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Whole grains (4.2) A series of observational studies revealed that people who typically consume 3 servings of whole grain per day have about a 20% lower risk of dying from heart disease and about a 15% lower risk of dying from stroke or cancer than those who consume no whole grains.

a. Explain how confounding makes it difficult to establish a cause-and-effect relationship between whole grain consumption and risk of dying from heart disease, stroke, or cancer, based on these studies.

b. Explain how researchers could establish a cause-and-effect relationship in this context.

Short Answer

Expert verified

Part (a) Amount of exercise could be confounded the results.

Part (b) The people in the treatment group eat 3 servings of whole grain per day, where the people in the control group do not eat whole grains at all.

Step by step solution

01

Part (a) Step 1: Given information

Persons who eat three servings of whole grains per day have a 20% reduced chance of dying from heart disease and a 15% lower risk of dying from stroke or cancer than people who don't eat any whole grains.

02

Part (a) Step 2: Explanation

When the effects of two variables on a response variable cannot be distinguished from one another, they are said to be confused.

It is impossible to establish a cause-and-effect relationship between whole grain consumption and the risk of dying from heart disease, stroke, or cancer since the results of the study could be contaminated by another variable.

For example, the quantity of exercise you get may affect your risk of dying from heart disease, stroke, or cancer, since if you exercise more, you'll be healthier and hence less likely to die from heart disease, stroke, or cancer. Although the amount of activity and whole grain consumption cannot be distinguished,

03

Part (b) Step 1: Explanation

If an experiment is used instead of observational research, a cause-and-effect link can be established in this setting. It may, for example, use a perfectly randomized experiment. In a truly randomized experiment, all individuals are assigned to a group at random. Choose a group of people and assign half of them to the treatment group, while the other half is assigned to the control group at random. The treatment group consumes three servings of whole grain per day, while the control group consumes no whole grains at all.

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Most popular questions from this chapter

The student newspaper at a large university asks an SRS of 250 undergraduates, "Do you favor eliminating the carnival from the term-end celebration?" All in all, 150 of the 250 are in favor. Suppose that (unknown to you) 55\% of all undergraduates favor eliminating the carnival. If you took a very large number of SRSs of size n=250 n=250 from this population, the sampling distribution of the sample proportion pโˆงp^would be

a. exactly Normal with mean 0.55 and standard deviation 0.03.

b. approximately Normal with mean 0.55 and standard deviation 0.03.

c. exactly Normal with mean 0.60 and standard deviation 0.03.

d. approximately Normal with mean 0.60 and standard deviation 0.03.

e. heavily skewed with mean 0.55 and standard deviation 0.03.

According to the U.S. Census, the proportion of adults in a certain county who owned their own home was 0.71. An SRS of 100 adults in a certain section of the county found that 65 owned their home. Which one of the following represents the approximate probability of obtaining a sample of 100 adults in which 65 or fewer own their home, assuming that this section of the county has the same overall proportion of adults who own their home as does the entire county?

a. (10065)(0.71)65(0.29)3510065(0.71)65(0.29)35

b. (10065)(0.29)65(0.71)3510065(0.29)65(0.71)35

c.

P(zโ‰ค0.65-0.71(0.71)(0.29)100)Pzโ‰ค0.65-0.71(0.71)(0.29)100

d.P(zโ‰ค0.65-0.71(0.65)(0.35)100)Pzโ‰ค0.65-0.71(0.65)(0.35)100

e.P(zโ‰ค0.65-0.71(0.71)(0.29)100)Pzโ‰ค0.65-0.71(0.71)(0.29)100

A study of rush-hour traffic in San Francisco counts the number of people in each car entering a freeway at a suburban interchange. Suppose that this count has mean 1.6 and standard deviation 0.75 in the population of all cars that enter at this interchange during rush hour.

a. Without doing any calculations, explain which event is more likely:

  • randomly selecting 1 car entering this interchange during rush hour and finding 2 or more people in the car
  • randomly selecting 35 cars entering this interchange during rush hour and finding an average of 2 or more people in the cars

b. Explain why you cannot use a Normal distribution to calculate the probability of the first event in part (a).

c. Calculate the probability of the second event in part (a).

The mean of this distribution (donโ€™t try to find it) will be

a. very close to the median.

b. greater than the median.

c. less than the median.

d. You canโ€™t say, because the distribution isnโ€™t symmetric.

e. You canโ€™t say, because the distribution isnโ€™t Normal.

Making auto parts Refer to Exercise 54 . How many axles would you need to sample if you wanted the standard deviation of the sampling distribution of x-xยฏto be 0.0005mm ? Justify your answer.

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