91影视

The frequency of 4 different kinds of seeds are recorded. The expected frequencies should occur in the ratio 9:3:3:1.

Assume that each experimental unit is selected randomly.

The test meets the requirements if the expected values are larger than 5.

Let O denote the observed frequencies of people of different races.

The following values are obtained:

\(\begin{aligned}{l}{O_1} = 307\\{O_2} = 77\\{O_3} = 98\\{O_4} = 18\end{aligned}\)

The sum of all observed frequencies is computed below:

\(\begin{aligned}{c}n = 307 + 77 + .... + 18\\ = 500\end{aligned}\)

Let E denote the expected frequencies.

Let the yellow smooth seed type be denoted by 1, green smooth seed type be denoted by 2, yellow wrinkled seed type be denoted by 3 and green wrinkled seed type be denoted by 4.

The expected frequencies will occur in the given ratio of 9:3:3:1.

Therefore, the expected frequencies are computed below:

\(\begin{aligned}{c}{E_1} = \frac{9}{{9 + 3 + 3 + 1}}\left( {500} \right)\\ = \frac{9}{{16}}\left( {500} \right)\\ = 281.25\end{aligned}\)

\(\begin{aligned}{c}{E_2} = \frac{3}{{9 + 3 + 3 + 1}}\left( {500} \right)\\ = \frac{3}{{16}}\left( {500} \right)\\ = 93.75\end{aligned}\)

\(\begin{aligned}{c}{E_3} = \frac{3}{{9 + 3 + 3 + 1}}\left( {500} \right)\\ = \frac{3}{{16}}\left( {500} \right)\\ = 93.75\end{aligned}\)

\[\begin{aligned}{c}{E_4} = \frac{1}{{9 + 3 + 3 + 1}}\left( {500} \right)\\ = \frac{1}{{16}}\left( {500} \right)\\ = 31.25\end{aligned}\]

Thus, the requirements are satisfied.

The hypotheses are,

\({H_0}:\)The observed frequencies of the 4 types of seeds are in the expected ratio as proposed by Mendel.

\({H_a}:\)The observed frequencies of the 4 types of seeds do not occur in the expected ratio as proposed by Mendel.

The table below shows the necessary calculations:

The value of the test statistic is equal to:

\(\begin{aligned}{c}{\chi ^2} = \sum {\frac{{{{\left( {O - E} \right)}^2}}}{E}} \\ = 2.357556 + 2.992667 + ...... + 5.618\\ = 11.161\end{aligned}\)

Thus,\({\chi ^2} = 11.161\).

Let k be the number of seed types, which is 4.

The degrees of freedom for\({\chi ^2}\)is computed below:

\(\begin{aligned}{c}df = k - 1\\ = 4 - 1\\ = 3\end{aligned}\)

The critical value of\({\chi ^2}\)at

\(\alpha = 0.05\)with 3 degrees of freedom is equal to 7.815.

The p-value is equal to 0.011.

Since the test statistic value is greater than the critical value and the p-value is less than 0.05, the null hypothesis is rejected.

There is enough evidence to conclude that theobserved frequencies of the 4 types of seeds do not occur in the expected ratio as proposed by Mendel.

Thus, results contradict Mendel鈥檚 theory at 0.05 level of significance.