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.