Answer: Let

• \(H_0\): eye colour is uniformly distributed between brown, blue and other, so each has probability \(\frac13\).

• \(H_1\): eye colour is not uniformly distributed.

Expected frequencies for a sample of 120 are \(40,40,40\).

The test statistic is

\(\chi^2=\frac{(38-40)^2}{40}+\frac{(52-40)^2}{40}+\frac{(30-40)^2}{40}=0.1+3.6+2.5=6.2\).

Degrees of freedom \(=3-1=2\).

At the 5% level, the critical value is \(5.991\).

Since \(6.2\gt 5.991\), reject \(H_0\).

There is sufficient evidence at the 5% significance level to conclude that eye colour is not uniformly distributed, so the scientist's claim is not supported by the data.

We use a \(\chi^2\) goodness of fit test.

The scientist claims the three eye-colour categories are equally likely, so under the null hypothesis the distribution is uniform.

• \(H_0\): the population proportions are brown : blue : other \(= \frac13 : \frac13 : \frac13\).

• \(H_1\): the population proportions are not all equal.

The observed frequencies are:

If the distribution is uniform, each category should contain \(\frac{120}{3}=40\) people.

So the expected frequencies are:

Now calculate the test statistic:

\(\chi^2=\sum \frac{(O-E)^2}{E}\)

\(\chi^2=\frac{(38-40)^2}{40}+\frac{(52-40)^2}{40}+\frac{(30-40)^2}{40}\)

\(=\frac{4}{40}+\frac{144}{40}+\frac{100}{40}\)

\(=0.1+3.6+2.5\)

\(=6.2\)

The number of degrees of freedom is

\(3-1=2\).

For \(2\) degrees of freedom at the 5% significance level, the critical value is \(5.991\).

Since \(6.2\gt 5.991\), the test statistic lies in the critical region, so we reject \(H_0\).

Therefore, there is sufficient evidence to suggest that the eye colours are not uniformly distributed in the population. The scientist's claim is not supported at the 5% level.