Question 1193379
**1. Set up Hypotheses**

* **Null Hypothesis (H0):** The variance of the drug weight in the sample is equal to the specified population variance. 
    * σ² = 0.36 kg² 
* **Alternative Hypothesis (H1):** The variance of the drug weight in the sample is not equal to the specified population variance.
    * σ² ≠ 0.36 kg² 

**2. Determine Test Statistic**

* We will use the **chi-square test statistic** for testing the variance of a normally distributed population. 
* The test statistic is calculated as: 
   χ² = (n - 1) * s² / σ² 
   where:
      * n is the sample size (15)
      * s² is the sample variance (0.05 kg²)
      * σ² is the population variance (0.36 kg²)

**3. Calculate Test Statistic**

χ² = (15 - 1) * 0.05 / 0.36 
χ² = 14 * 0.05 / 0.36 
χ² ≈ 1.94

**4. Determine Critical Values**

* **Significance Level (α):** 0.05
* **Degrees of Freedom (df):** n - 1 = 15 - 1 = 14
* **Critical Values:** 
    * Find the critical values from the chi-square distribution table for α/2 = 0.025 and 1 - α/2 = 0.975 with 14 degrees of freedom. 
    * Using a statistical software or table, we find:
        * Lower Critical Value (χ²_lower) ≈ 5.629
        * Upper Critical Value (χ²_upper) ≈ 26.119

**5. Decision Rule**

* **Reject H0** if the calculated chi-square statistic (χ²) falls outside the critical region (i.e., below χ²_lower or above χ²_upper).
* **Fail to reject H0** if the calculated chi-square statistic (χ²) falls within the critical region.

**6. Make a Decision**

* Our calculated χ² value (1.94) is less than the lower critical value (5.629). 
* Therefore, we **reject the null hypothesis (H0)**.

**Conclusion**

* There is sufficient evidence at the 0.05 significance level to conclude that the variance of the drug weight in the sample **differs significantly** from the specified population variance. 

**Interpretation**

* The quality engineer's claim that the variance of the drug weight does not differ significantly from the specified variance is **not supported by the sample data.** 
* This suggests that the drug manufacturing process might be experiencing more variability in drug weight than expected. 

**Note:**

* This analysis assumes that the sample is randomly selected and that the drug weights are normally distributed. 
* If these assumptions are not met, the results of the test may not be valid.

I hope this comprehensive explanation is helpful! Let me know if you have any further questions.