91Ó°ÊÓ

The goodness of fit test is the statistical method that helps determine the accuracy between the expected and observed data.The deviations between these two data sets are easily predicted from this test. It is used to determine the accuracy of the genetic crosses between two different parents and their possible outcomes.

The expected values of the unlinked genes are calculated as follows:

Expected number\( = \frac{{tota\ln umber}}{{no.of.offsprings}} = \frac{{900}}{4} = 225\)

The deviation value is calculated as follows:

\(o - e = 220 - 225 = - 5\)

\(o - e = 210 - 225 = - 15\)

\(o - e = 231 - 225 = 6\)

\(o - e = 239 - 225 = 14\)

The value of (o-e)²is estimated as follows:

\(\begin{aligned}{l}{\left( {{\bf{o}} - {\bf{e}}} \right)^{\bf{2}}} = {( - 5)^2} = 25\\{\left( {{\bf{o}} - {\bf{e}}} \right)^{\bf{2}}} = {( - 15)^2} = 225\\{\left( {{\bf{o}} - {\bf{e}}} \right)^{\bf{2}}} = {(6)^2} = 36\\{\left( {{\bf{o}} - {\bf{e}}} \right)^{\bf{2}}} = {(14)^2} = 196\end{aligned}\)

The value of (o-e)²/e is estimated as follows:

\(\begin{aligned}{l}\frac{{\begin{aligned}{*{20}{l}}{{{\left( {{\bf{o}} - {\bf{e}}} \right)}^{\bf{2}}}}\end{aligned}}}{e} = \frac{{25}}{{225}} = 0.11\\\frac{{\begin{aligned}{*{20}{l}}{{{\left( {{\bf{o}} - {\bf{e}}} \right)}^{\bf{2}}}}\end{aligned}}}{e} = \frac{{225}}{{225}} = 1\\\frac{{\begin{aligned}{*{20}{l}}{{{\left( {{\bf{o}} - {\bf{e}}} \right)}^{\bf{2}}}}\end{aligned}}}{e} = \frac{{36}}{{225}} = 0.16\\\frac{{\begin{aligned}{*{20}{l}}{{{\left( {{\bf{o}} - {\bf{e}}} \right)}^{\bf{2}}}}\end{aligned}}}{e} = \frac{{196}}{{225}} = 0.87\end{aligned}\)

The\({\chi ^2}\)is estimated as follows:

\({\chi ^2} = 0.11 + 1 + 0.16 + 0.87 = 2.14\)

The\({\chi ^2}\)is calculated by adding up all the estimated values in the last column and predicted as 2.14.

The values are calculated above and listed in the table by the given conditions.