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Bias in Clinical Trials? Researchers investigated the issue of race and equality of access to clinical trials. The following table shows the population distribution and the numbers of participants in clinical trials involving lung cancer (based on data from “Participation in Cancer Clinical Trials,” by Murthy, Krumholz, and Gross, Journal of the American Medical Association, Vol. 291, No. 22). Use a 0.01 significance level to test the claim that the distribution of clinical trial participants fits well with the population distribution. Is there a race/ethnic group that appears to be very underrepresented?

Race/ethnicity

White

non-Hispanic

Hispanic

Black

Asian/

Pacific

Islander

American Indian/

Alaskan Native

Distribution of

Population

75.6%

9.1%

10.8%

3.8%

0.7%

Number in Lung

Cancer Clinical Trials

3855

60

316

54

12

Short Answer

Expert verified

There is enough evidence to conclude that the participants are not distributed according to the population distribution.

American Indian/Alaskan Native ethnic group, and the Asian/Pacific Islander ethnic group, appear to be underrepresented.

Step by step solution

01

Given information

The number of participants in a clinical trial involving lung cancer are tabulated under different ethnic groups.

The expected population distribution under each ethnic group is also provided.

02

Check the requirements

Assume subjects are randomly selected.

Let O denote the observed frequencies of people of different races.

The following values are obtained:

\(\begin{aligned}{l}{O_1} = 3855\\{O_2} = 60\\{O_3} = 316\\{O_4} = 54\\{O_5} = 12\end{aligned}\)

The sum of all observed frequencies is computed below:

\[\begin{aligned}{c}n = 3855 + 60 + ... + 12\\ = 4297\end{aligned}\]

Let E denote the expected frequencies.

It is expected that the frequencies should fit well with the population distribution.

Therefore, the population distribution of each race is given as follows:

\(\begin{aligned}{c}{p_1} = \frac{{75.6}}{{100}}\\ = 0.756\\{p_2} = \frac{{9.1}}{{100}}\\ = 0.091\end{aligned}\)

\(\begin{aligned}{c}{p_3} = \frac{{10.8}}{{100}}\\ = 0.108\\{p_4} = \frac{{3.8}}{{100}}\\ = 0.038\end{aligned}\)

\(\begin{aligned}{c}{p_5} = \frac{{0.7}}{{100}}\\ = 0.007\end{aligned}\)

Now, the expected frequencies are computed below:

\(\begin{aligned}{c}{E_1} = n{p_1}\\ = 4297\left( {0.756} \right)\\ = 3248.532\end{aligned}\)

\(\begin{aligned}{c}{E_2} = n{p_2}\\ = 4297\left( {0.091} \right)\\ = 391.027\end{aligned}\)

\(\begin{aligned}{c}{E_3} = n{p_3}\\ = 4297\left( {0.108} \right)\\ = 464.076\end{aligned}\)

\(\begin{aligned}{c}{E_4} = n{p_4}\\ = 4297\left( {0.038} \right)\\ = 163.286\end{aligned}\)

\(\begin{aligned}{c}{E_5} = n{p_5}\\ = 4297\left( {0.007} \right)\\ = 30.079\end{aligned}\)

Also,the expected values are greater than 5.

Thus, the requirements of the test are satisfied.

03

State the hypotheses

The hypotheses is stated as follows:

\({H_o}:\)The distribution of observations fits the distribution of population

\({H_a}:\)The distribution of observations does not fit the distribution of the population.

The test is right-tailed.

04

Compute the test statistic

The table below shows the necessary calculations:

05

State the conclusion

There is enough evidence to conclude that the participants are not distributed according to the population distribution.

Since there are a few participants that belong to the American Indian/Alaskan Native ethnic group as well as the Asian/Pacific Islander ethnic group, it can be said that these two races are underrepresented.

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Most popular questions from this chapter

Flat Tire and Missed Class A classic story involves four carpooling students who missed a test and gave as an excuse a flat tire. On the makeup test, the instructor asked the students to identify the particular tire that went flat. If they really didn’t have a flat tire, would they be able to identify the same tire? The author asked 41 other students to identify the tire they would select. The results are listed in the following table (except for one student who selected the spare). Use a 0.05 significance level to test the author’s claim that the results fit a uniform distribution. What does the result suggest about the likelihood of four students identifying the same tire when they really didn’t have a flat tire?

Tire

Left Front

Right Front

Left Rear

Right Rear

Number Selected

11

15

8

16

One Big Bill or Many Smaller Bills In a study of the “denomination effect,” 150 women in China were given either a single 100 yuan bill or a total of 100 yuan in smaller bills. The value of 100 yuan is about $15. The women were given the choice of spending the money on specific items or keeping the money. The results are summarized in the table below (based on “The Denomination Effect,” by Priya Raghubir and Joydeep Srivastava, Journal of Consumer Research, Vol. 36). Use a 0.05 significance level to test the claim that the form of the 100 yuan is independent of whether the money was spent. What does the result suggest about a denomination effect?

Spent the Money

Kept the Money

Women Given a Single 100-Yuan Bill

60

15

Women Given 100 Yuan in Smaller Bills

68

7

Exercises 1–5 refer to the sample data in the following table, which summarizes the last digits of the heights (cm) of 300 randomly selected subjects (from Data Set 1 “Body Data” in Appendix B). Assume that we want to use a 0.05 significance level to test the claim that the data are from a population having the property that the last digits are all equally likely.

Last Digit

0

1

2

3

4

5

6

7

8

9

Frequency

30

35

24

25

35

36

37

27

27

24

If using a 0.05 significance level to test the stated claim, find the number of degrees of freedom.

Benford’s Law. According to Benford’s law, a variety of different data sets include numbers with leading (first) digits that follow the distribution shown in the table below. In Exercises 21–24, test for goodness-of-fit with the distribution described by Benford’s law.

Leading Digits

Benford's Law: Distributuon of leading digits

1

30.10%

2

17.60%

3

12.50%

4

9.70%

5

7.90%

6

6.70%

7

5.80%

8

5.10%

9

4.60%

Author’s Check Amounts Exercise 21 lists the observed frequencies of leading digits from amounts on checks from seven suspect companies. Here are the observed frequencies of the leading digits from the amounts on the most recent checks written by the author at the time this exercise was created: 83, 58, 27, 21, 21, 21, 6, 4, 9. (Those observed frequencies correspond to the leading digits of 1, 2, 3, 4, 5, 6, 7, 8, and 9, respectively.) Using a 0.01 significance level, test the claim that these leading digits are from a population of leading digits that conform to Benford’s law. Does the conclusion change if the significance level is 0.05?

Chocolate and Happiness In a survey sponsored by the Lindt chocolate company, 1708 women were surveyed and 85% of them said that chocolate made them happier.

a. Is there anything potentially wrong with this survey?

b. Of the 1708 women surveyed, what is the number of them who said that chocolate made them happier?

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