Answer:
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(27.356,9,TRUE)"
Since the p value is lower than the significance level assumed 0.05 we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.
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:
Amusement Parks Mexico Home Other Total
Freshman 23 21 43 21 108
Sophomore 34 23 14 26 97
Junior 25 30 23 26 104
Senior 27 33 17 19 96
Total 109 107 97 92 405
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is independence between the two random variables
H1: There is dependence between the two variables
The level os significance assumed for this case is
The statistic to check the hypothesis is given by:
The table given represent the observed values, we just need to calculate the expected values with the following formula
And the calculations are given by:
And the expected values are given by:
Amusement Parks Mexico Home Other Total
Freshman 29.07 28.53 25.87 24.53 108
Sophomore 26.11 25.63 23.23 22.03 97
Junior 27.99 27.48 24.91 23.62 104
Senior 25.84 25.36 22.99 21.81 96
Total 109 107 97 92 405
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(27.356,9,TRUE)"
Since the p value is lower than the significance level assumed 0.05 we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.
