A student carries out a genetics experiment with fruit flies to investigate the inheritance pattern of the white eye trait. The student crosses a homozygous white-eyed female with a wild-
type male and records observations about the flies in the F, generation. The student plans to use the F, data to perform a chi-square goodness-of-fit test for a model based on an X-linked
recessive pattern of inheritance. The student will use one degree of freedom and a significance level of p = 0.05. The setup for the student's chi-square goodness-of-fit test is presented in
Table 1
Table 1. Setup for the student's chl-square goodness-of-fit test
Phenotype
Red-eyed female
White-eyed male
Observed
53
47
Expected
50
50
The student calculates a chi-square value of 0.36. Which of the following statements best completes the student's chi-square goodness-of-fittest?

Chi square test helps to test null hypothesis. The option C is correct. The null hypothesis cannot be rejected.

Chi square Test:

The test is performed to show the statistical deference between the observed and expected value. It helps to test null hypothesis.

The formula of the Chi square test,

Where

Oi - The observed values

Ei - expected values.

Test for Red eyed female,

Oi -  53

Ei = 50

So, test results

For the White Eyed Male,

Oi = 47

Ei = 50

The test results,

Sum of the both experiments is 0.36.

The value for chi square test is less than 3.84, Hence, null hypothesis cannot be rejected.

Therefore, the option C is correct. The null hypothesis cannot be rejected.

To know more about Chi square test, refer to the link:

link

Full question attached

c

Explanation:

To carry out the Chi squared test the following formula is used:

(O - E)² / E

Where O is the observed values and E is the expected values. The sum of this equation for each phenotype is the Chi squared value

Observed: 53

Expected: 50

O - E= -3

(O - E)² = 9

(O - E)² / E = 9/ 50 = 0.18

Observed: 47

Expected: 50

O - E= 3

(O - E)² = 9

(O - E)² / E = 9/ 50 = 0.18

Sum = 0.36

The p value is dependent upon the number of degrees of freedom. The degrees of freedom is the number of categories in the experiment - 1. So here, there is one degree of freedom are there are two phenotypes being observed.

If we look up the critical values for each degree of freedom, we can see that for 1 degree of freedom the critical value is 3.841

Since the chi squared value is less than 3.84 (it is 0.36), we cannot reject the null hypothesis. So the answer is c

the temperature of the gas, in k, not celsius.