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Chapter 17: Problem 663

Given 4 pairs of observations \begin{tabular}{|c|c|c|c|c|} \hline \(\mathrm{X}\) & 6 & 7 & 4 & 3 \\ \hline \(\mathrm{Y}\) & 8 & 10 & 4 & 2 \\ \hline \end{tabular} compute and graph the least-squares regression line.

Short Answer

Expert verified
The least-squares regression line for the given dataset is \(Y = -\frac{10}{13}X + \frac{28}{13}\). To graph the line, plot the points (6,8), (7,10), (4,4), and (3,2), and then draw the line with the given equation that minimizes the residuals.

Step by step solution

01

Compute the sums and means

Firstly, sum each variable (X and Y) and their products (XY). Then, calculate the mean for X and Y by dividing their sums by the total number of observations (4, in this case). \[\sum X = 6 + 7 + 4 + 3 = 20\] \[\sum Y = 8 + 10 + 4 + 2 = 24\] \[\sum XY = 6*8 + 7*10 + 4*4 + 3*2 = 100\] Next, compute the mean of X and Y: \[\bar{X} = \frac{\sum X}{4} = \frac{20}{4} = 5\] \[\bar{Y} = \frac{\sum Y}{4} = \frac{24}{4} = 6\]
02

Compute the slope (b)

Now, calculate the slope (b) for the regression line using the following formula: \[b = \frac{\sum XY - n\bar{X}\bar{Y}}{\sum X^2 - n\bar{X}^2}\] where n is the number of observations and \(X^2\) represents the squared values of X. First, compute the sum of the squared values of X: \[\sum X^2 = 6^2 + 7^2 + 4^2 + 3^2 = 126\] Now, compute the slope: \[b = \frac{100 - 4*5*6}{126 - 4*5^2} = \frac{100 - 120}{126 - 100} = \frac{-20}{26} = -\frac{10}{13}\]
03

Compute the intercept (a)

Next, compute the intercept (a) using the following formula: \[a = \bar{Y} - b\bar{X}\] Substitute the values we computed earlier and get the intercept: \[a = 6 - (-\frac{10}{13})5 = 6 + \frac{50}{13} = \frac{28}{13}\] So, the least-squares regression line is: \[Y = -\frac{10}{13}X + \frac{28}{13}\]
04

Graph the regression line

To graph the least-squares regression line, plot all given data points and the regression equation: \[(6,8), (7,10), (4,4), (3,2)\] \[Y = -\frac{10}{13}X + \frac{28}{13}\] Once you plot the 4 data points, draw the regression line, making sure to indicate the equation of the line on the graph. Ensure that the line passes through the mean datapoint \((5, 6)\) and visually minimizes the residuals (vertical distances) between the given points and the line.

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Key Concepts

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

Regression Analysis
Regression analysis is a powerful statistical tool used to examine the relationship between two variables. It involves exploring how a dependent variable (also known as the outcome or Y) changes when an independent variable (also known as the predictor or X) varies. In the case of a simple linear regression, the goal is to fit a straight line to a set of data points to best describe this relationship.

This line, called the least-squares regression line, is determined in such a way that it minimizes the sum of the squares of the residuals鈥攖he differences between the observed values and the values predicted by the line. The formula of the line is often written as \( Y = a + bX \), where \( a \) is the intercept with the Y-axis, and \( b \) is the slope, reflecting the steepness of the line. Understanding the slope and intercept allows you to predict Y for any given value of X.
Slope Calculation
The slope of the regression line, denoted by \( b \), indicates the average change in the dependent variable (Y) for each unit increase in the independent variable (X). To calculate the slope, it's important to first gather the sums of X, Y, their products (XY), and the squared values of X (\(X^2\)).

The formula for calculating the slope is:\[ b = \frac{\sum XY - n\bar{X}\bar{Y}}{\sum X^2 - n\bar{X}^2} \]
Here, \( n \) represents the number of data points. A positive slope indicates a positive relationship between X and Y, while a negative slope suggests a negative relationship. In the given exercise, the slope was found to be negative, meaning that as X increases, Y tends to decrease.
Intercept Calculation
Once you have the slope, the next step is to calculate the intercept, represented by \( a \). The intercept is the point at which the regression line crosses the Y-axis, corresponding to the estimated value of Y when X is zero.

The formula for the intercept is straightforward:\[ a = \bar{Y} - b\bar{X} \]
It involves taking the mean of Y and subtracting the product of the slope and the mean of X. In the exercise, the intercept shows where the regression line intersects the Y-axis at around \( \frac{28}{13} \) when X is zero. The intercept, along with the slope, fully defines the least-squares regression line.
Sum of Squared Values
The sum of squared values plays a key role in regression analysis, particularly when calculating the slope of the regression line. It involves taking each value of X, squaring it, and then summing up all these squared numbers. The sum of \(X^2\) is used in the denominator of the slope formula:\[ b = \frac{\sum XY - n\bar{X}\bar{Y}}{\sum X^2 - n\bar{X}^2} \]

By squaring each deviation from the mean, we ensure that all values are positive and magnify larger differences. This sum emphasizes the variability of X and is important for understanding the strength of the relationship between the variables.

