Problem 3 - An Introduction to Mathematical Statistics and Its Applications

Question

Below are \(n=50\) computer-generated observations that are presumably a random sample from the exponential pdf, \(f_{Y}(y)=e^{-y}, y \geq 0\). Use Theorem 14.2.1 to test whether the difference between the sample median for these \(y_{i}\) 's \((=0.604)\) and the true median of \(f_{Y}(y)\) is statistically significant. Let \(\alpha=0.05\). \(\begin{array}{lllllll}0.27187 & 0.46495 & 0.19368 & 0.80433 & 1.25450 & 0.62962 & 1.88300 \\ 1.31951 & 2.53918 & 1.21187 & 0.95834 & 0.49017 & 0.87230 & 0.88571 \\ 1.41717 & 1.75994 & 0.60280 & 2.19654 & 0.00594 & 4.11127 & 0.24130 \\ 0.16473 & 0.08178 & 1.01424 & 0.60511 & 0.87973 & 0.06127 & 0.24758 \\ 0.54407 & 0.05267 & 0.75210 & 0.13538 & 0.42956 & 0.02261 & 1.20378 \\ 1.09271 & 1.88705 & 0.17500 & 0.50194 & 0.52122 & 0.02915 & 0.27348 \\ 0.08916 & 0.72997 & 0.37185 & 0.06500 & 1.47721 & 4.02733 & 0.64003 \\ 0.05603 & & & & & & \end{array}\)

Step-by-Step Solution

Answer

Based on the analysis and calculations, it is found that the p-value (0.209) is greater than the significance level (\( \alpha = 0.05 \)). Thus, there is not enough evidence to reject the null hypothesis that the difference between the sample median and the true median is statistically significant. Therefore, we would not say that the difference between the sample median and true median is significant.

Step 1: Calculation of True Median of Exponential Distribution

The formula to calculate the median of an exponential distribution is \(ln(2)/lambda\). Given that the probability density function \(f_Y(y)=e^{-y}\), we can deduce that \(\lambda\) is 1. Therefore, the true median of exponential distribution = \(ln(2)\) = 0.693.

Step 2: Calculation of Test Statistic

Following Theorem 14.2.1 and using the Wilcoxon Signed-Rank Test for medians, the test statistic is calculate as \((Sample Median - True Median) / (Standard Deviation / sqrt(n))\). However, in the given problem, we don't have standard deviation of the sample. The standard deviation for a single sample in an exponential distribution is equal to standard deviation / sqrt(n). As we know that for an exponential distribution, standard deviation = 1 / \( \lambda \) = 1 (since \( \lambda \) = 1), the standard deviation for the sample is equal to 1 / sqrt(50). Now, we can calculate the test statistic = (0.604 - 0.693) / (1 / sqrt(50)) = -1.256.

Step 3: P-Value Calculation and Result Interpretation

The P-value associated with this test statistic, assuming a normal distribution, can be found using the standard normal distribution table or using software. Suppose the P-value is found to be 0.209. As the P-value is greater than the level of significance (\( \alpha = 0.05 \)), we would not reject the null hypothesis that the difference between the sample median and the true median is statistically significant.

Key Concepts

Wilcoxon Signed-Rank Test

The Wilcoxon Signed-Rank Test is a non-parametric statistical test used to compare paired data. It helps determine if there is a significant difference between two medians from related groups. Unlike parametric tests, it does not assume a normal distribution and is ideal for ordinal data or continuous data that does not follow a normal distribution.
Here are some key attributes of the test:

• It assesses differences between paired observations.
• It ranks the absolute differences without signs, assigning average ranks if there are any ties.
• After ranking, the test considers the sum of ranks for positive and negative differences to compute the test statistic.

In the context of this problem, the Wilcoxon Signed-Rank Test is used because raw data may not satisfy the conditions required for parametric tests like the t-test.

Random Sample

A random sample is a subset of a population selected in such a way that each member of the subset has an equal probability of being chosen from the population. This method ensures that the sample is representative, allowing for generalizations to be made about the wider population.

• Enhances fairness: Each individual is chosen purely by chance, which reduces sample bias.
• Diversity: Reflects a wide variety of responses or outcomes within the entire group.

In our scenario, the 50 observations provided represent a random sample from an exponential distribution. This gives us a robust basis for further analysis using statistical tests to infer the characteristics of the population from which the sample is drawn.

Statistical Significance

Statistical significance is a statistical assessment that helps determine if the results of a study are likely to be true and not occurred by random chance. When we say that a result is statistically significant, it indicates that there is evidence to reject the null hypothesis.

• The null hypothesis usually states that there is no effect or no difference.
• A commonly used threshold for determining statistical significance is \(\alpha = 0.05\).
• If the P-value is less than \(\alpha\), the result is typically considered significant, meaning the observed effect is unlikely under the null hypothesis.

In the given problem, even though the sample median is slightly different from the true median, the analysis revealed no statistical significance due to a high P-value.

P-Value

The P-value measures the strength of evidence against the null hypothesis. It tells us the probability of observing our data, or something more extreme, if the null hypothesis is true.

• A small P-value (< \(\alpha\)) implies strong evidence against the null hypothesis, leading to its rejection.
• A large P-value suggests weak evidence against the null hypothesis, so it is not rejected.
• In many studies, \(\alpha = 0.05\) serves as the cutoff for determining significance.

For this exercise, the calculated P-value was 0.209. Because this is greater than \(\alpha = 0.05\), we do not have enough evidence to claim statistical significance of the difference between the sample and true medians.