Respuesta :
Answer:
[tex]\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=6-1=5[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{5} >6.7)=0.244[/tex]
Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely
Step-by-step explanation:
We need to conduct a chi square test in order to check the following hypothesis:
H0: There is no difference in the frequencies
H1: There is a difference in the frequencies
The level of significance assumed for this case is [tex]\alpha=0.01[/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 observed values are:
Value 1 2 3 4 5 6
Frequency 26 32 44 37 27 34
And the expected values are for this case the same [tex] E_i = \frac{200}{6}= 33.33[/tex]
And now we can calculate the statistic:
[tex]\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=6-1=5[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{5} >6.7)=0.244[/tex]
Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely
