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 633: 7

Answer

Step 1: $H_0$ : The types of meal purchased are independent of the restaurant. $H_1$: The types of meal purchased are dependent on the restaurant. Step 2: Since α=0.01, the critical value using Table G with (3-1)(3-1) = (2)(2) =4 degrees of freedom is 13.277. Step 3: Expected Value: $E_1,1$ = $\frac{(98)(100)}{(300)}$ = 32.67 $E_1,2$ = $\frac{(98)(100)}{(300)}$ = 32.67 $E_1,3$ = $\frac{(98)(100)}{(300)}$= 32.67 $E_2,1$ = $\frac{(113)(100)}{(300)}$ = 37.67 $E_2,2$ = $\frac{(113)(100)}{(300)}$ = 37.67 $E_2,3$ = $\frac{(113)(100)}{(300)}$ = 37.67 $E_3,1$ = $\frac{(89)(100)}{(300)}$ = 29.67 $E_3,2$ = $\frac{(89)(100)}{(300)}$ = 29.67 $E_3,3$ = $\frac{(89)(100)}{(300)}$ = 29.67 Test Value : χ2 = Σ $\frac{(O-E)^{2}}{E}$ = $\frac{(26-32.67)^{2}}{32.67}$ + $\frac{(29-32.67)^{2}}{32.67}$ + $\frac{(43-32.67)^{2}}{32.67}$ + $\frac{(53-37.67)^{2}}{37.67}$ + $\frac{(27-37.67)^{2}}{37.67}$ + $\frac{(33-37.67)^{2}}{37.67}$ + $\frac{(21-29.67)^{2}}{29.67}$ + $\frac{(44-29.67)^{2}}{29.67}$ + $\frac{(24-29.67)^{2}}{29.67}$ =1.361+0.412+3.269+6.242+3.021+0.578+2.532+6.925+1.082 =25.421 Step 4: Since 25.421 > 13.277, the decision is to reject the null hypothesis. Step 5: There is enough evidence to claim that the types of meal purchased by the patrons are dependent upon the restaurant.
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