To conduct a chi-square Goodness-of-Fit test, we follow a standard methodology:
1. Set up the hypotheses:
- Null hypothesis ([tex]\( H_0 \)[/tex]): The die has a uniform distribution.
- Alternative hypothesis ([tex]\( H_a \)[/tex]): The die does not have a uniform distribution.
2. Observed and expected frequencies:
- Observed counts: [tex]\([7, 10, 14, 16, 9, 22]\)[/tex]
- Expected counts: [tex]\([13, 13, 13, 13, 13, 13]\)[/tex]
3. Calculate the chi-square test statistic ([tex]\( \chi^2 \)[/tex]):
[tex]\[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\][/tex]
where [tex]\( O_i \)[/tex] is the observed frequency and [tex]\( E_i \)[/tex] is the expected frequency.
4. Compute the individual components:
- For outcome 1: [tex]\(\frac{(7 - 13)^2}{13} = \frac{36}{13}\)[/tex]
- For outcome 2: [tex]\(\frac{(10 - 13)^2}{13} = \frac{9}{13}\)[/tex]
- For outcome 3: [tex]\(\frac{(14 - 13)^2}{13} = \frac{1}{13}\)[/tex]
- For outcome 4: [tex]\(\frac{(16 - 13)^2}{13} = \frac{9}{13}\)[/tex]
- For outcome 5: [tex]\(\frac{(9 - 13)^2}{13} = \frac{16}{13}\)[/tex]
- For outcome 6: [tex]\(\frac{(22 - 13)^2}{13} = \frac{81}{13}\)[/tex]
5. Sum these components:
[tex]\[
\chi^2 = \frac{36}{13} + \frac{9}{13} + \frac{1}{13} + \frac{9}{13} + \frac{16}{13} + \frac{81}{13}
\][/tex]
[tex]\[
\chi^2 = \frac{36 + 9 + 1 + 9 + 16 + 81}{13} = \frac{152}{13} \approx 11.692
\][/tex]
So, the test statistic [tex]\( \chi^2 \)[/tex] rounded to three decimal places is:
[tex]\[
\boxed{11.692}
\][/tex]
