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.

Manufacturers are testing a die to make sure that it is fair (has a uniform distribution). They roll the die 72 times and record the outcomes in the table below. Find the test statistic, [tex]\chi_0^2[/tex], for the chi-square goodness-of-fit test. Round the final answer to three decimal places.

[tex]\[ \chi_0^2=\sum_k \frac{(O-E)^2}{E} \][/tex]

\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Outcome & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline Expected & 12 & 12 & 12 & 12 & 12 & 12 \\
\hline Observed & 8 & 12 & 17 & 10 & 11 & 14 \\
\hline
\end{tabular}

Provide your answer below:

chi-square test statistic [tex]=\square[/tex]


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].