Answer :
Let's go step-by-step through the problem and its solution.
### (a) State the null and alternative hypotheses
The hypotheses for the chi-square Goodness-of-Fit test are:
- [tex]\( H_0 \)[/tex] (Null hypothesis): The die has a uniform distribution.
- [tex]\( H_a \)[/tex] (Alternative hypothesis): The die does not have a uniform distribution.
### (b) Given information
- The chi-square test statistic, [tex]\(\chi_0^2\)[/tex], is 11.692.
### (c) Determine the critical value
To determine the critical value, we need to look at the chi-square table provided for the given significance level [tex]\(\alpha = 0.01\)[/tex] and degrees of freedom (df).
- Degrees of freedom (df): For a fair die with 6 faces, all equally likely, we would typically have [tex]\(df = 6 - 1 = 5\)[/tex]. However, based on the problem setup, the degrees of freedom provided for the test is 3.
From the given chi-square table at [tex]\(\alpha = 0.01\)[/tex] for 3 degrees of freedom:
- The [tex]\(\chi^2\)[/tex] critical value is 11.345.
### Decision Rule
Compare the test statistic [tex]\(\chi_0^2\)[/tex] to the critical value:
- If [tex]\(\chi_0^2 \)[/tex] > critical value: Reject [tex]\(H_0\)[/tex].
- If [tex]\(\chi_0^2 \)[/tex] ≤ critical value: Fail to reject [tex]\(H_0\)[/tex].
Given:
- [tex]\(\chi_0^2 = 11.692\)[/tex]
- Critical value = 11.345
### Decision
Since [tex]\(11.692\)[/tex] (the test statistic) is greater than [tex]\(11.345\)[/tex] (the critical value):
- We reject the null hypothesis, [tex]\(H_0\)[/tex].
### Conclusion
Based on the chi-square Goodness-of-Fit test with a significance level of 0.01 and 3 degrees of freedom, the test statistic of 11.692 exceeds the critical value of 11.345. Therefore, there is sufficient evidence to reject the null hypothesis. We conclude that the die does not have a uniform distribution.
### (a) State the null and alternative hypotheses
The hypotheses for the chi-square Goodness-of-Fit test are:
- [tex]\( H_0 \)[/tex] (Null hypothesis): The die has a uniform distribution.
- [tex]\( H_a \)[/tex] (Alternative hypothesis): The die does not have a uniform distribution.
### (b) Given information
- The chi-square test statistic, [tex]\(\chi_0^2\)[/tex], is 11.692.
### (c) Determine the critical value
To determine the critical value, we need to look at the chi-square table provided for the given significance level [tex]\(\alpha = 0.01\)[/tex] and degrees of freedom (df).
- Degrees of freedom (df): For a fair die with 6 faces, all equally likely, we would typically have [tex]\(df = 6 - 1 = 5\)[/tex]. However, based on the problem setup, the degrees of freedom provided for the test is 3.
From the given chi-square table at [tex]\(\alpha = 0.01\)[/tex] for 3 degrees of freedom:
- The [tex]\(\chi^2\)[/tex] critical value is 11.345.
### Decision Rule
Compare the test statistic [tex]\(\chi_0^2\)[/tex] to the critical value:
- If [tex]\(\chi_0^2 \)[/tex] > critical value: Reject [tex]\(H_0\)[/tex].
- If [tex]\(\chi_0^2 \)[/tex] ≤ critical value: Fail to reject [tex]\(H_0\)[/tex].
Given:
- [tex]\(\chi_0^2 = 11.692\)[/tex]
- Critical value = 11.345
### Decision
Since [tex]\(11.692\)[/tex] (the test statistic) is greater than [tex]\(11.345\)[/tex] (the critical value):
- We reject the null hypothesis, [tex]\(H_0\)[/tex].
### Conclusion
Based on the chi-square Goodness-of-Fit test with a significance level of 0.01 and 3 degrees of freedom, the test statistic of 11.692 exceeds the critical value of 11.345. Therefore, there is sufficient evidence to reject the null hypothesis. We conclude that the die does not have a uniform distribution.