SOLUTION: The side effects of a new drug are being tested against a placebo. A simple random sample of 565 patients yields the results below. At a significance level of a= 0.10, is there eno

Question 1171900: The side effects of a new drug are being tested against a placebo. A simple random sample of 565 patients yields the results below. At a significance level of a= 0.10, is there enough evidence to conclude that the treatment is independent of the side effect of nausea?
Result Drug Placebo Total
Nausea 36 13 49
No nausea 254 262 516
Total 290 275 565
The null hypothesis H_∅ :
The alternative hypothesis H_a :
The level of significance a=
The degrees of freedom:
The test statistic:
The critical value:
Decision:

Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
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.

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