Respuesta :
Answer:
a)[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]
[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(17.03,4,TRUE)"
Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.
b)
P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714
P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375
P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5
P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357
P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375
P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2
P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714
P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125
P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3
And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.
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:
Quality management     Excellent    Good   Fair   Total
Excellent                 40         35     25    100
Good                    25         35     10     70
Fair                     5          10      15     30
Total                    70         80     50    200
Part a
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is independence between the two categorical variables
H1: There is association between the two categorical variables
The level of 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 with the following formula [tex]E_i = \frac{total col * total row}{grand total}[/tex]
And the calculations are given by:
[tex]E_{1} =\frac{70*100}{200}=35[/tex]
[tex]E_{2} =\frac{80*100}{200}=40[/tex]
[tex]E_{3} =\frac{50*100}{200}=25[/tex]
[tex]E_{4} =\frac{70*70}{200}=24.5[/tex]
[tex]E_{5} =\frac{80*70}{200}=28[/tex]
[tex]E_{6} =\frac{50*70}{200}=17.5[/tex]
[tex]E_{7} =\frac{70*30}{200}=10.5[/tex]
[tex]E_{8} =\frac{80*30}{200}=12[/tex]
[tex]E_{9} =\frac{50*30}{200}=7.5[/tex]
And the expected values are given by:
Quality management     Excellent    Good   Fair    Total
Excellent                 35        40      25     100
Good                    24.5      28      17.5     85
Fair                     10.5       12      7.5     30
Total                    70         80     65     215
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(40-35)^2}{35}+\frac{(35-40)^2}{40}+\frac{(25-25)^2}{25}+\frac{(25-24.5)^2}{24.5}+\frac{(35-28)^2}{28}+\frac{(25-17.5)^2}{17.5}+\frac{(5-10.5)^2}{10.5}+\frac{(10-12)^2}{12}+\frac{(15-7.5)^2}{7.5} =17.03[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(rows-1)(cols-1)=(3-1)(3-1)=4[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{4} >17.03)=0.0019[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(17.03,4,TRUE)"
Since the p value is lower than the significance level we can reject the null hypothesis at 5% of significance, and we can conclude that we have association or dependence between the two variables.
Part b
We can find the probabilities that Quality of Management and the Reputation of the Company would be the same like this:
Let's define some notation first.
E= Quality Management excellent   Ex=Reputation of company excellent
G= Quality Management good   Gx=Reputation of company good
F= Quality Management fait   Ex=Reputation of company fair
P(EΛ Ex) =40/215=0.186
P(GΛ Gx) =35/215=0.163
P(FΛ Fx) =15/215=0.0697
If we have dependence then the conditional probabilities would be higher values.
P(E|Ex)= P(EΛEx )/ P(Ex) = (40/215)/ (70/215)= 40/70=0.5714
P(E|Gx)= P(EΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375
P(E|Fx)= P(EΛFx )/ P(Fx) = (25/215)/ (50/215)= 25/50=0.5
P(G|Ex)= P(GΛEx )/ P(Ex) = (25/215)/ (70/215)= 25/70=0.357
P(G|Gx)= P(GΛGx )/ P(Gx) = (35/215)/ (80/215)= 35/80=0.4375
P(G|Fx)= P(GΛFx )/ P(Fx) = (10/215)/ (50/215)= 10/50=0.2
P(F|Ex)= P(FΛEx )/ P(Ex) = (5/215)/ (70/215)= 5/70=0.0714
P(F|Gx)= P(FΛGx )/ P(Gx) = (10/215)/ (80/215)= 10/80=0.125
P(F|Fx)= P(FΛFx )/ P(Fx) = (15/215)/ (50/215)= 15/50=0.3
And that's what we see here almost all the conditional probabilities are higher than 0.2 so then the conclusion of dependence between the two variables makes sense.
