Question

A new technology for the production of integrated circuits is being mastered. The technology will be rejected if the percentage of the defective integrated circuits exceeds 2%. How many integrated circuits have to be checked to make sure that the technology gives a sufficiently small amount of defects with a probability of 0.5% if the actual probability is p = 0.04? Use a = 0.05. Use the z-values rounded to two decimal places to obtain the answer. n = the absolute tolerance is +/-1

Answer 1

To ensure the new integrated circuit technology is acceptable, we must calculate the sample size needed to achieve a test with high confidence and low type II error. This involves using the actual defect probability, z-values, and normal approximation methods to find the required number of circuits to test.

Step-by-step explanation:

The question relates to determining the sample size needed to check the new integrated circuit production process, ensuring that the percentage of defects does not exceed a specified limit with a certain probability. The actual probability of a defect (p) is 0.04, and the technology will be rejected if the percentage of defective integrated circuits exceeds 2%, with a confidence level (1-α) of 95% (where α = 0.05) and a probability of making a type II error of 0.5%. The z-values and standard deviation calculations for sample size estimation are critical in this scenario.

Using the given z-value of ±2 for the two-tailed test and the standard deviation of the proportion p = 0.04, we can estimate the minimum sample size required. The formula for sample size calculation with the given parameters and using the normal approximation for binomial distribution would provide us with the desired sample size to ensure that the defect rate does not exceed the rejection threshold with the specified testing confidence and power. A key step is to identify the corresponding z-value for the given α, which can be obtained from standard normal distribution tables or statistical software.