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Solved: Section I: Multiple Choice Select the best answer for each question. Exercises
Chapter 11, Problem T11.8(choose chapter or problem)
All current carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate the relationship between current configuration and type of cancer, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC).
Here are the data:
Computer software was used to analyze the data. The output included the value x2 = 0.435.
The expected count of cases with lymphoma in homes with an HCC is
(a) \(\frac{79 \cdot 31}{215}\) . (b) \(\frac{10 \cdot 21}{215}\) . (c) \(\frac{79 \cdot 31}{10}\) . (d) \(\frac{136 \cdot 31}{215}\). (e) None of these
Questions & Answers
QUESTION:
All current carrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate the relationship between current configuration and type of cancer, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC).
Here are the data:
Computer software was used to analyze the data. The output included the value x2 = 0.435.
The expected count of cases with lymphoma in homes with an HCC is
(a) \(\frac{79 \cdot 31}{215}\) . (b) \(\frac{10 \cdot 21}{215}\) . (c) \(\frac{79 \cdot 31}{10}\) . (d) \(\frac{136 \cdot 31}{215}\). (e) None of these
ANSWER:Step 1 of 2
The expected count is the product of the row total and the column total, divided by the sample size .