Question

Let \(X_{1}, X_{2}, \ldots, X_{9}\) be a random sample of size 9 from a distribution that is \(N\left(\mu, \sigma^{2}\right)\) (a) If \(\sigma\) is known, find the length of a \(95 \%\) confidence interval for \(\mu\) if this interval is based on the random variable \(\sqrt{9}(\bar{X}-\mu) / \sigma\) (b) If \(\sigma\) is unknown, find the expected value of the length of a \(95 \%\) confidence interval for \(\mu\) if this interval is based on the random variable \(\sqrt{9}(\bar{X}-\mu) / S\). Hint: \(\quad\) Write \(E(S)=(\sigma / \sqrt{n-1}) E\left[\left((n-1) S^{2} / \sigma^{2}\right)^{1 / 2}\right]\). (c) Compare these two answers.

Short answer

The length of the confidence interval when σ is known is 2 * 1.96 * (σ / 3), and when σ is unknown, it is 2 * 2.306 * (σ / (3*sqrt(2/4))). When σ is unknown, the confidence interval is wider reflecting the greater uncertainty.

Step-by-step solution

Part (a)

Since σ is known, the confidence interval for the population mean μ can be calculated from the formula which involves the Z-score. The length of this interval is given by the distance 2 * Z * (σ / √n) from the sample mean to either end of the interval. Here, Z is the Z-score that corresponds to the desired level of confidence, n is the sample size (n=9 in this case). For a confidence level of 95%, Z=1.96. So, the length of the confidence interval becomes 2 * 1.96 * (σ / √9) = 2 * 1.96 * (σ / 3).

Part (b)

When σ is unknown, a t-score is used instead of a Z-score. The expected length of the confidence interval can be calculated similarly as in part (a), but replacing Z by t (with degree of freedom n-1) and σ by S. The formula provided hints that we need to account for the expectation of S. Starting with this formula and using the fact that a chi-squared distribution with n-1 degrees of liberty has expected value n-1, you should get E(S) = σ·sqrt((2/(n-1)). Then, using a t-distribution table or a calculator, find the t-score corresponding to a confidence level of 95% with 8 degrees of freedom, which is approximately 2.306. So, the expected length of the confidence interval is 2 * 2.306 * ((σ*sqrt(2/8))/ √9) = 2 * 2.306 * (σ / (3*sqrt(2/4))).

Part (c)

Comparing these answers, we see that the expected length of the confidence interval is greater when σ is unknown and is estimated from the sample. This makes sense, as more uncertainty (not knowing σ, having to estimate it from the sample) should lead to a wider confidence interval.

Key concepts

Statistical Inference

Statistical inference is a cornerstone of data analysis, allowing us to make predictions or decisions about a population based on sample data. With only a subset of the data from a larger population, statistical inference empowers us to estimate population parameters, test hypotheses, and model relationships using probability.

In the exercise, we're dealing with inference about the population mean, \( \mu \). A key part of this is estimating the precision of our inference, which is where confidence intervals come in. They provide a range of values within which we believe the population mean lies, based on our sample data, with a certain level of confidence. The '95%' confidence level indicates there is a 95% probability that the interval contains the true mean \( \mu \).

Population Mean Estimation

Understanding how to estimate a population mean involves grasping the concept of the confidence interval, an integral tool in statistics. To construct a confidence interval for the population mean, you need a point estimate (in most cases, the sample mean), a standard measure of error (standard deviation for known variance or standard error for unknown variance), and a critical value from a statistical distribution.

For part (a) of the exercise, where the population standard deviation is known, the steps follow a straightforward application of these components. The formula used is grounded in the normal distribution, and the known standard deviation facilitates a clear-cut computation of the interval's length. But in reality, we don't always have the luxury of known variances, which is why part (b) turns to a t-score and introduces an expected value operation for the sample standard deviation, accounting for additional uncertainty.

Z-score vs. t-score

Selecting between a Z-score and a t-score is crucial when estimating confidence intervals. This boils down to one question: is the population standard deviation (\( \sigma \)) known or not? When \( \sigma \)) is known, and the distribution of the sample mean is normally distributed or the sample size is large, the Z-score is used; it's based on the standard normal distribution.

The exercise showcases that with a known \( \sigma \) and a confidence level of 95%, we use the Z-score of approximately 1.96. When \( \sigma \) is unknown and the sample size is small, a t-score from the t-distribution is necessary – this accounts for the extra uncertainty and results in a wider confidence interval. This can be observed in the contrast between steps 1 and 2 of the solution, where part (b) uses a t-score (approximately 2.306 for 8 degrees of freedom) instead of a Z-score.