Question 1171900
Absolutely! Let's break down this hypothesis test step-by-step.

**1. Setting up the Hypotheses**

* **Null Hypothesis (H₀):** The treatment is independent of the side effect of nausea. In other words, there is no association between the drug and nausea.
* **Alternative Hypothesis (H₁):** The treatment is not independent of the side effect of nausea. There is an association between the drug and nausea.

**2. Level of Significance**

* α = 0.10 (given)

**3. Degrees of Freedom**

* Degrees of freedom (df) = (rows - 1) * (columns - 1)
* In this case, df = (2 - 1) * (2 - 1) = 1 * 1 = 1

**4. Calculating the Test Statistic (Chi-Square)**

We'll use the chi-square (χ²) test for independence.

* **Observed Values (O):** The values in the table.
* **Expected Values (E):** We need to calculate the expected values for each cell.
    * Expected value = (row total * column total) / grand total

Here's how we calculate the expected values:

* E(Nausea, Drug) = (49 * 290) / 565 ≈ 25.15
* E(Nausea, Placebo) = (49 * 275) / 565 ≈ 23.85
* E(No nausea, Drug) = (516 * 290) / 565 ≈ 264.85
* E(No nausea, Placebo) = (516 * 275) / 565 ≈ 251.15

Now, we calculate the chi-square statistic:

χ² = Σ [(O - E)² / E]

χ² = [(36 - 25.15)² / 25.15] + [(13 - 23.85)² / 23.85] + [(254 - 264.85)² / 264.85] + [(262 - 251.15)² / 251.15]

χ² ≈ [(10.85)² / 25.15] + [(-10.85)² / 23.85] + [(-10.85)² / 264.85] + [(10.85)² / 251.15]

χ² ≈ (117.7225 / 25.15) + (117.7225 / 23.85) + (117.7225 / 264.85) + (117.7225 / 251.15)

χ² ≈ 4.680 + 4.936 + 0.444 + 0.469

χ² ≈ 10.529

**5. Finding the Critical Value**

* Using a chi-square distribution table with df = 1 and α = 0.10, the critical value is 2.706.

**6. Decision**

* Compare the test statistic (10.529) with the critical value (2.706).
* Since 10.529 > 2.706, the test statistic falls in the rejection region.
* Therefore, we reject the null hypothesis.

**Summary**

* **H₀:** The treatment is independent of nausea.
* **H₁:** The treatment is not independent of nausea.
* **α:** 0.10
* **df:** 1
* **Test statistic (χ²):** 10.529
* **Critical value:** 2.706
* **Decision:** Reject the null hypothesis.

**Conclusion**

There is enough evidence at the α = 0.10 significance level to conclude that the treatment is not independent of the side effect of nausea.