Question

A die is thrown 60 times and number of times the following faces were obtained  
 

Faces	1	2	3	4	5	6
No. of times	14	7	5	8	10	16

Can the die be regarded as fair? Given \[ \chi^2_{0.05,5}=11.07 \]

• A. The die is not fair
• B. The die is fair
• C. no conclusion
• D. experiment is conducted again

Hint:

In a Goodness-of-Fit test with uniform expected frequencies, the calculation can be simplified to:
\[ \chi^2 = \frac{\sum O_i^2}{E} - N \] Here: \( \sum O_i^2 = 14^2 + 7^2 + 5^2 + 8^2 + 10^2 + 16^2 = 196 + 49 + 25 + 64 + 100 + 256 = 690 \).
Thus: \( \chi^2 = \frac{690}{10} - 60 = 69 - 60 = 9 \). This is much faster and less prone to arithmetic errors!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem uses a Chi-Square (\( \chi^2 \)) Goodness-of-Fit Test to determine if an observed frequency distribution of die faces aligns with the theoretical distribution of a fair die.
Key Formula or Approach:
The Chi-Square test statistic is calculated as:
\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \] where \( O_i \) is the observed frequency and \( E_i \) is the expected frequency under the null hypothesis.

Step 2: Detailed Explanation:
Let us perform the statistical test step-by-step:
- Hypotheses:
\( H_0 \): The die is fair (all faces are equally likely with probability \( p_i = 1/6 \)).
\( H_1 \): The die is not fair.
- Expected Frequencies (\( E_i \)):
Since a die has 6 faces and is thrown 60 times, the expected frequency for each face under \( H_0 \) is:
\[ E_i = 60 \times \frac{1}{6} = 10 \quad \text{for } i = 1, 2, \dots, 6 \] - Calculate the Chi-Square Statistic:
The observed frequencies are \( O = [14, 7, 5, 8, 10, 16] \).
\[ \chi^2 = \frac{(14-10)^2}{10} + \frac{(7-10)^2}{10} + \frac{(5-10)^2}{10} + \frac{(8-10)^2}{10} + \frac{(10-10)^2}{10} + \frac{(16-10)^2}{10} \] \[ \chi^2 = \frac{4^2 + (-3)^2 + (-5)^2 + (-2)^2 + 0^2 + 6^2}{10} \] \[ \chi^2 = \frac{16 + 9 + 25 + 4 + 0 + 36}{10} \] \[ \chi^2 = \frac{90}{10} = 9.0 \] - Compare with Critical Value:
We are given the critical value at the \( 5% \) significance level with \( 5 \) degrees of freedom:
\[ \chi^2_{\text{critical}} = \chi^2_{0.05, 5} = 11.07 \] Since our calculated test statistic:
\[ \chi^2_{\text{calculated}} = 9.0 < 11.07 \] We fail to reject the null hypothesis \( H_0 \).
Therefore, we conclude that the differences between observed and expected frequencies are not statistically significant, and the die can be regarded as fair.

Step 3: Final Answer:
The die can be regarded as fair.
Therefore, the correct choice is Option (B).