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The goodness of fit is measured by\({\chi ^{^2}}\). This statistic measures the amounts by which the observed values differ from their respective predictions to indicate how closely the two sets of values match. The formula for calculating this value is

\({\chi ^{}} = \sum \frac{{{{\left( {o - e} \right)}^2}}}{e}\)

Where o=observed and e= expected. Calculate the\({\chi ^{^2}}\)value for the data using the table below. Fill out the table, carrying out the operations indicated in the top row. Then add up the entries in the last column to find the\({\chi ^{^2}}\)value.

Testcross Offspring

Expected

(e)

Observed

(o)

Deviation

(o-e)

(o-e)2

(o-e)2/e

(A-B-)

220

(aaB-)

210

(A-bb)

231

(aabb)

239

\({\chi ^2}\) =sum

Short Answer

Expert verified

The expected number of offspring is estimated at 225. The deviation values are calculated as -5,-15, 4,16. The value of (o-e)2 is calculated as 25, 225, 36, 196. The value of (o-e)2/e is calculated as 0.11,1, .16, and .87 , respectively. The sum is estimated as 2.14.

Step by step solution

01

Goodness of fit

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.

02

Calculation of \({\chi ^{^2}}\)value

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.

03

Tabulation with the expected number results

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

Testcross Offspring

Expected

(e)

Observed

(o)

Deviation

(o-e)

(o-e)2

(o-e)2/e

(A-B-)

225

220

-5

25

0.11

(aaB-)

225

210

-15

225

1

(A-bb)

225

231

6

36

0.16

(aabb)

225

239

14

196

0.87

\({\chi ^2}\) =

2.14

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