Respuesta :
Answer:
Step-by-step explanation:
From the given information:
Null and alternative hypothesis is:
[tex]\mathbf{H_o: \text{The die is not loaded i.e. six numbbers are equally alike}}[/tex]
[tex]\mathbf{H_a: \text{The die is loaded i.e. six numbbers are not equally alike}}[/tex]
Numbers Observed Expected (O - E) (O-E)^2 (O-E)^2/E
Frequency (O) Frequency (E)
1 31 30 1 1 0.03
2 34 30 4 16 0.53
3 26 30 -4 16 0.53
4 16 30 -14 196 6.53
5 32 30 2 4 0.13
6 41 30 11 121 4.03
Total 180 [tex]X^2= \sum (\dfrac{O-E}{E})^2=11.78[/tex]
degree of freedom = n - 1
= 6 - 1
= 5
Critical value at [tex]X^2_{0.05/2,5} =11.07[/tex]
Since the calculated [tex]X^2 \ is \ > X^2_{0.025/5}[/tex] , then we reject [tex]H_o[/tex]
Conclusion: Accept the alternative hypothesis.
The information provided gives sufficient evidence for us to conclude that the given die is loaded.