IDNLearn.com provides a comprehensive solution for all your question and answer needs. Get the information you need from our community of experts who provide accurate and comprehensive answers to all your questions.
Sagot :
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].
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Trust IDNLearn.com for all your queries. We appreciate your visit and hope to assist you again soon.