Question

In a clinical trial of the effectiveness of echinacea for preventing

colds, the results in the table below were obtained (based on data from “An Evaluation of Echinacea Angustifoliain Experimental Rhinovirus Infections,” 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

Short answer

Getting a cold is independent of the treatment group. Thus, Echinacea is not effective to prevent colds.

Step-by-step solution

Given information

The data forthe effectiveness of Echinacea for preventing colds is provided.

The level of significance is 0.05.

Compute the expected frequencies

Assume that random selections are done for subjects, and each subject is assigned randomly to each group.

Theexpected frequency formulais computed as,

\(E = \frac{{\left( {row\;total} \right)\left( {column\;total} \right)}}{{\left( {grand\;total} \right)}}\)

The row and column total for the observed frequencies is represented as,

	Placebo	Echinacea: 20% Extract	Echinacea: 60% Extract	Row Total
Got a Cold	88	48	42	178
Did Not Get a Cold	15	4	10	29
Column Total	103	52	52	207

Theexpected frequency tableis represented as,

	Placebo	Echinacea: 20% Extract	Echinacea: 60% Extract
Got a Cold	88.5700	44.7150	44.7150
Did Not Get a Cold	14.4300	7.2850	7.2850

Each expected value is greater than 5.

Thus, the requirements for the test are satisfied.

State the null and alternate hypothesis

The hypotheses are stated as:

\({H_0}:\)Getting a cold is independent of the treatment group.

\({H_1}:\)Getting a cold is dependent on the treatment group.

Compute the test statistic

The value of the test statistic is computed as,

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = \frac{{{{\left( {88 - 88.5700} \right)}^2}}}{{88.5700}} + \frac{{{{\left( {48 - 44.7150} \right)}^2}}}{{44.7150}} + ... + \frac{{{{\left( {10 - 7.2850} \right)}^2}}}{{7.2850}}\\ = 2.925\end{aligned}\)

Therefore, the value of the test statistic is 2.925.

Compute the degrees of freedom

The degrees of freedomare computed as,

\(\begin{aligned}{c}\left( {r - 1} \right)\left( {c - 1} \right) = \left( {2 - 1} \right)\left( {3 - 1} \right)\\ = 2\end{aligned}\)

Therefore, the degrees of freedom are 2.

Compute the critical value

From the chi-square table, the critical value for the row corresponding to 2 degrees of freedom and at 0.05 level of significance 5.991.

Therefore, the critical value is 5.991.

The P-value is obtained as 0.232.

State the decision

Since the critical value (5.991) is greater than the value of the test statistic (2.925). In this case, the null hypothesis fails to be rejected.

Therefore, the decision is that null hypothesis is failed to be rejected.

The P-value is obtained as 0.2316.

State the conclusion

There issufficient evidence to support the claimthat getting a cold is independent of the treatment group.

Thus, it can be said that Echinacea is not effective to prevent colds.