ = 73 $E_1,2$ = $\frac{(219)(96)}{(288)}$ = 73 $E_1,3$ = $\frac{(219)(96)}{(288)}$= 73 $E_2,1$ = $\frac{(69)(96)}{(288)}$ = 23 $E_2,2$ = $\frac{(69)(96)}{(288)}$ = 23 $E_2,3$ = $\frac{(69)(96)}{(288)}$ = 23 Test Value : χ2 = Σ $\frac{(O-E)^{2}}{E}$ = $\frac{(77-73)^{2}}{73}$ + $\frac{(74-73)^{2}}{73}$ + $\frac{(68-73)^{2}}{73}$ + $\frac{(19-23)^{2}}{23}$ + $\frac{(22-23)^{2}}{23}$ + $\frac{(28-23)^{2}}{23}$ =0.219+0.014+0.342+0.696+0.043+1.087 =2.401 Step 4: Since 2.401 < 4.605, the decision is to reject the null hypothesis. Step 5: There is not enough evidence to claim that the proportions of the customers in the three stores who made a list before going shopping and the stores are dependent upon each other.
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