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Chapter 9: Q 96. (page 610)

Preventing colds A medical experiment investigated whether taking the herb echinacea could help prevent colds. The study measured 50 different response variables usually associated with colds, such as low-grade fever, congestion, frequency of coughing, and so on. At the end of the study, those taking echinacea displayed significantly better responses at theα=0.05 level than those taking a placebo for 3 of the 50 response variables studied. Should we be convinced that echinacea helps prevent colds? Why or why not?

Short Answer

Expert verified

There isn't enough evidence to establish that Echinacea is helpful in preventing colds.

Step by step solution

01

Given information

α=0.05

02

Calculation

To conclude that Echinacea is useful, there is insufficient evidence. The substantial results for the three variables are almost certainly coincidental.

When the null hypothesis is true 5%of the time, there is a type Ierror getting a significant result at the α=0.05level.

If 50t−tests are run at this level of significance and all 50null hypotheses are true, the average Type Ierror is 0.05×50=2.5

The three most important findings Such errors are very likely to occur in this setting.

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Most popular questions from this chapter

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a loud noise as stimulus. The appropriate hypotheses for the significance test are

a. H0:μ=18;Ha:μ≠18

b. H0:μ=18;Ha:μ>18

c. H0:μ<18;Ha:μ=18

d. H0:μ=18;Ha:μ<18

e. H0:x¯=18;Ha:x¯<18

pg559¯

No homework Refer to Exercises 1 and 9. What conclusion would you make at theα=0.05α=0.05level?

Proposition XA political organization wants to determine if there is convincing evidence that a majority of registered voters in a large city favor Proposition X. In an SRS of 1000registered voters, 482favor the proposition. Explain why it isn’t necessary to carry out a significance test in this setting.

Side effects A drug manufacturer claims that less than 10%of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300out of 5000Alzheimer’s patients whose families have given informed consent for the patients to participate in the study. In all, 25of the subjects experience nausea.

a. Describe a Type I error and a Type II error in this setting, and give a possible

consequence of each.

b. Do these data provide convincing evidence for the drug manufacturer’s claim?

A95%confidence interval for the proportion of viewers of a certain reality television

show who are over 30 years old is (0.26,0.35). Suppose the show's producers want to est the hypothesis \H0:p=0.25against Ha: Ha:p≠0.25. Which of the following is an appropriate conclusion for them to draw at the α=0.05

a. Fail to reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old equals 0.25

b. Fail to reject H0there is not convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from0.25.

c. Reject H0; there is not convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from 0.25

. d. Reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old is greater than 0.25.

e. Reject H0; there is convincing evidence that the true proportion of viewers of this reality TV show who are over 30 years old differs from0.25.
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