SOLUTION: An optical firm buys glass slabs to be ground into lenses, and it is known that the variance of the refractive index of the glass slabs is to be no more than 1.04 × 10^−3. The f

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Question 1178502: An optical firm buys glass slabs to be ground into lenses, and it is known that the variance of the refractive index of the glass slabs is to be no more than 1.04 × 10^−3. The firm rejects a shipment of glass slabs if the sample variance of 16 pieces selected at random exceeds 1.15 × 10^−3. Assuming that the sample values may be looked on as a random sample from a normal population, what is the probability that a shipment will be rejected even though σ^2 = 1.04 × 10^−3?


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