Question 1193650
**1. Set up Hypotheses**

* **Null Hypothesis (H0):** p = 1/6 (Matt has no special relationship with the number 2; the probability of rolling a 2 is 1/6)
* **Alternative Hypothesis (H1):** p > 1/6 (Matt has a special relationship with the number 2; the probability of rolling a 2 is greater than 1/6)

**2. Calculate the Test Statistic**

* **Sample Proportion (p-hat):** 
    * p-hat = Number of successes / Number of trials = 11/50 = 0.22

* **Expected Number of Successes (under H0):**
    * Expected Successes = n * p0 = 50 * (1/6) = 8.33

* **Standard Error:**
    * Standard Error = √(p0 * (1 - p0) / n) = √((1/6) * (5/6) / 50) ≈ 0.0373

* **Test Statistic (z-score):**
    * z = (p-hat - p0) / Standard Error = (0.22 - 1/6) / 0.0373 ≈ 3.46

**3. Determine the P-value**

* **P-value:** The probability of observing a z-score as extreme or more extreme than the calculated z-score (3.46) under the null hypothesis.
* **Using a standard normal distribution table or statistical software:**
    * P(Z > 3.46) ≈ 0.0003 

**Interpretation**

* The P-value of 0.0003 is very small. This means that if the null hypothesis (p = 1/6) were true, it would be extremely unlikely to observe 11 or more twos in 50 rolls.

**Conclusion**

* Since the P-value is very small (typically, a significance level of 0.05 is used), we have strong evidence to reject the null hypothesis. 
* There is strong evidence to suggest that Matt may have a special relationship with the number 2, and he rolls it more often than expected by chance.

**Note:** This analysis assumes that the rolls of the die are independent and that the die is fair except for the potential bias towards rolling a 2.