Respuesta :
Answer:
[tex]E_{A} =0.25*160=40[/tex]
[tex]E_{B} =0.25*160=40[/tex]
[tex]E_{C} =0.4*160=64[/tex]
[tex]E_{D} =0.05*160=8[/tex]
[tex]E_{F} =0.05*160=8[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25[/tex]
The answer would be:
c. 5.25
Step-by-step explanation:
The observed values are given by:
A: 36
B: 42
C: 60
D: 14
E: 8
Total =160
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is no difference in the proportions for the final grades
H1: There is a difference in the proportions for the final grades
The level of significance assumed for this case is [tex]\alpha=0.01[/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]
Now we just need to calculate the expected values with the following formula [tex]E_i = \% * total[/tex]
And the calculations are given by:
[tex]E_{A} =0.25*160=40[/tex]
[tex]E_{B} =0.25*160=40[/tex]
[tex]E_{C} =0.4*160=64[/tex]
[tex]E_{D} =0.05*160=8[/tex]
[tex]E_{F} =0.05*160=8[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25[/tex]
The answer would be:
c. 5.25
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(categories-1)=(5-1)=4[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{4} >5.25)=0.263[/tex]
The p value is higher than the significance so we have enough evidence to FAIL to reject the null hypothesis
