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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 鈥淧articipation in Cancer Clinical Trials,鈥 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

Critical Thinking: Was Allstate wrong? The Allstate insurance company once issued a press release listing zodiac signs along with the corresponding numbers of automobile crashes, as shown in the first and last columns in the table below. In the original press release, Allstate included comments such as one stating that Virgos are worried and shy, and they were involved in 211,650 accidents, making them the worst offenders. Allstate quickly issued an apology and retraction. In a press release, Allstate included this: 鈥淎strological signs have absolutely no role in how we base coverage and set rates. Rating by astrology would not be actuarially sound.鈥

Analyzing the Results The original Allstate press release did not include the lengths (days) of the different zodiac signs. The preceding table lists those lengths in the third column. A reasonable explanation for the different numbers of crashes is that they should be proportional to the lengths of the zodiac signs. For example, people are born under the Capricorn sign on 29 days out of the 365 days in the year, so they are expected to have 29/365 of the total number of crashes. Use the methods of this chapter to determine whether this appears to explain the results in the table. Write a brief report of your findings.

Zodiac sign

Dates

Length(days)

Crashes

Capricorn

Jan.18-Feb. 15

29

128,005

Aquarius

Feb.16-March 11

24

106,878

Pisces

March 12-April 16

36

172,030

Aries

April 17-May 13

27

112,402

Taurus

May 14-June 19

37

177,503

Gemini

June 20-July 20

31

136,904

Cancer

July21-Aug.9

20

101,539

Leo

Aug.10-Sep.15

37

179,657

Virgo

Sep.16-Oct.30

45

211,650

Libra

Oct.31-Nov 22

23

110,592

Scorpio

Nov. 23-Nov. 28

6

26,833

Ophiuchus

Nov.29-Dec.17

19

83,234

Sagittarius

Dec.18-Jan.17

31

154,477

In Exercises 5鈥20, conduct the hypothesis test and provide the test statistic and the P-value and , or critical value, and state the conclusion.

Testing a Slot Machine The author purchased a slot machine (Bally Model 809) and tested it by playing it 1197 times. There are 10 different categories of outcomes, including no win, win jackpot, win with three bells, and so on. When testing the claim that the observed outcomes agree with the expected frequencies, the author obtained a test statistic of\({\chi ^2} = 8.185\). Use a 0.05 significance level to test the claim that the actual outcomes agree with the expected frequencies. Does the slot machine appear to be functioning as expected?

Cybersecurity The table below lists leading digits of 317 inter-arrival Internet traffic times for a computer, along with the frequencies of leading digits expected with 叠别苍蹿辞谤诲鈥檚 law (from Table 11-1 in the Chapter Problem).

a. Identify the notation used for observed and expected values.

b. Identify the observed and expected values for the leading digit of 2.

c. Use the results from part (b) to find the contribution to the\({\chi ^2}\)test statistic from the category representing the leading digit of 2.

Leading Digit

1

2

3

4

5

6

7

8

9

叠别苍蹿辞谤诲鈥檚

Law

30.1%

17.6%

12.5%

9.7%

7.9%

6.7%

5.8%

5.1%

4.6%

Leading Digits

of Inter-Arrival

Traffic Times

76

62

29

33

19

27

28

21

22

In Exercises 1鈥4, use the following listed arrival delay times (minutes) for American Airline flights from New York to Los Angeles. Negative values correspond to flights that arrived early. Also shown are the SPSS results for analysis of variance. Assume that we plan to use a 0.05 significance level to test the claim that the different flights have the same mean arrival delay time.

Flight 1

-32

-25

-26

-6

5

-15

-17

-36

Flight 19

-5

-32

-13

-9

-19

49

-30

-23

Flight 21

-23

28

103

-19

-5

-46

13

-3

P-Value If we use a 0.05 significance level in analysis of variance with the sample data given in Exercise 1, what is the P-value? What should we conclude? If a passenger abhors late flight arrivals, can that passenger be helped by selecting one of the flights?

In a clinical trial of the effectiveness of echinacea for preventing

colds, the results in the table below were obtained (based on data from 鈥淎n Evaluation of Echinacea Angustifoliain Experimental Rhinovirus Infections,鈥 by Turner et al., NewEngland Journal of Medicine,Vol. 353, No. 4). Use a 0.05 significance level to test the claim that getting a cold is independent of the treatment group. What do the results suggest about the

effectiveness of echinacea as a prevention against colds?

Treatment Group


Placebo

Echinacea:

20% Extract

Echinacea:

60% Extract

Got a Cold

88

48

42

Did Not Get a Cold

15

4

10

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