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

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.

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.

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.

The table below shows the necessary calculations:

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.