Answer :
To find the chi-square test statistic for the chi-square goodness-of-fit test, we follow these steps:
1. Identify the observed frequencies (O): These are given as:
- [tex]\(O_1 = 8\)[/tex]
- [tex]\(O_2 = 12\)[/tex]
- [tex]\(O_3 = 17\)[/tex]
- [tex]\(O_4 = 10\)[/tex]
- [tex]\(O_5 = 11\)[/tex]
- [tex]\(O_6 = 14\)[/tex]
2. Identify the expected frequencies (E): Since we are testing the fairness of the die, each outcome (1 to 6) should ideally occur 72/6 = 12 times (given directly in the table):
- [tex]\(E_1 = 12\)[/tex]
- [tex]\(E_2 = 12\)[/tex]
- [tex]\(E_3 = 12\)[/tex]
- [tex]\(E_4 = 12\)[/tex]
- [tex]\(E_5 = 12\)[/tex]
- [tex]\(E_6 = 12\)[/tex]
3. Apply the chi-square formula:
[tex]\[ \chi_0^2 = \sum_{k=1}^{6} \frac{(O_k - E_k)^2}{E_k} \][/tex]
We calculate each term in the sum separately:
[tex]\[ \frac{(O_1 - E_1)^2}{E_1} = \frac{(8 - 12)^2}{12} = \frac{(-4)^2}{12} = \frac{16}{12} = 1.333 \][/tex]
[tex]\[ \frac{(O_2 - E_2)^2}{E_2} = \frac{(12 - 12)^2}{12} = \frac{0^2}{12} = 0 \][/tex]
[tex]\[ \frac{(O_3 - E_3)^2}{E_3} = \frac{(17 - 12)^2}{12} = \frac{5^2}{12} = \frac{25}{12} = 2.083 \][/tex]
[tex]\[ \frac{(O_4 - E_4)^2}{E_4} = \frac{(10 - 12)^2}{12} = \frac{(-2)^2}{12} = \frac{4}{12} = 0.333 \][/tex]
[tex]\[ \frac{(O_5 - E_5)^2}{E_5} = \frac{(11 - 12)^2}{12} = \frac{(-1)^2}{12} = \frac{1}{12} = 0.083 \][/tex]
[tex]\[ \frac{(O_6 - E_6)^2}{E_6} = \frac{(14 - 12)^2}{12} = \frac{2^2}{12} = \frac{4}{12} = 0.333 \][/tex]
4. Sum the calculated terms:
[tex]\[ \chi_0^2 = 1.333 + 0 + 2.083 + 0.333 + 0.083 + 0.333 = 4.166 \][/tex]
5. Round the final answer to three decimal places:
[tex]\[ \chi_0^2 \approx 4.167 \][/tex]
Therefore, the chi-square test statistic [tex]\(\chi_0^2 = 4.167\)[/tex].
1. Identify the observed frequencies (O): These are given as:
- [tex]\(O_1 = 8\)[/tex]
- [tex]\(O_2 = 12\)[/tex]
- [tex]\(O_3 = 17\)[/tex]
- [tex]\(O_4 = 10\)[/tex]
- [tex]\(O_5 = 11\)[/tex]
- [tex]\(O_6 = 14\)[/tex]
2. Identify the expected frequencies (E): Since we are testing the fairness of the die, each outcome (1 to 6) should ideally occur 72/6 = 12 times (given directly in the table):
- [tex]\(E_1 = 12\)[/tex]
- [tex]\(E_2 = 12\)[/tex]
- [tex]\(E_3 = 12\)[/tex]
- [tex]\(E_4 = 12\)[/tex]
- [tex]\(E_5 = 12\)[/tex]
- [tex]\(E_6 = 12\)[/tex]
3. Apply the chi-square formula:
[tex]\[ \chi_0^2 = \sum_{k=1}^{6} \frac{(O_k - E_k)^2}{E_k} \][/tex]
We calculate each term in the sum separately:
[tex]\[ \frac{(O_1 - E_1)^2}{E_1} = \frac{(8 - 12)^2}{12} = \frac{(-4)^2}{12} = \frac{16}{12} = 1.333 \][/tex]
[tex]\[ \frac{(O_2 - E_2)^2}{E_2} = \frac{(12 - 12)^2}{12} = \frac{0^2}{12} = 0 \][/tex]
[tex]\[ \frac{(O_3 - E_3)^2}{E_3} = \frac{(17 - 12)^2}{12} = \frac{5^2}{12} = \frac{25}{12} = 2.083 \][/tex]
[tex]\[ \frac{(O_4 - E_4)^2}{E_4} = \frac{(10 - 12)^2}{12} = \frac{(-2)^2}{12} = \frac{4}{12} = 0.333 \][/tex]
[tex]\[ \frac{(O_5 - E_5)^2}{E_5} = \frac{(11 - 12)^2}{12} = \frac{(-1)^2}{12} = \frac{1}{12} = 0.083 \][/tex]
[tex]\[ \frac{(O_6 - E_6)^2}{E_6} = \frac{(14 - 12)^2}{12} = \frac{2^2}{12} = \frac{4}{12} = 0.333 \][/tex]
4. Sum the calculated terms:
[tex]\[ \chi_0^2 = 1.333 + 0 + 2.083 + 0.333 + 0.083 + 0.333 = 4.166 \][/tex]
5. Round the final answer to three decimal places:
[tex]\[ \chi_0^2 \approx 4.167 \][/tex]
Therefore, the chi-square test statistic [tex]\(\chi_0^2 = 4.167\)[/tex].