Questions: PHENOTYPES, OBSERVE, EXPECTED, O-E, (O-E)^2 Dark Brown eyes, 125, , , Light brown cyes, 78, , , Light brown eyes, 87, , , White cyes, 25, , , TOTAL, , , ,

Solution Steps

The expected frequencies for each phenotype, assuming equal distribution, are calculated as follows:

\[ E = \frac{N}{k} = \frac{315}{4} = 78.75 \]

where \(N\) is the total number of observations (315) and \(k\) is the number of phenotypes (4). Thus, the expected frequencies are:

\[ E = [78.75, 78.75, 78.75, 78.75] \]

The Chi-Square test statistic (\(\chi^2\)) is calculated using the formula:

\[ \chi^2 = \sum_i \frac{(O_i - E_i)^2}{E_i} \]

Substituting the observed (\(O\)) and expected (\(E\)) values:

\[ \chi^2 = \frac{(125 - 78.75)^2}{78.75} + \frac{(78 - 78.75)^2}{78.75} + \frac{(87 - 78.75)^2}{78.75} + \frac{(25 - 78.75)^2}{78.75} \]

Calculating this gives:

\[ \chi^2 = 64.7206 \]

The degrees of freedom (\(df\)) for this test is given by:

\[ df = k - 1 = 4 - 1 = 3 \]

Using a significance level of \(\alpha = 0.05\), the critical value for the Chi-Square distribution with 3 degrees of freedom is:

\[ \chi^2(0.95, 3) = 7.8147 \]

The p-value associated with the calculated Chi-Square statistic is:

\[ P = P(\chi^2 > 64.7206) = 0.0 \]

Final Answer