Question 1178502
Here's how to solve this problem step-by-step:

**1. Understand the Problem**

* We are dealing with a hypothesis test about the variance of a normal population.
* The null hypothesis (H0) is that the population variance (σ²) is 1.04 × 10⁻³.
* The firm rejects the shipment if the sample variance (s²) exceeds 1.15 × 10⁻³.
* We need to find the probability of rejecting H0 when it is actually true (Type I error).

**2. Use the Chi-Square Distribution**

* The chi-square distribution is used to test hypotheses about variances.
* The test statistic is: χ² = (n - 1) * s² / σ²
    * where:
        * n is the sample size (16)
        * s² is the sample variance (1.15 × 10⁻³)
        * σ² is the population variance (1.04 × 10⁻³)

**3. Calculate the Chi-Square Test Statistic**

* χ² = (16 - 1) * (1.15 × 10⁻³) / (1.04 × 10⁻³)
* χ² = 15 * 1.15 / 1.04
* χ² = 16.6346

**4. Determine the Degrees of Freedom**

* Degrees of freedom (df) = n - 1 = 16 - 1 = 15

**5. Find the Probability**

* We need to find the probability that χ² > 16.6346 with 15 degrees of freedom.
* This is a right-tailed test.
* We can use a chi-square table or a calculator/software to find this probability.

Using a chi-square calculator or software:

* The probability P(χ² > 16.6346) with 15 degrees of freedom is approximately 0.338.

**6. Conclusion**

* The probability that the shipment will be rejected even though the population variance is 1.04 × 10⁻³ is approximately 0.338.

Therefore, there is a 33.8% chance that the firm will reject a good shipment.