Question

a trucking firm has a large inventory of spare parts that have been in storage for a long time. it knows that some proportion of these spare parts will have deteriorated to the point where they are no longer usable. in order to estimate this proportion, the firm tests 60 parts and finds that 25% of them are not usable. find the 95% (asymptotic) confidence interval for the population proportion of parts that are not usable.

Answer 1

The 95% confidence interval for the population proportion of parts that are not usable is approximately 0.140 to 0.360.

How to find confidence interval?

Standard Error:

The formula is given by:

`\[ \hat{p} \pm Z * \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]`

Where:

`\( \hat{p} \)` = sample proportion of non-usable parts.

n = sample size.

Z = Z-score corresponding to the desired confidence level (for 95%, Z ≈ 1.96).

`\[ \text{Standard Error} = \sqrt{(0.25 * (1 - 0.25))/(60)} \\= \sqrt{(0.25 * 0.75)/(60)} \\= \sqrt{(0.1875)/(60)} \\\approx 0.0559 \]`

Confidence Interval:

Using the Z-score for a 95% confidence level (1.96):

Lower Bound:

`\[ 0.25 - 1.96 * 0.0559 \approx 0.25 - 0.1096 \\= 0.1404 \]`

Upper Bound:

`\[ 0.25 + 1.96 * 0.0559 \approx 0.25 + 0.1096 \\= 0.3596 \]`

Therefore, the 95% confidence interval for the proportion of parts that are not usable is approximately from 0.1404 (14.04%) to 0.3596 (35.96%).