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Let's break down this problem step by step.

1. **Test to be used**:
   - The test required for this problem is a **Goodness-of-Fit Test**, specifically the **Chi-Square Goodness-of-Fit Test**. This is used to determine if the observed frequencies differ from the expected frequencies under the null hypothesis.

2. **Null and alternative hypotheses**:
   - \( H_0 \) (null hypothesis): The die is fair, meaning each face has an equal probability \( p_1 = p_2 = p_3 = p_4 = p_5 = p_6 = \frac{1}{6} \).
   - \( H_1 \) (alternative hypothesis): At least one of the probabilities differs from \( \frac{1}{6} \), i.e., the die is not fair.

3. **Expected Frequencies**:
   - Total number of rolls = 105
   - Since each side of a fair die should appear with probability \( \frac{1}{6} \), the expected frequency for each face is:
   \[
   \text{Expected frequency} = \frac{1}{6} \times 105 = 17.50
   \]
   Thus, the expected frequency for each side (1 through 6) is 17.50.

4. **Degree of freedom**:
   - The formula for degrees of freedom (df) is:
     \[
     \text{df} = (\text{number of categories} - 1) = 6 - 1 = 5
     \]

5. **Chi-Square Test Statistic**:
   The test statistic is calculated using the formula:
   \[
   \chi^2 = \sum \frac{(\text{Observed} - \text{Expected})^2}{\text{Expected}}
   \]
   Let's calculate this step by step for each category (side of the die):

   - For side 1: \( \frac{(17 - 17.5)^2}{17.5} = \frac{(-0.5)^2}{17.5} = \frac{0.25}{17.5} = 0.0143 \)
   - For side 2: \( \frac{(16 - 17.5)^2}{17.5} = \frac{(-1.5)^2}{17.5} = \frac{2.25}{17.5} = 0.1286 \)
   - For side 3: \( \frac{(26 - 17.5)^2}{17.5} = \frac{(8.5)^2}{17.5} = \frac{72.25}{17.5} = 4.1286 \)
   - For side 4: \( \frac{(8 - 17.5)^2}{17.5} = \frac{(-9.5)^2}{17.5} = \frac{90.25}{17.5} = 5.1571 \)
   - For side 5: \( \frac{(18 - 17.5)^2}{17.5} = \frac{(0.5)^2}{17.5} = \frac{0.25}{17.5} = 0.0143 \)
   - For side 6: \( \frac{(20 - 17.5)^2}{17.5} = \frac{(2.5)^2}{17.5} = \frac{6.25}{17.5} = 0.3571 \)

   Summing these values gives:
   \[
   \chi^2 = 0.0143 + 0.1286 + 4.1286 + 5.1571 + 0.0143 + 0.3571 = 9.7999 \approx 9.800
   \]

6. **P-value**:
   Using a chi-square distribution table or a calculator with the chi-square distribution, you can find the p-value for \( \chi^2 = 9.8 \) with 5 degrees of freedom. This gives:
   \[
   \text{p-value} \approx 0.0817
   \]

7. **Conclusion**:
   - The significance level \( \alpha = 0.05 \).
   - Since the p-value (0.0817) is greater than \( \alpha \), we **fail to reject the null hypothesis**. There is insufficient evidence to suggest that the die is not fair.

### Summary of Answers:
1. **Test**: Chi-Square Goodness-of-Fit Test
2. **Expected Frequencies**: 17.50 for each side
3. **Degrees of Freedom**: 5
4. **Test Statistic**: 9.800
5. **P-value**: 0.0817
6. **Conclusion**: Fail to reject the null hypothesis.

Would you like to explore any specific step further or go into more detail?

Here are 5 related questions you might consider:
1. How would the test change if the die had more sides?
2. What would happen if we used a different significance level, such as \( \alpha = 0.01 \)?
3. Can you explain how to interpret the degrees of freedom in chi-square tests?
4. How is the chi-square distribution related to the normal distribution?
5. What is the importance of ensuring the sample size is large enough in such tests?

**Tip**: In any hypothesis test, always compare the p-value with the significance level to determine if you reject or fail to reject the null hypothesis.

Your statistics instructor would like to use a 6-sided die to choose the next quiz question. You suspect that the die is not fair, and you are conducting a multinomial Goodness-of-Fit hypothesis test at a significance level of 0.05. You drew 105 samples and recorded the observed frequencies.

Learn how to perform a Chi-Square Goodness-of-Fit Test to determine whether a 6-sided die is fair. This problem involves calculating the expected frequencies, test statistic, p-value, and making conclusions based on the hypothesis test.

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