Respuesta :
Answer:
a) [tex]\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6[/tex]
b) [tex]df=Categories-1=10-1=9[/tex]
Step-by-step explanation:
We assume the following info:
Favorite Subject Number of students
English 25
Math 30
Science 30
Art/Music 15
Total 100
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".
Part a
The system of hypothesis on this case are:
H0: There is no difference with the distribution proposed
H1: There is a difference with the distribution proposed
The level os significance assumed for this case is [tex]\alpha=0.05[/tex]
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]
The table given represent the observed values, we just need to calculate the expected values are 25 for each category.
And the calculations are given by:
[tex]E_{English} =25[/tex]
[tex]E_{Math} =25[/tex]
[tex]E_{Science} =25[/tex]
[tex]E_{Music} =25[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(25-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(30-25)^2}{25}+\frac{(15-25)^2}{25}=6[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=Categories-1=4-1=3[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{3} >6)=0.112[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(6,3,TRUE)"
Part b
For this case we have this formula:
[tex]df=Categories-1=10-1=9[/tex]
