A restaurant is interested in the number of tables served during the breakfast, lunch, and dinner hours. The manager anticipates that twice as many people come to the restaurant for lunch as for breakfast, and that twice as many people come for dinner as for lunch. She takes a random sample of 100 receipts from a week, and determines whether the receipt corresponds with breakfast, lunch, or dinner hours. A total of 22 receipts were for breakfast, 30 were for lunch, and 48 were for dinner. Do the data provide convincing evidence at the Alpha = 0.01 level that the manager’s predicted proportions were incorrect? A. Since Chi-square = 5.7 and the P-value = 0.0578, there is convincing evidence that the manager’s predicted proportions were correct. B. Since Chi-square = 5.7 and the P-value = 0.0578, there is not convincing evidence that the manager’s predicted proportions were incorrect. C. Since Chi-square = 6.027 and the P-value = 0.0491, there is not convincing evidence that the manager’s predicted proportions were incorrect. D. Since Chi-square = 5.7 and the P-value ≈ 1, there is not convincing evidence that the manager’s predicted proportions were incorrect. E. Since Chi-square = 4.514 and the P-value = 0.1047, there is not convincing evidence that the manager’s predicted proportions were incorrect.
A restaurant is interested in the number of tables served during the breakfast, lunch, and dinner hours. The manager anticipates that twice as many people come to the restaurant for lunch as for breakfast, and that twice as many people come for dinner as for lunch. She takes a random sample of 100 receipts from a week, and determines whether the receipt corresponds with breakfast, lunch, or dinner hours. A total of 22 receipts were for breakfast, 30 were for lunch, and 48 were for dinner. Do the data provide convincing evidence at the Alpha = 0.01 level that the manager’s predicted proportions were incorrect?
A. Since Chi-square = 5.7 and the P-value = 0.0578, there is convincing evidence that the manager’s predicted proportions were correct.
B. Since Chi-square = 5.7 and the P-value = 0.0578, there is not convincing evidence that the manager’s predicted proportions were incorrect.
C. Since Chi-square = 6.027 and the P-value = 0.0491, there is not convincing evidence that the manager’s predicted proportions were incorrect.
D. Since Chi-square = 5.7 and the P-value ≈ 1, there is not convincing evidence that the manager’s predicted proportions were incorrect.
E. Since Chi-square = 4.514 and the P-value = 0.1047, there is not convincing evidence that the manager’s predicted proportions were incorrect.
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