Answer:
A. We should reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
Step-by-step explanation:
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
We need to conduct a chi square test in order to check the following hypothesis:
H0: The student answers have the uniform distribution.
H1: The student answers do not have the uniform distribution.
The level os significance assumed for this case is ![\alpha=0.05](https://tex.z-dn.net/?f=%5Calpha%3D0.05)
The statistic to check the hypothesis is given by:
![\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}](https://tex.z-dn.net/?f=%5Cchi%5E2%20%3D%5Csum_%7Bi%3D1%7D%5En%20%5Cfrac%7B%28O_i%20-E_i%29%5E2%7D%7BE_i%7D)
The table given represent the observed values, we just need to calculate the expected values with the following formula ![E_i = \frac{total col * total row}{grand total}](https://tex.z-dn.net/?f=E_i%20%3D%20%5Cfrac%7Btotal%20col%20%2A%20total%20row%7D%7Bgrand%20total%7D)
On this case we assume that the calculated statistic is given by:
Statistic calculated
![\chi^2_{calc}=13.167](https://tex.z-dn.net/?f=%5Cchi%5E2_%7Bcalc%7D%3D13.167)
P value
Assuming the we have 2 rows and 4 columns on the contingency table.
Now we can calculate the degrees of freedom for the statistic given by:
![df=(rows-1)(cols-1)=(2-1)(4-1)=3](https://tex.z-dn.net/?f=df%3D%28rows-1%29%28cols-1%29%3D%282-1%29%284-1%29%3D3)
We can calculate the critical value with this formula in excel:" =CHISQ.INV(0.95,3)" On this case we got that the critical value is:
![\chi^2_{crit}=7.815](https://tex.z-dn.net/?f=%5Cchi%5E2_%7Bcrit%7D%3D7.815)
Since our calculated value is higher than the cirtical value we have enough evidence to reject the null hypothesis at the significance level of 5%.
And we can also calculate the p value given by:
![p_v = P(\chi^2_{3} >13.167)=0.0043](https://tex.z-dn.net/?f=p_v%20%3D%20P%28%5Cchi%5E2_%7B3%7D%20%3E13.167%29%3D0.0043)
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(13.167,3,TRUE)"
Since the p value is lower than the significance level we reject the null hypothesis at 5% of significance.
A. We should reject H0.
C. At the 5% significance level, there is sufficient evidence to conclude that the students were not guessing.
The reason why we select option C is because if we reject the null hypothesis of uniform distribution then we are rejecting the claim that the students are guessing.