Respuesta :
Answer:
[tex]df=categor-1=6-1=5[/tex]
The critical value can be founded with the following Excel formula:
=CHISQ.INV(1-0.05,5)
And we got [tex] \chi^2_{critc}= 11.0705[/tex]
a. 11.070
And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance
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".
Solution to the problem
For this case we want to test:
H0: Absenteeism is distributed evenly throughout the week
H1: Absenteeism is NOT distributed evenly throughout the week
We have the following data:
Monday Tuesday Wednesday Thursday Friday Saturday Total
12 9 11 10 9 9 60
The level of significance assumed for this case is [tex]\alpha=0.05[/tex]
The statistic to check the hypothesis is given by:
[tex]\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{60}{6}= 10[/tex] and the expected value is the same for all the days since that's what we want to test.
now we can calculate the statistic:
[tex]\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8[/tex]
Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:
[tex]df=categor-1=6-1=5[/tex]
The critical value can be founded with the following Excel formula:
=CHISQ.INV(1-0.05,5)
And we got [tex] \chi^2_{critc}= 11.0705[/tex]
a. 11.070
And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance
