EX 3.32 power supply units

• Scenario:

  • An electronics manufacturer claims that at most $10\%$ of power supply units need service during the warranty period.
  • Technicians purchase 20 units for accelerated testing.
  • Let $p$ represent the probability that a unit needs repair.

• Hypothesis:

  • Null hypothesis ($H_0$): $p\le 0.10$
  • Alternative hypothesis ($H_a$): $p>0.10$

• Decision rule:

  • Reject $H_0$ in favor of $H_a$ if $x\ge 5$ (where $x$ is the observed number of units needing repair).
  • Consider the claim plausible if $x\le 4$.

• Binomial distribution:

  • The number of units needing repair follows:
  $$X\sim \mathrm{Bin}(20,p)$$

• Incorrect conclusion probability:

  • Calculate the probability of rejecting the claim when $p=0.10$:
    • This is the probability of observing $x\ge 5$ when $p=0.10$.
    • Use the cumulative distribution function (CDF):
  $$P(X\ge 5)=1-P(X\le 4)$$

• Calculation of $P(X\le 4)$:

  • Use the binomial probability formula:
  $$P(X=k)=\left(\genfrac{}{}{0ex}{}{20}{k}\right)(0.10{)}^k(0.90{)}^{20-k}$$
  • Sum for $k=0,1,2,3,4$:
  $$P(X\le 4)=\sum _{k=0}^4\left(\genfrac{}{}{0ex}{}{20}{k}\right)(0.10{)}^k(0.90{)}^{20-k}$$

• Final expression:

  • Thus, the probability of rejecting the claim incorrectly when $p=0.10$ is:
  $$P(\text{Reject }{H}_0\mid p=0.10)=1-P(X\le 4)$$

$$\begin{array}{rl}P(X\ge 5\text{ when }p=0.10)& =1-B(4;20,0.1)\\ & =1-0.957\\ & =0.043\end{array}$$

• Incorrect conclusion analysis:

  • Consider a new scenario where $p=0.20$.

• Probability that the claim is not rejected:

  • We want to find:
  $$P(X\le 4\text{ when }p=0.20)$$
  • This is computed as:
  $$P(X\le 4)=B(4;20,0.20)$$
  • Result:
  $$P(X\le 4)=0.630$$

• Interpretation of the results:

  • The first probability (rejecting the claim when $p=0.10$) is relatively small.
  • The second probability (not rejecting the claim when $p=0.20$) is large and deemed intolerable:
    • When the true probability of needing service is $p=0.20$, 63% of samples will incorrectly support the manufacturer’s claim.

• Conclusion:

  • The decision rule, as stated, may lead to significant errors in judgment, particularly when the actual proportion needing service is higher than claimed.

• Considerations for adjusting the decision rule:

  • The current cutoff value for rejecting the claim is 5.

  • Potential impact of changing the cutoff:

    • Replacing the cutoff of 5 with a smaller number:

  • Would likely reduce the probability of not rejecting the claim when $p=0.20$ (making it smaller than 0.630).

  • However, this would increase the probability of rejecting the claim incorrectly when $p=0.10$.

• Balancing error probabilities:

  • The challenge is to make both types of erroneous conclusions (Type I and Type II errors) small:
    • Type I error: Rejecting the claim when it is true.
    • Type II error: Not rejecting the claim when it is false.

• Recommendation for reducing both error probabilities:

  • The only effective way to minimize both error probabilities is to base the decision rule on a larger sample size.
  • Increasing the sample size provides more data, which helps improve the accuracy of the decision-making process.

• Conclusion:

  • A more robust experimental design with a larger number of units will lead to better reliability in assessing the manufacturer’s claim.