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Calculate the sums of squares and cross-products, \(S_{x x}\) and \(S_{x x}\) $$\begin{array}{c|ccc}x & 1 & 3 & 2 \\\\\hline y & 6 & 2 & 4\end{array}$$

Short Answer

Expert verified
In this exercise, we calculated the sum of squares (Sₓₓ) and the sum of cross-products (Sₓᵧ) using the given data. The means of x and y were found to be 2 and 4, respectively. We calculated the differences between each value and their respective means and used these to compute Sₓₓ and Sₓᵧ. The final result is that the sum of squares (Sₓₓ) is 2, and the sum of cross-products (Sₓᵧ) is -4.

Step by step solution

01

Calculate the means of x and y

First, calculate the mean of each variable, \(x\) and \(y\). This is done by finding the sum of the values in each column and dividing it by the number of observations, which is 3 in this case. Mean of \(x\) (\(\bar{x}\)): \(\bar{x} = \frac{1 + 3 + 2}{3} = 2\) Mean of \(y\) (\(\bar{y}\)): \(\bar{y} = \frac{6 + 2 + 4}{3} = 4\)
02

Calculate the differences between each value and their respective means

Now, we need to find the differences between each value of \(x\) and the mean of \(x\), as well as the differences between each value of \(y\) and the mean of \(y\). Differences for x: \(1 - 2 = -1\) \(3 - 2 = 1\) \(2 - 2 = 0\) Differences for y: \(6 - 4 = 2\) \(2 - 4 = -2\) \(4 - 4 = 0\)
03

Calculate the sum of squares \(S_{xx}\)

\(S_{xx}\) is the sum of the squared differences between each value of \(x\) and the mean of \(x\). From step 2, we got the differences as \(-1\), \(1\), and \(0\). We will square these and sum them up. \(S_{xx} = (-1)^2 + (1)^2 + (0)^2 = 1 + 1 + 0 = 2\)
04

Calculate the sum of cross-products \(S_{xy}\)

\(S_{xy}\) is the sum of the product of the differences between each value of \(x\) and the mean of \(x\), and each value of \(y\) and the mean of \(y\). Using the differences we got from step 2 (\(x\) differences: \(-1\), \(1\), \(0\), \(y\) differences: \(2\), \(-2\), \(0\)), we will find the product of each pair of \(x\) and \(y\) differences and sum them up. \(S_{xy} = (-1)(2) + (1)(-2) + (0)(0) = -2 - 2 + 0 = -4\)
05

Conclusion

The sum of squares \(S_{xx}\) is 2, and the sum of cross-products \(S_{xy}\) is -4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Statistical Mean
The statistical mean is a central concept in mathematics and statistics, often referred to simply as the 'average'. It is calculated by adding up all the numbers in a data set and then dividing by the count of those numbers. For example, if we have a set of numbers such as 1, 3, and 2, the sum is 6 and there are 3 numbers. Dividing the sum by the count, the mean would be \( \bar{x} = \frac{6}{3} = 2 \).

The mean gives us a measure of central tendency, or an idea of where the middle of the data set lies. It's extremely useful in various fields, including economics, social sciences, and engineering, for summarizing a set of data with a single value. But remember, the mean is only a representative value; it does not account for the spread or variance within the data.
Sum of Squared Differences
The sum of squared differences is a measure of variability in a data set. It's the total of each data point's deviation from the mean, squared. Squaring the differences does two things: it eliminates negative values and it gives more weight to larger differences. We use the formula \( S_{xx} = \sum(x_i - \bar{x})^2 \), where \( x_i \) represents each data point and \( \bar{x} \) is the mean.

For instance, if we have the values 1, 3, and 2 with a mean of 2, the squared differences would be (-1)^2, (1)^2, and (0)^2. The sum of these squared differences is 1 + 1 + 0 = 2. This measure is fundamental in calculating variance and standard deviation, which are key statistics for understanding the dispersion in a data set.
Sum of Cross-Products
When working with bivariate data, which involves pairs of related data points (like height and weight of individuals), we often need to calculate the sum of cross-products. This term refers to the sum of the product of paired differences from their respective means. It is expressed as \( S_{xy} = \sum((x_i - \bar{x})(y_i - \bar{y})) \).

The measure helps us understand the relationship between two variables. If the sum of cross-products is positive, the variables tend to move together, while a negative sum suggests they move in opposite directions. In the given example, we found the sum of cross-products \( S_{xy} \) to be -4, indicating an inverse relationship between x and y in this sample.
Bivariate Data Analysis
Bivariate data analysis involves the exploration and understanding of relationships between two different variables. We often look at how one variable may predict or affect the other. This type of analysis includes plotting data on scatter plots, calculating correlation coefficients, and modeling relationships with regression lines.

Tools like the sum of squares and sum of cross-products play a critical role in such analysis. They are used to calculate coefficients of determination (\( r^2 \) for correlation) and slopes in regression equations, aiding predictions and interpretations about the interconnectedness of the variables under study. Remember, while these statistical measures can show relationships, they don't necessarily imply causation.

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Most popular questions from this chapter

Refer to Exercise \(11 .\) The sample correlation coefficient \(r\) for the stride rate and the average acceleration rate for the 69 skaters was . \(36 .\) Do the data provide sufficient evidence to indicate a correlation between stride rate and average acceleration for the skaters? Use the \(p\) -value approach.

What diagnostic plot can you use to determine whether the data satisfy the normality assumption? What should the plot look like for normal residuals?

Use the data given in Exercises 6-7 (Exercises 17-18, Section 12.1). Construct the ANOVA table for a simple linear regression analysis, showing the sources, degrees of freedom, sums of squares, and mean sauares. $$\begin{aligned}&\text { Six points have these coordinates: }\\\&\begin{array}{l|llllll}x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 9.7 & 6.5 & 6.4 & 4.1 & 2.1 & 1.0\end{array}\end{aligned}$$ a. Find the least-squares line for the data. b. Plot the six points and graph the line. Does the line appear to provide a good fit to the data points? c. Use the least-squares line to predict the value of \(y\) when \(x=3.5\) d. Fill in the missing entries in the MS Exce/ analysis of variance table.

Independent and Dependent Variables Identify which of the two variables in Exercises \(10-14\) is the independent variable \(x\) and which is the dependent variable \(y .\)Speed of a wind turbine and the amount of electricity generated by the turbine.

Plot the data points given in Exercises 4-5. Based on the graph, what will be the sign of the correlation coefficient? Then calculate the correlation coefficient, \(r\), and the coefficient of determination, \(r^{2} .\) Is the sign of \(r\) as you expected? $$\begin{array}{l|llllll}x & 1 & 2 & 3 & 4 & 5 & 6 \\\\\hline y & 7 & 5 & 5 & 3 & 2 & 0\end{array}$$

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