Answer:
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
The statistic to check the hypothesis is given by:
The observed values are given:
The expected values are given by:
And now we can calculate the statistic:
Now we can calculate the degrees of freedom for the statistic given by:
And we can calculate the p value given by:
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.
