Ming, a quality assurance analyst at a bottling factory, wants to use a one-sample z-interval to estimate the proportion of 500 mL bottles that are underfilled. She requires the margin of error to be no more than ±4% at a 90% confidence level.

Ming, a quality assurance analyst at a bottling factory, wants to use a one-sample z-interval to estimate the proportion of 500 mL bottles that are underfilled. She requires the margin of error to be no more than ±4% at a 90% confidence level.

Required:
Determine the smallest sample size needed to achieve the desired margin of error.

The Correct Answer and Explanation is:

To determine the smallest sample size required for Ming’s estimate, we use the formula for the margin of error in a one-sample z-interval for a population proportion:

Formula:

ME=z∗⋅p^(1−p^)nME = z^* \cdot \sqrt{\frac{\hat{p}(1 – \hat{p})}{n}}

Where:

  • ME is the margin of error
  • z∗z^* is the critical value corresponding to the confidence level
  • p^\hat{p} is the estimated proportion (we use 0.5 for maximum variability when no prior estimate is available)
  • n is the sample size

Step-by-step Calculation:

  1. Confidence Level = 90%, so the critical value z∗z^* ≈ 1.645
  2. Desired Margin of Error (ME) = 0.04
  3. Assume p^=0.5\hat{p} = 0.5 to ensure the maximum required sample size

0.04=1.645⋅0.5(1−0.5)n0.04 = 1.645 \cdot \sqrt{\frac{0.5(1 – 0.5)}{n}} 0.04=1.645⋅0.25n0.04 = 1.645 \cdot \sqrt{\frac{0.25}{n}} 0.041.645=0.25n\frac{0.04}{1.645} = \sqrt{\frac{0.25}{n}} (0.041.645)2=0.25n\left(\frac{0.04}{1.645}\right)^2 = \frac{0.25}{n} n=0.25(0.041.645)2n = \frac{0.25}{\left(\frac{0.04}{1.645}\right)^2} n≈0.25(0.0243)2≈0.250.000591≈423n \approx \frac{0.25}{(0.0243)^2} \approx \frac{0.25}{0.000591} \approx 423


Answer:

The smallest sample size needed is approximately 423.


Explanation

Ming aims to estimate the proportion of underfilled 500 mL bottles in her bottling factory with a margin of error no greater than ±4% and a confidence level of 90%. To achieve this precision and confidence, determining the correct sample size is crucial.

In estimating a population proportion, the one-sample z-interval is appropriate when the population is large, and the sampling distribution of the sample proportion is approximately normal. This is generally true when both npnp and n(1−p)n(1-p) are at least 10. Since the true proportion is unknown, Ming must assume the worst-case scenario for variability—this occurs when the sample proportion p^=0.5\hat{p} = 0.5, which maximizes the product p^(1−p^)\hat{p}(1 – \hat{p}).

The formula for the margin of error for a population proportion involves the critical value from the standard normal distribution, the estimated proportion, and the sample size. For a 90% confidence level, the critical value is approximately 1.645.

By rearranging the formula to solve for the sample size nn, we find that a sample of at least 423 bottles is required. This ensures that the estimate of the proportion of underfilled bottles will be within ±4% of the true proportion, 90% of the time. Using this approach helps ensure statistically valid and reliable quality control in the factory’s production process.

Scroll to Top