Random Sample Of 500 Receipts From A Population

A Random Sample Of 500 Receipts Was Taken From A Population With An Un

A random sample of 500 receipts was taken from a population with an unknown proportion of errors. The confidence interval for the true proportion of receipts that contain an error is given as (4.22%, 10.18%). Based on this information, the following questions are addressed:

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Introduction

Sampling and statistical inference play essential roles in quality control and operational efficiency within various industries, including retail. The analysis of error proportions in receipts can inform management about the accuracy of transaction processing, the effectiveness of internal controls, and the need for process improvements. When a random sample is taken from a large population of receipts, and a confidence interval for the error proportion is calculated, it provides an estimated range within which the true error rate is likely to fall with a certain level of confidence. The process involves understanding point estimates, constructing confidence intervals, and determining sample sizes necessary for desired precision levels.

Part (a): Estimating the Number of Error Receipts in the Sample

The confidence interval provided, (4.22%, 10.18%), has its center at a point estimate of the population proportion of receipts with errors. This point estimate, commonly denoted as p̂, is at the midpoint of the interval and is calculated as the average of the lower and upper bounds:

p̂ = (4.22% + 10.18%) / 2 = (0.0422 + 0.1018) / 2 ≈ 0.072

This means that the best estimate of the proportion of receipts with errors in the sample is approximately 7.2%. Given a sample size of 500 receipts, the estimated number of receipts with errors is:

Number of error receipts = p̂ × 500 ≈ 0.072 × 500 ≈ 36

Therefore, about 36 receipts out of the 500 sampled were found to contain errors.

Part (b): Approximate the Confidence Level

The confidence level related to a confidence interval is the probability that the interval will contain the true population parameter upon repeated sampling. To approximate this, we analyze the margin of error (ME), which is half the width of the confidence interval:

ME = (10.18% - 4.22%) / 2 = (0.1018 - 0.0422) / 2 ≈ 0.0298, or about 2.98%

Assuming a normal distribution approximation for the binomial proportion, the margin of error relates to the standard error (SE) and the z-value corresponding to the confidence level:

ME = z* × SE

where

SE = sqrt[(p̂(1 - p̂))/n] = sqrt[(0.072)(0.928)/500] ≈ sqrt(0.067/500) ≈ 0.0116

Thus, the z-value is:

z* = ME / SE ≈ 0.0298 / 0.0116 ≈ 2.57

A z-value of approximately 2.57 corresponds closely to a confidence level of about 99%. This suggests the original confidence interval was constructed with roughly 99% confidence.

Part (c): Determining the Required Sample Size for a Reduced Margin of Error

To reduce the margin of error to at most 1% (0.01), we use the formula for the margin of error in the context of a confidence interval:

ME = z* × sqrt[(p(1 - p))/n]

Since the problem asks us to be conservative—meaning we should not assume the sample proportion as the best estimate—but rather use the confidence level derived previously (~99%), we will consider the most conservative estimate for p as 0.5, because it maximizes the variance p(1 - p).

The z-value for a 99% confidence level is approximately 2.58.

n = [z* / ME]^2 × p̄(1 - p̄)

Plugging in the values:

n = (2.58 / 0.01)^2 × 0.5 × 0.5 ≈ (258)^2 × 0.25 ≈ 66,564 × 0.25 ≈ 16,640.97

Therefore, a sample size of approximately 16,641 receipts is needed to ensure a margin of error of at most 1% with 99% confidence, assuming the most conservative estimate of p.

Conclusion

In summary, based on the provided confidence interval, roughly 36 receipts out of the sampled 500 contain errors. The approximate confidence level used for this interval is about 99%. To achieve a higher precision with a margin of error of 1%, a significantly larger sample size of approximately 16,641 receipts is necessary. These findings are instrumental for quality assurance managers aiming to monitor and improve receipt processing accuracy effectively.

References

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