Suppose, as in Exercise 12.1 , that two populations have...

Suppose, as in Exercise 12.1 , that two populations have respective variances $\sigma_{1}^{2}=9$ and $\sigma_{2}^{2}=25$ Find the smallest sample size and the corresponding sample allocation that will yield a $95 \%$ confidence interval for $\mu_{1}-\mu_{2}$ that is 2 units in length.

Key Concepts

Confidence Intervals for the Difference of Means

This concept involves constructing an interval estimate around the difference between two population means based on sample data. It uses sampling distributions, typically assuming normality or relying on the central limit theorem, and allows one to infer about the population parameter with a stated level of confidence. In this context, the interval’s length is governed by the estimated standard errors from both samples and the critical value from the appropriate distribution.

Margin of Error and Its Relationship to Sample Size

The margin of error quantifies the maximum expected difference between the true population parameter and the sample estimate, and is directly affected by the sample size. A smaller margin of error (or a desired fixed length of the confidence interval) can be achieved by increasing the sample size, given fixed variances and confidence level. Determining the smallest sample size necessary for a given margin of error is an essential aspect of planning experiments and surveys.

Optimal Allocation in Sampling

When sampling from two or more populations that have different variances, optimal allocation refers to the strategy of distributing samples among the populations so as to minimize overall variance or cost while achieving a desired level of precision. A common method, such as Neyman allocation, assigns more samples to the population with the higher variance. This ensures that the overall precision of the estimate, here the confidence interval for the difference of means, meets the required specifications with the smallest total sample size.

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