Question

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1 pts Question 4 In one of Mendel's dihybrid crosses of peas, he observed round yellow, round green, wrinkled yellow, and wrinkled green F2 plants. He used the Chi Square test to see if the number of plants he observed in each category matched the expected 9:3:3:1 ratio. The Chi Square value he calculated was 6.15. Use the probability table below to determine the p-value. How should he interpret this p-value? Degrees of Probability (p-value) Freedom 0.95 0.90 0.50 0.80 0.70 0.30 0.10 0.001 0.20 0.05 0.01 0.004 0.02 0.06 0.15 0.46 1.07 2.71 3.84 6.64 10.83 1.64 0.21 0.10 0.45 1.39 2.41 0.71 3.22 5.99 9.21 13.82 4.60 0.35 0.58 3 1.01 1.42 2.37 3.66 4.64 7.82 6.25 11.34 16.27 0.71 1.06 1.65 2.20 3.36 4.88 5.99 9.49 7.78 13.28 18.47 Any variation of his data from the 9:3:3: 1 ratio may be expected by chance and are not significant; do not reject the hypothesis. These data are significantly different from the 9:3:3: 1 ratio; reject the hypothesis.

Answer 1

In Chi-square test Degree of freedom is a critical factor and it helps to determine the number of independent variables involved in an experiment. In the above example number of phenotypes observed are as 4- round yellow, round green, wrinkled yellow and wrinkled green. Hence the degree of freedom in a dihybrid cross is calculated as follows:

Degree of Freedom= No.of. observed phenotypes -1

Degree of Freedom = 4-1 =3.

$\chi$² Value calculated in the example= 6.15

So in the table we locate for the the value nearest to 6.15 for degree of freedom =3.

The nearest $\chi$² value in the table is 6.25 for degree of freedom three. In the same column we move up the rows to check the P -value. So according to the table provided for $\chi$²= 6.25 and Degree of Freedom =3, p value = 0.1.

This value is nearest to the calculated $\chi$² value= 6.15.

Hence we can extrapolate that for $\chi$²= 6.15 and Degree of Freedom=3, p value will be = 0.1.

p values > = 0.05 are not statistically significant. p value <= 0.05 means that there is a 5% chance that observed result is wrong.

Hence for the above example for p= 0.1, the data are significantly different from the 9:3:31 ration; hence hypothesis is rejected.