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Explain the three categories of Quality Control and differentiate between efficiency and sufficiency properties of a good estimator

Business/Management

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5(a) Explain the three categories of Quality Control?

The three main categories of Quality Control are:

  • Acceptance Sampling: This involves inspecting a sample of products from a larger batch (lot) to decide whether to accept or reject the entire lot. It's used for incoming materials, in-process goods, or final products.
  • Process Control: Also known as Statistical Process Control (SPC), this category focuses on monitoring a production process while it is happening. It uses tools like control charts to detect variations and ensure the process remains stable and within desired limits, preventing defects.
  • Measurement System Analysis (MSA): This category involves evaluating the quality and reliability of the measurement system itself. It ensures that the data collected for quality control is accurate and precise, as faulty measurements can lead to incorrect decisions about product quality or process stability.

5(b) Differentiate between efficiency and sufficiency properties of a good estimator

  • Efficiency: An estimator is considered efficient if it has the smallest possible variance among all unbiased estimators. This means that, on average, the estimates produced by an efficient estimator will be closer to the true population parameter compared to other unbiased estimators, making it more precise.
  • Sufficiency: An estimator is sufficient if it captures all the information about the population parameter that is contained in the sample data. In other words, no other estimator could extract more information from the sample about the parameter than a sufficient estimator.

5(c) Hypothesis test for the mean volume of soft drink

  1. State the Hypotheses

    The null hypothesis (H0H_0) states that the mean volume is 200 ml. The alternative hypothesis (H1H_1) states that the mean volume is less than 200 ml.

    H0:μ=200 mlH_0: \mu = 200 \text{ ml} H1:μ<200 mlH_1: \mu < 200 \text{ ml}
  2. Choose the Test Statistic

    Since the population standard deviation (σ\sigma) is known and the sample size (n=50n = 50) is large, we use a z-test for the mean.
    The formula for the z-test statistic is:

    Z=xˉ−μ0σ/nZ = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}}
  3. Calculate the Test Statistic

    Given:
    Sample mean (xˉ\bar{x}) = 198 ml
    Hypothesized population mean (μ0\mu_0) = 200 ml
    Population standard deviation (σ\sigma) = 10 ml
    Sample size (nn) = 50

    Z=198−20010/50Z = \frac{198 - 200}{10 / \sqrt{50}} Z=−210/7.07106Z = \frac{-2}{10 / 7.07106} Z=−21.41421Z = \frac{-2}{1.41421} Z≈−1.414Z \approx -1.414
  4. Determine the Critical Value

    For a one-tailed (left-tailed) test at a 5% level of significance (α=0.05\alpha = 0.05), the critical z-value is found from the standard normal distribution table.

    Zcritical=−1.645Z_{\text{critical}} = -1.645
  5. Make a Decision

    Compare the calculated Z-statistic with the critical Z-value.
    Calculated Z=−1.414Z = -1.414
    Critical Z=−1.645Z = -1.645
    Since −1.414>−1.645-1.414 > -1.645, the calculated Z-statistic does not fall into the rejection region. Therefore, we fail to reject the null hypothesis.

  6. State the Conclusion

    There is not enough evidence at the 5% level of significance to conclude that the mean volume of soft drink per bottle is less than 200 ml.

    Fail to reject H0. There is no sufficient evidence that the mean volume is less than 200 ml.\boxed{\text{Fail to reject } H_0\text{. There is no sufficient evidence that the mean volume is less than 200 ml.}}
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