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Interpret the results of the chi-square test.A die is rolled 180 times and the following data are obtained.Number Frequency 1 31 2 34 3 26 4 16 5 32 6 41A chi-square test was conducted to determine, at the 5% significance level, whether or not the die is loaded (i.e., that the six numbers are not equally likely).Carry out the hypothesis test.

Sagot :

ajeigbeibraheem

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.

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