Question

You self-pollinate a heterozygous purple flower. The following were your observed phenotypic ratio for 100 flowers: 70 purple flowers and 30 white flowers. Note: Purple is the dominant phenotype. The critical value for 95% confidence is 3.841 Answer the following questions: 1. What is your null hypothesis? 2. What is your degree of freedom? 3. What is your chi squared value? 4. With 95% confidence, is your null hypothesis supported?! Null hypothesis: There is statically significant difference between our expected and observed values. Degree of freedom: 1 Chi squared: 1.333 Null hypothesis is supported since our chi squared value is larger than our critical value, there are no statically significant difference. Null hypothesis: There is no statically significant difference between our expected and observed values. Degree of freedom: 1 Chi squared: 1.333 Null hypothesis is supported since our chi squared value is lower than our critical value, there are no statically significant difference Null hypothesis: There is no statically significant difference between our expected and observed values. Degree of freedom: 1 Chi squared: 3.98 Null hypothesis is supported since our chi squared value is lower than our critical value, there are no statically significant difference Null hypothesis: There is statically significant difference between our expected and observed values. Degree of freedom: 2 Chi squared: 1333 Null hypothesis is supported since our chi squared value is lower than our critical value, there are no statically significant difference.

Answer 1

Null hypothesis: There is no statistically significant difference between our expected and observed values.

Degree of freedom: 1

Chi squared: 1.333

Null hypothesis is supported since our chi squared value is lower than our critical value, there are no statistically significant difference

Step-by-step explanation:

The Null hypothesis is a hypothesis that stipulates that there is no difference between the population being compared during statistical analyses/tests. Hence, in this case, the Null hypothesis posits that there is no significant difference between the expected and the observed phenotypic ratio.

Degree of freedom is calculated as n-1, where n = number of phenotypes in this case. Hence, Degree of freedom = 2 - 1 which is equal to 1.

Chi square (
`X^(2)`) =
`(Observed frequency - Expected frequency)^(2)/Expected frequency`

Since purple is dominant over white, the phenotype of the offspring should be 3:1 purple:white according to law of segregation. Hence, since there are 100 offspring in total;

Expected frequencies: purple 75, white 25

Observed frequencies: purple 70, white 30

(
`X^(2)`) white =
`(30-25)^(2)/25` = 1

(
`X^(2)`) purple =
`(70-75)^(2)/75` = 0.33

Hence, total (
`X^(2)`) = 1.33

The decision rule is such that;

• If calculated (
  `X^(2)`) value is greater than critical value, the Null hypothesis is supported. If it is otherwise, the Null hypothesis is supported.

In this case, the calculated (
`X^(2)`) value is 1.33 and the critical value is 3.841, hence calculated value is less than the critical value. The Null hypothesis is supported.