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.

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

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 () =

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

() white = = 1

() purple = = 0.33

Hence, total () = 1.33

The decision rule is such that;

In this case, the calculated () 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.

measuring physical properties lab reportinstructions: in the physical properties lab, you will test the physical properties of different building materials in order to pick the appropriate materials for a new home in tiny world. record your observations and test measurements in the lab report below. youwill submit your completed report.name and title: lyric lawton, mr.herbert, 10/13/17, physical properties lab. objective(s): i need to parvus build his house out of the right material, so he will stay cool in the mountains.hypothesis: if we use wood to build the roof of the house, then the temperature inside the house will remain cooler.if we use clay brick to build the walls and floors of the house, then the temperature inside the house will remain cooler.if we use aluminium to bring electricity into the home, then it will allow electricity to flow into the home at a faster rate.if we use iron to construct the latches on the windows and doors, then the magnetism will keep the latches secure.

on island x the species needed to adapt because they were being subject to predators which didn't inhabit island y.