Question

6. A study is conducted to test a new drug claimed to reduce diastolic blood pressure in adults with a history of coronary heart disease. What is the most efficient study to test whether the drug reduces diastolic blood pressure? Justify.

9. The following table shows the distribution of BMI in children living in United States and European urban neighborhoods. (The data are in millions.) What is the probability that a child selected at random is overweight? What is the probability that a child living in a U.S. urban neighborhood is overweight? What is the probability that a child living in a European urban neighborhood is overweight? What is the probability that a child lives in a U.S. urban neighborhood and is obese? What is the probability that a child is overweight and neighborhood location independent? Justify briefly.

Neighborhood	Normal Weight	Overweight	Obese
United States	125	50	40
Europe	101	42	21

10. Suppose that the probability that a child living in an urban area in the United States is obese is 20%. If a social worker sees 15 children living in urban areas, answer the following: What is the probability that none are obese? What is the probability that 5 are obese?

Answer 1

6:

To test whether the drug is drug is effective or not we can take a fairly large random sample of women with a history of coronary heart disease. Then divided women in two groups control and experimental. In the control group women were on the medication that they already taking and in experimental group women will take new drug. Then we need to compare the results to test whether drug is effective or not.

9:

Following table shows the row total and column total:

Neighborhood	Normal Weight	Overweight	Obese	Total
United States	125	50	40	215
Europe	101	42	21	164
Total	226	92	61	379

The probability that a child selected at random is overweight is

P(overweight) = 92 / 379 = 0.2427

The  probability that a child living in a U.S. urban neighborhood is overweight is

P(overweight | United States) = 50 / 215 = 0.2326

The probability that a child living in a European urban neighborhood is overweight is

P(overweight | Europe) = 42 / 164 = 0.2561

The probability that a child lives in a U.S. urban neighborhood and is obese

P(overweight and United States) = 40 / 379 = 0.1055

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Since probability P(overweight | United States) is not equal to P(overweight) so these events are not independent.

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10:

Here we need to use binomial distribution with following parameters

n=15 and p=0.20

The probability that none are obese is

The probability that 5 are obese is