Answer:
![\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860](https://tex.z-dn.net/?f=%5Cchi%5E2%20%3D%20%5Cfrac%7B%2827-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2831-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2842-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2840-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2828-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2832-33.33%29%5E2%7D%7B33.33%7D%20%3D5.860)
![p_v = P(\chi^2_{5} >5.860)=0.32](https://tex.z-dn.net/?f=p_v%20%3D%20P%28%5Cchi%5E2_%7B5%7D%20%3E5.860%29%3D0.32)
Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.
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".
Assume the following dataset:
Number: 1, 2 , 3 , 4 , 5 ,6
Frequency: 27, 31, 42, 40, 28, 32
We need to conduct a chi square test in order to check the following hypothesis:
H0: The outcomes are equally likely.
H1: The outcomes are not equally likely.
The level of 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 observed values are given:
![O_{2}=31](https://tex.z-dn.net/?f=O_%7B2%7D%3D31)
![O_{4}=40](https://tex.z-dn.net/?f=O_%7B4%7D%3D40)
![O_{6}=32](https://tex.z-dn.net/?f=O_%7B6%7D%3D32)
The expected values are given by:
![E_{2} =\frac{1}{6}*200=33.33](https://tex.z-dn.net/?f=E_%7B2%7D%20%3D%5Cfrac%7B1%7D%7B6%7D%2A200%3D33.33)
![E_{4} =\frac{1}{6}*200=33.33](https://tex.z-dn.net/?f=E_%7B4%7D%20%3D%5Cfrac%7B1%7D%7B6%7D%2A200%3D33.33)
![E_{6} =\frac{1}{6}*200=33.33](https://tex.z-dn.net/?f=E_%7B6%7D%20%3D%5Cfrac%7B1%7D%7B6%7D%2A200%3D33.33)
And now we can calculate the statistic:
![\chi^2 = \frac{(27-33.33)^2}{33.33}+\frac{(31-33.33)^2}{33.33}+\frac{(42-33.33)^2}{33.33}+\frac{(40-33.33)^2}{33.33}+\frac{(28-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33} =5.860](https://tex.z-dn.net/?f=%5Cchi%5E2%20%3D%20%5Cfrac%7B%2827-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2831-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2842-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2840-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2828-33.33%29%5E2%7D%7B33.33%7D%2B%5Cfrac%7B%2832-33.33%29%5E2%7D%7B33.33%7D%20%3D5.860)
Now we can calculate the degrees of freedom for the statistic given by:
![df=Categories-1=6-1=5](https://tex.z-dn.net/?f=df%3DCategories-1%3D6-1%3D5)
And we can calculate the p value given by:
![p_v = P(\chi^2_{5} >5.860)=0.32](https://tex.z-dn.net/?f=p_v%20%3D%20P%28%5Cchi%5E2_%7B5%7D%20%3E5.860%29%3D0.32)
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(5.860,5,TRUE)"
Since the p value is higher than the significance level assumed 0.05 we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that we can assume that we have equally like results.