Answer:
[tex] df = n-1=10-1=9[/tex]
a. 9
Step-by-step explanation:
Previous concepts
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 assume that we have the following system of hypothesis:
H0: The data follows the distribution proposed
H1: The data not follows the distribution proposed
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
This statistic have a Chi Square distribution distribution with k-1 degrees of freedom, where n represent the number of categories on this case k=10. And if we find the degrees of freedom we got:
[tex] df = k-1=10-1=9[/tex]
a. 9