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Most popular questions from this chapter

A data set relates proportional limit and tensile strength in certain alloys of gold collected for presentation at a Dentistry Convention. (Proportional limit is the load in psi at which the elongation of a sample no longer obeys Hooke's Law.) Let \(\left(\mathrm{X}_{i}, \mathrm{Y}_{\mathrm{i}}\right)\) be an observed ordered pair consisting of \(\mathrm{X}_{\mathrm{i}}\), an observed tensile strength, and \(\mathrm{Y}_{\mathrm{i}}\) an observed proportional limit, each measured in pounds per square inch (psi). After 25 observations of this sort the following summary statistics are: $$ \begin{array}{ll} \dot{\sum} \mathrm{X}_{\mathrm{i}}=2,991,300, & \underline{\mathrm{X}}=119,652 \\\ \sum \mathrm{X}_{\mathrm{i}}^{2}=372,419,750,000 . & \\ \sum \mathrm{Y}_{\mathrm{i}}=2,131,200, & \underline{\mathrm{Y}}=85,248 \\ \sum \mathrm{Y}_{\mathrm{i}}^{2}=196,195,960,000 & \\ \sum \mathrm{X}_{\mathrm{i}} \mathrm{Y}_{\mathrm{i}}=269,069,420,000 & \end{array} $$ Compute the regression coefficients relating proportional limit and tensile strength.

An investigator has data on 1,000 individuals who have been in psychotherapy for five years. Variable \(\mathrm{X}\) tells the score the individual received on a personality test. Variable \(\mathrm{Y}\) tells his improvement in psychotherapy over the five years. Here are the data: \(\sum \mathrm{XY}=30,000 . \sum \mathrm{X}=3,000 . \sum \mathrm{X}^{2}=14,000\) \(\sum \mathrm{Y}=5,000\) (a) Compute the value of \(\mathrm{b}^{\wedge}\) in the regression equation. (b) What is the regression equation? (c) Predict the improvement score of a subject with \(\mathrm{X}=4\). (d) If \(\mathrm{S}_{\mathrm{y}}=10\), compute r.

There is a complex system of relationships in the business world. As an example, the number of new movies which appear in the course of a week has an appreciable effect on the weekly change in the Dow-Jones Industrial Average." This is the opinion of a certain armchair economist. This fellow hires you as a consultant and expects you to test his theory. In the first 5 weeks you observe the following: \begin{tabular}{|c|c|c|c|c|c|} \hline \(\mathrm{X}\), number of new movies & 1 & 2 & 4 & 5 & 5 \\ \hline Y, change in Dow-Jones Industrial Average & \(-2\) & 4 & \(-5\) & 7 & \(-8\) \\\ \hline \end{tabular} What is the correlation between \(\mathrm{X}\) and \(\mathrm{Y}\) ? What implications does this have for the theory?

The heights of fathers, \(X\), and the heights of their oldest sons when grown, \(\mathrm{Y}\), are given as measurements to the nearest inch. \begin{tabular}{|l|l|l|l|l|l|l|} \hline \(\mathrm{X}\) & 68 & 64 & 70 & 72 & 69 & 74 \\ \hline \(\mathrm{Y}\) & 67 & 68 & 69 & 73 & 66 & 70 \\ \hline \end{tabular} (a) Construct a scattergram. (b) Find the equation of the least squares regression line. (c) Compute the standard error of estimate. (e) Compute the coefficient of correlation \(\mathrm{r}\).

Suppose a new method is designed for determining the amount of magnesium in sea water. If the method is a good one, there will be a strong relation between the true amount of magnesium in the water and the amount indicated by this new method. 10 samples of "sea water" are prepared, each sample containing a known amount of magnesium. The samples are then tested by the new method. The data from this experiment is present in the form of the summary statistics. \(\mathrm{X}\) represents the true amount of \(\mathrm{Mg}\) present and \(\mathrm{Y}\) the corresponding amount determined by the new method. The data is $$ \begin{array}{lll} \sum X_{i}=311 & \sum X_{i}^{2}=10,100 \quad \sum X Y=10,074 \\ \sum Y_{i}=310.1 & \text { and } \sum Y_{i}^{2}=10,055.09 \end{array} $$ Find the regression equation, the standard error \(\sigma\) and \(\mathrm{r}^{2}\), the coefficient of variation. Develop \(95 \%\) confidence intervals for \(\alpha\) and \(\beta^{\wedge}\) and test the hypotehsis that the true equation has parameters \(\alpha=0\) and \(\beta=1\).
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