Question

An engineer is constructing a confidence interval to estimate the population proportion p of cracked glass panels supplied by company X. To better inform his boss, any such interval is allowed a total width of 10% only (e.g., "percentage of cracked panels is estimated to be between 22% to 32%"), regardless of the sample size or confidence level used. (a) If there is a prior rough estimate of p available at 0.32 while a 99% confidence level is needed, what is the minimum required sample size?

Answer 1

The minimum required sample size for a 99% confidence interval with a total width of 10% and a prior estimate of p = 0.32 is 512.

Step-by-step explanation:

Confidence interval width: We need to achieve a total width of 10% for the confidence interval. Since the confidence level is 99%, we need a margin of error (half the width) of 5% on each side.

Sample size formula: For proportions, the sample size formula for a confidence interval with margin of error E and confidence level 1 - α (where α is the significance level) is:

n = (Z^2_α / E^2) * p * (1 - p)

where:

Z_α is the z-score for the desired confidence level (1.96 for 99% confidence)

E is the margin of error (0.05)

p is the prior estimate of the population proportion (0.32)

Calculating sample size: Plugging in the values:

n = (1.96^2 / 0.05^2) * 0.32 * (1 - 0.32) ≈ 511.76

Rounding up: Since we need a whole number of samples, we round up to the nearest integer: n = 512.

Therefore, the engineer needs to collect a minimum sample size of 512 cracked glass panels to construct a 99% confidence interval with a total width of 10% using the prior estimate of p = 0.32.

Answer 2

To determine the minimum required sample size, we can use the formula for the confidence interval for a proportion.

Step-by-step explanation:

To determine the minimum required sample size, we can use the formula for the confidence interval for a proportion:

`n = (Z / E)^2 * p * (1-p)`

Here, Z is the z-value corresponding to the desired confidence level (99% confidence level corresponds to a z-value of approximately 2.58), E is the maximum desired margin of error (10% or 0.1), and p is the prior rough estimate of p (0.32).

Substituting these values into the formula, we can solve for n:

`n = (2.58 / 0.1)^2 * 0.32 * (1-0.32)`

= 672.71

Therefore, the minimum required sample size is 673.