Question

Calculate the sums of squares and cross-products, \(S_{x x}\) and \(S_{x x}\) $$(3,6) \quad(5,8) \quad(2,6) \quad(1,4) \quad(4,7) \quad(4,6)$$

Short answer

In this exercise, we calculated the sums of squares and cross-products, \(S_{xx}\) and \(S_{xy}\), for a given dataset of paired values (x, y). We found the mean values of x and y, calculated the deviation of each data point from the mean values, and used the deviations to find the sums of squares and cross-products. The final answers are: $$S_{xx} = \frac{320}{36}$$ $$S_{xy} = -\frac{29}{36}$$

Step-by-step solution

Calculate the mean values of x and y

Using the given data points, calculate the mean values for \(x\) and \(y\): $$\bar{x} = \frac{3 + 5 + 2 + 1 + 4 + 4}{6} = \frac{19}{6}$$ $$\bar{y} = \frac{6 + 8 + 6 + 4 + 7 + 6}{6} = \frac{37}{6}$$

Calculate the deviation of each data point from the mean values

Now, calculate the deviation of each data point from the mean values of \(x\) and \(y\): $$(x_1-\bar{x}, y_1-\bar{y}) = (3-\frac{19}{6}, 6-\frac{37}{6}) = (-\frac{1}{6}, -\frac{7}{6})$$ $$(x_2-\bar{x}, y_2-\bar{y}) = (5-\frac{19}{6}, 8-\frac{37}{6}) = (\frac{11}{6}, \frac{1}{6})$$ $$(x_3-\bar{x}, y_3-\bar{y}) = (2-\frac{19}{6}, 6-\frac{37}{6}) = (-\frac{7}{6}, -\frac{7}{6})$$ $$(x_4-\bar{x}, y_4-\bar{y}) = (1-\frac{19}{6}, 4-\frac{37}{6}) = (-\frac{13}{6}, -\frac{19}{6})$$ $$(x_5-\bar{x}, y_5-\bar{y}) = (4-\frac{19}{6}, 7-\frac{37}{6}) = (\frac{5}{6}, \frac{5}{6})$$ $$(x_6-\bar{x}, y_6-\bar{y}) = (4-\frac{19}{6}, 6-\frac{37}{6}) = (\frac{5}{6}, -\frac{7}{6})$$

Calculate \(S_{xx}\) and \(S_{xy}\)

Calculate the sums of squares and cross-products using the deviation values found in the previous step: $$S_{xx} = (-\frac{1}{6})^2 + (\frac{11}{6})^2 + (-\frac{7}{6})^2 + (-\frac{13}{6})^2 + (\frac{5}{6})^2 + (\frac{5}{6})^2 = \frac{320}{36}$$ $$S_{xy} = (-\frac{1}{6} \cdot -\frac{7}{6}) + (\frac{11}{6} \cdot \frac{1}{6}) + (-\frac{7}{6} \cdot -\frac{7}{6}) + (-\frac{13}{6} \cdot -\frac{19}{6}) + (\frac{5}{6} \cdot \frac{5}{6}) + (\frac{5}{6} \cdot -\frac{7}{6}) = -\frac{29}{36}$$ So, the sums of squares and cross-products are: $$S_{xx} = \frac{320}{36}$$ $$S_{xy} = -\frac{29}{36}$$

Key concepts

Statistics

Statistics play a critical role in summarizing and analyzing data to draw conclusions about a given population or sample. The field broadly encompasses various measures and calculations designed to capture different characteristics of data.

In the context of the exercise, one important computation is the sum of squares, denoted as
\(S_{xx}\). The sum of squares is a key calculation in statistical analyses, particularly in variance and regression analysis. It measures the total squared deviation of each observation from the mean, providing insight into the data's variability. Higher values indicate a greater spread around the mean.

Calculating the sum of squares is crucial when conducting hypothesis tests or creating a model to predict trends. Understanding the underlying variation within data allows for more accurate predictions and sensible conclusions, making the sum of squares a fundamental concept in statistics.

Mean Deviation

Mean deviation, also known as the mean absolute deviation, is a descriptive statistic that provides a measure of the 'average' distance between each data point and the mean of the data set. It's a way to quantify the amount of variation or dispersion in a set of values.

To calculate mean deviation, you take the absolute value of the difference between each data point and the mean, and then average those differences. Its importance in statistics stems from its use as a measure of dispersion that is less sensitive to outliers compared to the variance or standard deviation.

While mean deviation was not calculated directly in our exercise, understanding this concept helps us appreciate the role of the deviations computed in step 2 of the solution. The deviations, albeit not absolute in this case, are stepping stones to our final calculations and indicate how each data point relates to the dataset's average position.

Bivariate Data

Bivariate data involves pairs of linked numerical observations—for example, the height and weight of a group of individuals. Bivariate analysis seeks to understand the relationship between these two variables. Concepts like correlation and regression come into play, which help in studying the strength and direction of the relationship.

In our exercise, the data points given are examples of bivariate data, where each data point consists of an
\(x\)-value and a
\(y\)-value. The calculation of the cross-product term
\(S_{xy}\) reflects the relationship between the two variables across the dataset. If the
\(S_{xy}\) is positive, it generally signifies that as one variable increases, the other does likewise, and a negative value suggests an inverse relationship.

For students analyzing bivariate data, it's important to recognize patterns and relationships revealed by these calculations, as they provide foundational insights for more complex statistical modeling and interpretation.