The problem "The Fake Quality Report" typically involves testing a claim about the average quality score of products using sample data. Below is a step-by-step solution based on a common scenario for such problems. If the exact problem details differ, adjust the values accordingly. A factory claims that the average quality score of its products is at least 90. An inspector tests 36 randomly selected products and finds:

• Sample mean ((\bar{x})) = 87
• Sample standard deviation ((s)) = 8
• Significance level ((\alpha)) = 5%

Assume the population is normally distributed. Test the factory's claim.

Solution:

Step 1: State the Hypotheses

• Null Hypothesis ((H_0)): (\mu \geq 90) (factory's claim is true).
• Alternative Hypothesis ((H_1)): (\mu < 90) (factory's claim is false; inspector's suspicion).

This is a left-tailed test.

Step 2: Determine the Test Statistic

Since the population standard deviation ((\sigma)) is unknown and the sample size ((n = 36)) is large, use the t-test:
[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} ] where:

• (\bar{x} = 87),
• (\mu_0 = 90),
• (s = 8),
• (n = 36).

Calculate:
[ t = \frac{87 - 90}{8 / \sqrt{36}} = \frac{-3}{8 / 6} = \frac{-3}{1.333} \approx -2.25 ]

Step 3: Find the Critical Value

• Degrees of freedom ((df)) = (n - 1 = 35).
• For (\alpha = 0.05) (left-tailed test), the critical t-value ((t_{\alpha, df})) is -1.690 (from t-distribution table).

Step 4: Compare Test Statistic with Critical Value

• Test statistic: (t = -2.25)
• Critical value: (t_{\text{crit}} = -1.690)

Since (-2.25 < -1.690), the test statistic falls in the rejection region.

Step 5: Calculate the p-value (Optional but Recommended)

• For (t = -2.25) and (df = 35), the p-value is 015 (using t-distribution calculator/table).
• Since (p\text{-value} = 0.015 < \alpha = 0.05), reject (H_0).

Step 6: Conclusion

Reject the null hypothesis ((H_0)). There is sufficient evidence at the 5% significance level to conclude that the average quality score is less than 90. The factory's claim is likely fake.

Key Notes:

1. Why t-test?
   Population standard deviation is unknown, and the sample size is large ((n > 30)), but the t-test is appropriate (especially since (n) is not extremely large).

2. Assumptions:

   • Population is normally distributed (given).
   • Sample is random and independent.

3. Real-World Implication:
   The inspector's sample suggests the factory's quality report may be falsified, as the actual average quality is below the claimed threshold.

If your problem has different values (e.g., sample size, mean, standard deviation, or significance level), substitute them into the formulas above.