Question

A quality-control engineer wants to estimate the fraction of defectives in a large lot of film cartridges. From previous experience, he feels that the actual fraction of defectives should be somewhere around .05. How large a sample should he take if he wants to estimate the true fraction to within .01, using a \(95 \%\) confidence interval?

Short answer

Answer: The engineer should sample at least 865 film cartridges to achieve the desired level of precision.

Step-by-step solution

Identify the known values and symbols

We are given the following information: - Anticipated proportion of defects (from previous experience): \(p \approx 0.05\) - Desired level of error in estimation: \(E = 0.01\) - Confidence interval: \(95 \%\). Let \(n\) represent the required sample size.

Calculate the Z-score for the given confidence interval

For a \(95 \%\) confidence interval, we can calculate the Z-score by finding the critical value corresponding to a probability of \(0.975\) (since we are looking at both tails) in a standard normal distribution table. The Z-score is denoted as \(Z_{\alpha/2}\). \(Z_{\alpha/2} = 1.96\)

Use the formula for margin of error estimation in proportions

The formula for the margin of error (the maximum difference between the observed proportion and the true proportion) in proportion estimation is given by: \(E = Z_{\alpha/2} \sqrt{\frac{p(1-p)}{n}}\) We need to solve for \(n\): \(E = 0.01 = 1.96 \sqrt{\frac{0.05(1-0.05)}{n}}\)

Solve for n

Square both sides of the equation: \((0.01)^2 = (1.96)^2 \frac{0.05(1-0.05)}{n}\) Divide both sides by \((1.96)^2\): \(\frac{(0.01)^2}{(1.96)^2} = \frac{0.05(1-0.05)}{n}\) Now find \(n\): \(n = \frac{0.05(1-0.05)}{\frac{(0.01)^2}{(1.96)^2}}\)

Calculate the required sample size

Plug in the values and calculate \(n\): \(n = \frac{0.05(0.95)}{\frac{0.0001}{3.8416}}\) \(n \approx 864.25\) Since we cannot have a fraction of a sample, round up to the nearest whole number: \(n = 865\) So, the quality-control engineer should take a sample of at least \(865\) film cartridges to estimate the true fraction of defectives to within \(0.01\) using a \(95 \%\) confidence interval.

Key concepts

Confidence Interval

When a quality-control engineer or a researcher is working to understand a characteristic of a large population, it's often impractical to examine every individual. Instead, they take a sample and make inferences about the population. A confidence interval (CI) is a type of estimate computed from the statistics of the observed data. This gives a range of values for an unknown parameter (like the proportion of defective items) and is associated with a confidence level that quantifies the level of certainty in the estimate.

The confidence level is expressed as a percentage, such as 95%, and indicates how often the true parameter would fall within the calculated interval if the same population were sampled repeatedly. In our exercise, the engineer wants to use a 95% CI, meaning there is a 95% chance that the true fraction of defectives lies within the boundaries set by the interval.

Margin of Error

Margin of error is a statistic expressing the amount of random sampling error in a survey's results. It represents how much you can expect survey results to reflect the views of the actual population. The smaller the margin of error, the closer you can expect the sample's estimate to be to the true value of the population parameter.

In the given exercise, the quality-control engineer wants the margin of error to be 0.01 for the estimated fraction of defectives, which means the true fraction should be no more than 0.01 away from the observed proportion, in either direction. The margin of error is wrapped up in the calculation of the confidence interval, and it's determined by factors such as the size of the sample and the variance in the population—which, in this case, is estimated based on past data.

Proportion Estimation

Proportion estimation is a statistical technique used to infer the value of a particular proportion—like the the proportion of defective items in a batch—of the total population. Estimating a proportion involves selecting a sample from the population and using the number of times the characteristic of interest appears to estimate the proportion for the entire population.

The exercise in question centers around proportion estimation, where the engineer is aiming to estimate the true fraction of defectives in a lot. The anticipated proportion, based on past experience, is about 0.05 or 5%, but the engineer wants to understand this proportion with a higher degree of certainty by determining an appropriate sample size.

Z-score

The Z-score, also known as a standard score, indicates how many standard deviations an element is from the mean. In the context of confidence intervals and sample size determination, the Z-score helps in quantifying the number of standard deviations a data point (like the proportion of defects) is expected to fall from the population proportion. When setting up a confidence interval, the Z-score is used to represent the desired confidence level.

For a 95% confidence interval, this 'critical value' ensures that 95% of the time, the true proportion lies within the estimated range. In our example, a Z-score of 1.96 is used, corresponding to the critical value for a 95% confidence interval (reflecting the 95% probability within the two tails of the normal distribution), and it is a crucial factor in determining the required sample size to achieve the desired precision.