Answer:
Step-by-step explanation:
Hello!
Given the variables
X: Curfew of middle school students (Categorized Yes/No)
Y: Average grade of middle school students. (Categorized: A, B, C, and D)
The objective is to test if there is an association between both variables, you have to conduct a Chi-Square test of independence.
In the null hypothesis, you state that both variables are independent vs. the alternative hypothesis that the variables are dependent.
The hypotheses for this test are:
H₀: Pij= Pi. * P.j i= (1)Yes, (2)No; j= (1)A, (2)B, (3)C, (4)D
H₁: Te variables are not independent.
α: 0.05
[tex]X^2= sum\frac{(O_{ij}-E_{ij})^2}{E_{ij}} ~~X_{(r-1)*(c-1)}[/tex]
Where
Oij is the observed frequency for the i-row and j-column
Eij is the expected frequency for the i-row and j-column
r= number of categories in the rows
c= number of categories in the columns
[tex]X^2_{H_0}= 1.4796703[/tex]
p-value: 0.687
The p-value is greater than the significance level, so the decision is to not reject the null hypothesis. So using a 5% significance level, you can conclude that having a curfew and the average grades of middle schoolers are two independent variables.
Answer:
Nothing, because the conditions for use of the chi-square test are not met.
Step-by-step explanation:
Correct. The conditions for the chi-square test are not met. Not all of the expected counts are 5 or greater, so the chi-square model is not a good fit for the distribution of test statistics. In this case, we cannot trust the accuracy of the P-value.
