Answer :
Answer:
[tex]\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295[/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} >18.295)=0.00038[/tex]
Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year
Step-by-step explanation:
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is no difference of birthdates distributed throughout the year
H1: There is a difference between birthdates distributed throughout the year
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{total}{4}[/tex]
And replacing we got:
[tex]E_{1} =\frac{67+56+30+37}{4}=47.5[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295[/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} >18.295)=0.00038[/tex]
Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the year