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.