To analyze this situation, we need to determine what the company can conclude based on the simulation results, particularly in relation to the null and alternative hypotheses.
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Null Hypothesis (H0): The proportion of defective products is greater than or equal to 3 out of 1000 (or 0.003), which implies that 3 or more defects would be expected frequently if H0 were true.
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Alternative Hypothesis (H1): The proportion of defective products is less than 3 out of 1000.
From the simulation results, we have the following defects and their frequencies:
- 0 defects: 12 times
- 1 defect: 50 times
- 2 defects: 20 times
- 3 defects: 14 times
- 4 defects: 4 times
The total number of simulations is 100. To assess the p-value, we should consider how many of the simulations resulted in 3 or more defects:
- Number of simulations with 3 defects: 14
- Number of simulations with 4 defects: 4
- Total simulations with 3 or more defects: \(14 + 4 = 18\)
This means there were 18 out of 100 simulations (or 18%) where the number of defects was 3 or more.
To compute the p-value for the alternative hypothesis, we can assess how unusual it would be to observe 18 or more defects (3 or more) under the null hypothesis. If we find that 82 simulations had fewer than 3 defects (0, 1, or 2), this shows a significant number of simulations demonstrating that defects are less than 3.
Given that 82% of the trials resulted in fewer than 3 defects, it indicates that it is very likely that the proportion of defective products is indeed less than 3 out of 1000, leading us to reject the null hypothesis.
Thus, we conclude that the p-value is small (close to 0), suggesting strong evidence against the null hypothesis.
Therefore, the correct conclusion is:
C) The p-value is close to 0. Reject the null hypothesis.