Question

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.

Answer 1

Explanation:

From the given information:

Null and alternative hypothesis is:

`\mathbf{H_o: \text{The die is not loaded i.e. six numbbers are equally alike}}`

`\mathbf{H_a: \text{The die is loaded i.e. six numbbers are not equally alike}}`

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
`X^2= \sum ((O-E)/(E))^2=11.78`

degree of freedom = n - 1

= 6 - 1

= 5

Critical value at
`X^2_(0.05/2,5) =11.07`

Since the calculated
`X^2 \ is \ > X^2_(0.025/5)` , then we reject
`H_o`

Conclusion: Accept the alternative hypothesis.

The information provided gives sufficient evidence for us to conclude that the given die is loaded.