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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 the伪=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

伪=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 伪=0.05level.

If 50t鈭�tests are run at this level of significance and all 50null hypotheses are true, the average Type Ierror is 0.05脳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

A company that manufactures classroom chairs for high school students

claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0:渭=300Ha:渭<300where 渭 is the true mean breaking strength of this company鈥檚 classroom chairs.

a. The power of the test to detect that 渭=294 based on a random sample of 30

chairs and a significance level of 伪=0.05 is 0.71. Interpret this value.

b. Find the probability of a Type I error and the probability of a Type II error for the test in part (a).

c. Describe two ways to increase the power of the test in part (a).

Teen drivers Refer to Exercise 51.

a. Construct and interpret a 95% confidence interval for the true proportion p of all teens in the state who passed their driving test on the first attempt. Assume that the conditions for inference are met.

b. Explain why the interval in part (a) provides more information than the test in Exercise 51.

Paying high prices? A retailer entered into an exclusive agreement with a supplier who guaranteed to provide all products at competitive prices. To be sure the supplier honored the terms of the agreement, the retailer had an audit performed on a random sample of 25 invoices. The percent of purchases on each invoice for which an alternative supplier offered a lower price than the original supplier was recorded.17 For example, a data value

of 38 means that the price would be lower with a different supplier for 38% of the items on the invoice. A histogram and some numerical summaries of the data are shown here. The retailer would like to determine if there is convincing evidence that the mean percent of purchases for which an alternative supplier offered lower prices is greater than 50% in the population of this company鈥檚 invoices.

a. State appropriate hypotheses for the retailer鈥檚 test. Be sure to define your parameter.

b. Check if the conditions for performing the test in part (a) are met.

An opinion poll asks a random sample of adults whether they favor banning ownership of handguns by private citizens. A commentator believes that more than half of all adults favor such a ban. The null and alternative hypotheses you would use to test this claim are

邪.H0:p鈭�=0.5;Ha:p鈭�>0.5H0:p^=0.5;Ha:p^>0.5.

b. H0:p=0.5;Ha:p>0.5H0:p=0.5;Ha:p>0.5.

c. H0:p=0.5;Ha:p<0.5H0:p=0.5;Ha:p<0.5.

d. H0:p=0.5;Ha:p鈮�0.H0:p=0.5;Ha:p鈮�0.5.

e. H0:p>0.5;Ha:p=0.5H0:p>0.5;Ha:p=0.5.

How much juice? Refer to Exercises 3 and 11 .

a. What conclusion would you make at the 伪=0.10伪=0.10level?

b. Would your conclusion from part (a) change if a 5 \% significance level was used instead? Explain your reasoning.
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