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

The level of significance is 0.05.

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,

Theexpected frequency tableis represented as,

Each expected value is greater than 5.

Thus, the requirements for the test are satisfied.

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.

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.

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.

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.

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.

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.