Problem 18 - Calculus for Biology and Medicine

Question

Suppose that \(X\) is exponentially distributed with mean \(1 .\) A computer generates the following sample of independent observations from the population \(X\) : $$ \begin{array}{l} 0.3169,0.5531,2.376,1.150,0.6174 \\ 0.1563,2.936,1.778,0.7357,0.1024 \end{array} $$ Find the sample mean and the sample variance, and compare them with the corresponding population parameters.

Step-by-Step Solution

Answer

Sample mean: 1.07278, Sample variance: 0.78360. Both are close to the population parameters, especially the mean.

Step 1: Understand the Exponential Distribution

Given that the random variable \(X\) is exponentially distributed with a mean of 1, its probability density function is defined as \( f(x) = e^{-x}, \ x \geq 0\). Note that for an exponential distribution \(X \sim \text{Exp}(\lambda)\), the mean is \(1/\lambda\). Since the mean is 1, we have \(\lambda = 1\).

Step 2: Calculate the Sample Mean

The sample mean \(\bar{x}\) is the average of the observed values. Compute it as follows: \[\bar{x} = \frac{0.3169 + 0.5531 + 2.376 + 1.150 + 0.6174 + 0.1563 + 2.936 + 1.778 + 0.7357 + 0.1024}{10}\]Calculating, we get \(\bar{x} \approx 1.07278\).

Step 3: Calculate the Sample Variance

The sample variance \(s^2\) is calculated using the formula: \[s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2\]This gives:\[s^2 = \frac{1}{9} \left[ (0.3169 - 1.07278)^2 + (0.5531 - 1.07278)^2 + \cdots + (0.1024 - 1.07278)^2 \right]\]Calculating each squared difference, we find \(s^2 \approx 0.78360\).

Step 4: Compare Sampling Results with Population Parameters

For an exponential distribution with \(\lambda = 1\), the theoretical mean is 1 and the variance is also 1. Comparing, the sample mean \( \bar{x} \approx 1.07278\) is close to the population mean; the sample variance \(s^2 \approx 0.78360\) is somewhat less than the population variance.

Key Concepts

Probability Density Function

The probability density function (PDF) is a fundamental concept in statistics that describes the likelihood of a random variable taking on a particular value. For the exponential distribution, the PDF is expressed as \( f(x) = e^{-x}, \ x \geq 0 \). This simple formula tells us that as \( x \) increases, the probability of drawing a value from the distribution decreases exponentially. Hence, the distribution is right-skewed, with more small values and fewer large ones.
This function is essential in understanding how data behaves within this distribution. In our exercise, the exponential distribution with a mean of 1 means that we are looking at events that happen at a constant average rate over time. The PDF plays a critical role in comparing individual data points to the expected behavior in such distributions. By examining the PDF, we can predict which values are most likely and understand the randomness and variability typically seen in an exponential distribution.

Sample Mean

The sample mean is a measure of central tendency that indicates the average value from a set of observations. It is an estimator of the population mean, providing insight into the expected value of the dataset.
To find the sample mean \( \bar{x} \), we sum all the observed values and divide by the number of observations. In this exercise, the calculated sample mean is \( \bar{x} \approx 1.07278 \). This is obtained by taking the sum of all sample data points and dividing by 10:

• Sum of observations = 0.3169 + 0.5531 + 2.376 + 1.150 + 0.6174 + 0.1563 + 2.936 + 1.778 + 0.7357 + 0.1024
• Number of observations (n) = 10
• Sample mean \( \bar{x} = \frac{sum}{n} \approx 1.07278 \)

This result is then compared to the population mean of 1, showing how close the sample mean is to the expected value of the population. In practice, such comparisons help determine the accuracy of the sample estimate and its potential as a representation of the larger population.

Sample Variance

The sample variance measures the spread of data points around the sample mean. It quantifies variability within a data set and is crucial for understanding the diversity of observations. To calculate the sample variance \( s^2 \), we use the formula:\[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \]The sample variance for our exercise is roughly \( s^2 \approx 0.78360 \). Here's how it's computed:

• Calculate each squared deviation from the sample mean.
• Sum these squared deviations.
• Divide by \( n-1 \), where \( n \) is the sample size.

The sample variance is slightly lower than the population variance of 1 in this exercise. Understanding this difference helps in assessing the adequacy of the sample in representing population variability. Smaller variance indicates data points are closely packed around the mean, whereas larger variance suggests more spread.

Population Parameters

Population parameters describe attributes of the entire group from which a sample is drawn. These are fixed values that give insights into the characteristics of a population. In the case of an exponential distribution where the population parameter \( \lambda = 1 \), the theoretical population mean and variance are both 1.
These parameters establish a baseline for comparison with sample statistics. For instance, the population mean indicates the average of the entire dataset if we could observe every possible outcome. Similarly, the population variance measures how much each data point would stray from the mean if we had access to the entire population's data.
In this exercise, we see that both the sample mean \( \bar{x} \approx 1.07278 \) and sample variance \( s^2 \approx 0.78360 \) provide a close approximation of the population parameters. While exact matches are rare due to chance and sample size, these close values illustrate the reliability and accuracy of the sample in reflecting population characteristics. Understanding these relationships is key to statistical inference, where we use sample data to make educated guesses about a broader population.