We aim to determine the approximate sample size ($n$) required to construct a 99% confidence interval for the probability of a chip being acceptable ($p$) with a length of approximately 0.05. Here's the step-by-step process:

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Step 1: Confidence Interval Formula

The confidence interval for $p$ in this situation is:

$CI = p \pm z_{\alpha/2} \cdot \sqrt{\frac{p(1-p)}{n}}$

The length of the confidence interval is given by:

$\text{Length} = 2 \cdot z_{\alpha/2} \cdot \sqrt{\frac{p(1-p)}{n}}$

Where:

• $z_{\alpha/2}$ is the critical value for a 99% confidence interval.
• $p$ is the sample proportion of acceptable chips.
• $n$ is the sample size.

We want the length to be approximately $0.05$. Hence:

$0.05 = 2 \cdot z_{\alpha/2} \cdot \sqrt{\frac{p(1-p)}{n}}$

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Step 2: Estimate $z_{\alpha/2}$

For a 99% confidence interval, $z_{\alpha/2} \approx 2.576$.

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Step 3: Estimate $p$

The initial sample gives $24$ acceptable chips out of $30$. Hence, the sample proportion is:

$p = \frac{24}{30} = 0.8$

The variance term is:

$p(1-p) = 0.8 \cdot (1 - 0.8) = 0.8 \cdot 0.2 = 0.16$

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Step 4: Rearrange the Formula for $n$

Rearranging the formula for the length of the confidence interval:

$n = \frac{(2 \cdot z_{\alpha/2})^2 \cdot p(1-p)}{\text{Length}^2}$

Substituting the known values:

• $z_{\alpha/2} = 2.576$,
• $p(1-p) = 0.16$,
• $\text{Length} = 0.05$,

$n = \frac{(2 \cdot 2.576)^2 \cdot 0.16}{0.05^2}$

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Step 5: Simplify the Expression

$n = \frac{(5.152)^2 \cdot 0.16}{0.0025}$

First, calculate $5.152^2$:

$5.152^2 = 26.525$

Then:

$n = \frac{26.525 \cdot 0.16}{0.0025} = \frac{4.244}{0.0025} = 1697.6$

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Final Answer

The required sample size is approximately:

$n \approx 1698$

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Do you want more details or have any questions?

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Related Questions

1. How does increasing the confidence level affect the required sample size?
2. Why does the approximation $p \approx 0.8$ work for this calculation?
3. What is the impact of reducing the desired confidence interval length further?
4. How would this calculation change for a two-sided test with different sample proportions?
5. What assumptions are made about the distribution of acceptable and unacceptable chips?

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Tip:

Always verify whether the sample size is appropriate after calculation, especially if the population size is limited.