Problem 21 - An Introduction to Mathematical Statistics and Its Applications

Question

Let \(\bar{Y}\) and \(S\) denote the sample mean and sample standard deviation, respectively, based on a set of \(n=\) 20 measurements taken from a normal distribution with \(\mu=90.6\). Find the function \(k(S)\) for which $$ P[90.6-k(S) \leq \bar{Y} \leq 90.6+k(S)]=0.99 $$

Step-by-Step Solution

Answer

The function \(k(S) = 2.575 \cdot (S/\sqrt{20})\).

Step 1: Understand the formula for the range containing the sample mean

The function given in the exercise describes the range within which the sample mean falls with a 99% confidence. The equation can be expressed as \(90.6 - k(S) \leq \bar{Y} \leq 90.6 + k(S)\), where \(k(S)\) is the 'flexibility' we're adding/subtracting to/from the mean in order to ascertain the range.

Step 2: Extract necessary details from the exercise

From the exercise, we know that \(\bar{Y}\) is the sample mean, \(S\) is the sample standard deviation, \(n\) is the sample size (which is 20), and \(\mu\) is the population mean (which is 90.6). These details are going to guide the substitutions and calculations.

Step 3: Infer the formula for \(k(S)\)

We know that \(\bar{Y}\) follows a Normal distribution with mean \(\mu\) and variance \(S^2/n\). From the normal standard distribution table, for a 99% confidence interval (which refers to a critical value of 2.575), the range condition becomes: \(\mu - 2.575(S/\sqrt{n}) \leq \bar{Y} \leq \mu + 2.575(S/\sqrt{n})\). By comparing this with the initial range equation, we can conclude that \(k(S) = 2.575 \cdot (S/\sqrt{n})\).

Key Concepts

Sample Mean

The sample mean, denoted as \(\bar{Y}\), plays a crucial role in statistical analyses as it represents the average of a group of observations. It's the sum of all observations divided by the number of observations. In the context of confidence intervals, we're often interested in how this sample mean compares to the population mean (\(\mu\)) which is an unknown constant.

For instance, if we take 20 measurements from a given population, the sample mean is calculated by adding up all 20 measurements and then dividing by 20. It's essential to know that the sample mean is a random variable itself, and it has its own distribution, which is particularly important when we're making inferences about the population mean. The closer the sample mean is to the population mean, the more accurate our inferences can be.

Sample Standard Deviation

The sample standard deviation (\(S\)) gauges the variability or dispersion of the data points in a sample. It indicates how much individual measurements deviate from the sample mean. A low standard deviation suggests that the data points tend to be close to the mean, whereas a high standard deviation indicates that the data points are spread out over a wider range of values.

The formula to calculate the sample standard deviation involves taking the square root of the sum of the squared differences between each measurement and the sample mean, divided by \(n-1\), where \(n\) is the sample size-1. This calculation is crucial for constructing confidence intervals because it helps determine the spread of the sample mean distribution around the population mean. While dealing with the concept in our example, understanding how the standard deviation influences the range of the sample mean is key to interpreting the confidence interval.

Normal Distribution

A normal distribution is a bell-shaped curve that is symmetrical about the mean. It is defined by its mean (\(\mu\)) and standard deviation (\(\sigma\)), which determine the location and spread of the distribution. In a normal distribution, most of the observations cluster around the central peak, and the probabilities for values further away from the mean taper off symmetrically in both directions.

Understanding the normal distribution is fundamental when we discuss confidence intervals. In the context of the given exercise, the sample mean distribution can be approximated to be normally distributed if the sample size is large enough or the population from which the sample is drawn is normally distributed. This assumption allows us to use z-scores and the standard normal distribution to calculate the probability that the sample mean falls within a specific range. For a 99% confidence interval, as in our exercise, we want to find the range where there's a 99% chance that the true population mean lies within it. This is crucial because it provides a high level of certainty about our estimate of the population mean.