Question

For a numerical example of the \(F\) -test derived in Exercise \(6.5 .7\), here are two generated data sets. The first was generated by the \(\mathrm{R}\) call \(\operatorname{rexp}(10,1 / 20)\), i.e., 10 observations from a \(\Gamma(1,20)\) -distribution. The second was generated by \(\operatorname{rexp}(12,1 / 40)\). The data are rounded and can also be found in the file genexpd. rda. (a) Obtain comparison boxplots of the data sets. Comment. (b) Carry out the F-test of Exercise 6.5.7. Conclude in terms of the problem at level \(0.05\) $$ \begin{aligned} &\mathrm{x}: 11.1 .11 .7 & 12.7 & 9.6 & 14.7 & 1.6 & 1.756 .13 .3 & 2.6 \\ &\mathrm{y}: 55.6 & 40.5 & 32.7 & 25.6 & 70.6 & 1.4 & 51.5 & 12.6 & 16.9 & 63.3 & 5.6 & 66.7 \end{aligned} $$

Short answer

The boxplots will visually showcase the differences in spread and central tendencies among the two data sets. The outcome of the F-test will then determine whether these differences are statistically significant at the 0.05 level. The specific results will depend on the calculated F statistic and its comparison to the critical F value.

Step-by-step solution

Generate boxplots to compare the data sets

To generate the boxplots, you can use a statistical software or tool that supports boxplot creation (like R). The aim is to visualize the range, interquartile range (IQR), median, and possible outliers of the two data sets (x and y). This will provide insight into how the two data sets compare.

Compute the means and variances

Before carrying out the F-test, calculate the sample means and variances for both data sets. This will provide necessary input for the F-test. Use the formulas \(\bar{x}, \ \bar{y} = \frac{1}{n} \sum x_i, \ \frac{1}{m} \sum y_i\) for means and \(s_{x}^{2}, s_{y}^{2} = \frac{1}{n-1} \sum (x_i - \bar{x})^{2}, \ \frac{1}{m-1} \sum (y_i - \bar{y})^{2}\) for variances where \(n, m\) are the number of observations in each data set.

Conduct the F-test

After calculating the sample variances, compute the F statistic as the ratio of the two variances \(s_{x}^{2}/s_{y}^{2}\). Then, compare this calculated F-statistic with the critical F-value obtained from the F-distribution table for degrees of freedom \((n-1, m-1)\) and significance level 0.05. If the calculated F-statistic is greater than the critical value, we reject the null hypothesis that the variances are equal.

Interpret the results

Finally, draw conclusions based on the results of the test. If the null hypothesis was rejected, then it may be concluded that the two data sets have significantly different variances. If the null hypothesis was not rejected, this means that there is no sufficient evidence to suggest that the variances are different.

Key concepts

Boxplot Comparison

Boxplots, also known as whisker diagrams, are an excellent graphical method for comparing the distribution of numerical data across different categories or groups. By representing data in terms of quartiles, they show the median, the lower and upper quartiles, and the range of the data, including any outliers.

When comparing two data sets, such as the generated set \(x\) and \(y\) in our example, boxplots can quickly reveal differences in central tendency (median), spread (interquartile range), as well as the presence of outliers, which can be critical for subsequent statistical tests.

Interpreting Boxplots

In our exercise, contrasting the boxplots of sets \(x\) and \(y\) would likely show contrasting medians, range, and variability. If one plot has a longer range or more outliers, it can suggest different underlying processes generating these two sets. Disparities in these aspects of the boxplots inform the analyst that the sample means and variances need closer examination, leading us to the next concept.

Sample Means and Variances

Understanding sample means and variances is quintessential for comparing two sets of data. In statistical terms, the mean provides a measure of central tendency, while the variance measures the spread - how much the data points differ from the mean.

The formula to calculate sample mean \( \bar{x} \) or \( \bar{y} \) involves summing all data points (\( \sum x_i \) or \( \sum y_i \) respectively) and dividing by the number of points (\( n \) or \( m \) respectively). To find the variance (\( s_{x}^{2} \) or \( s_{y}^{2} \) respectively), we sum the squares of the differences between each data point and the sample mean, then divide by the number of points minus one.

Why Variance Matters

For statistical tests like the F-test, the variances of two samples are specifically compared. This comparison assumes that if the processes generating the data are similar, then the variances should not differ significantly. Hence, a significant difference in variances can signal that the datasets may not be from the same population or process.

F-distribution

The F-distribution is a fundamental aspect of comparing variances across different samples. It arises when the ratio of two independent variances, each divided by their respective degrees of freedom (the number of observations minus one), is taken.

This ratio follows the F-distribution, with the two sets of degrees of freedom as its parameters. In the F-test, we use this distribution to find a critical value that becomes the threshold to decide whether the variances are significantly different.

Critical Values in F-distribution

For our example, we would look up the F-distribution table for our degrees of freedom - based on the number of observations in each sample - and our chosen significance level, typically 0.05. If our calculated F statistic is greater than the critical value from the table, it indicates a statistically significant difference between variances of the two samples at the 5% significance level.

Hypothesis Testing

Hypothesis testing is a statistical method that allows us to make inferences about a population parameter by analyzing sample data. In the context of an F-test, we use hypothesis testing to determine if there are significant differences in variances between two samples.

The null hypothesis (\( H_0 \)) would state that there is no variance difference between the datasets. The alternative hypothesis (\( H_a \)) claims that a difference does exist. An F-test analyzes whether the evidence is strong enough to reject the null hypothesis in favor of the alternative.

Making a Decision

Upon computing the F-statistic and comparing it against the critical F-value of the F-distribution, we make a decision. If the statistic exceeds the critical value, we reject the null hypothesis; if not, we do not have enough evidence to do so. This rejection or failure to reject directs us towards conclusions about the population variances being equal or not at the chosen level of significance.