91Ó°ÊÓ

The stock market Some people think that the behavior of the stock market in January predicts its behavior for the rest of the year. Take the explanatory variable \(x\) to be the percent change in a stock market index in January and the response variable \(y\) to be the change in the index for the entire year. We expect a positive correlation between \(x\) and \(y\) because the change during January contributes to the full year's change. Calculation from data for an 18 -year period gives $$ \begin{array}{c}{\overline{x}=1.75 \% \quad s_{x}=5.36 \% \quad \overline{y}=9.07 \%} \\ {s_{y}=15.35 \% \quad r=0.596}\end{array} $$ (a) Find the equation of the least-squares line for predicting full-year change from January change. Show your work. (b) The mean change in January is \(\overline{x}=1.75 \%\) . Use your regression line to predict the change in the index in a year in which the index rises 1.75\(\%\) in January. Why could you have given this result (up to roundoff error) without doing the calculation?

Step by step solution

01

Understand the Context

We are given that the explanatory variable \( x \) represents the percent change in a stock market index in January, while the response variable \( y \) is the percent change for the entire year. The correlation coefficient \( r = 0.596 \), the mean values \( \overline{x} = 1.75\% \) and \( \overline{y} = 9.07\% \), and the standard deviations \( s_x = 5.36\% \) and \( s_y = 15.35\% \) have been provided. We are to find the least-squares regression line equation and make predictions based on it.

02

Calculate the Slope (b) of the Regression Line

The formula for the slope \( b \) of the least-squares regression line is given by:\[ b = r \left( \frac{s_y}{s_x} \right) \]Substituting the given values:\[ b = 0.596 \left( \frac{15.35}{5.36} \right) \]\[ b \approx 1.708 \]

03

Calculate the Y-Intercept (a) of the Regression Line

The formula for the y-intercept \( a \) is:\[ a = \overline{y} - b \overline{x} \]Substitute the calculated \( b = 1.708 \) and the given means into the equation:\[ a = 9.07 - 1.708 \times 1.75 \]\[ a \approx 6.05 \]

04

Write the Equation of the Least-squares Line

The equation of the least-squares line is:\[ y = a + bx \]Substitute \( a = 6.05 \) and \( b = 1.708 \):\[ y = 6.05 + 1.708x \]

05

Predict the Yearly Change for January Rise of 1.75%

Substitute \( x = 1.75 \) into the regression equation to predict \( y \):\[ y = 6.05 + 1.708 \times 1.75 \]\[ y \approx 9.05 \% \]Note that this result is close to \( \overline{y} = 9.07 \% \), which makes sense because predicting at the mean value of \( x \) usually yields \( \overline{y} \).

06

Conclusion on the Result without Calculation

Since the January change \( x = \overline{x} = 1.75\% \), substituting \( \overline{x} \) into the regression line will yield \( \overline{y} = 9.07\% \), due to the linear relationship centered around their means. Thus, up to roundoff errors, the prediction is inherently equal to the mean yearly change.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Correlation Coefficient

The correlation coefficient, often denoted by \( r \), is a numerical measure that describes the strength and direction of a linear relationship between two variables. In our stock market example, the value of \( r = 0.596 \) indicates a moderate positive correlation between the January change () in a stock market index and the entire year's change. This means as January's change increases, there is a tendency for the year's change to increase as well.- A correlation coefficient can range from -1 to 1. - An \( r \) of 1 indicates a perfect positive linear relationship. - An \( r \) of -1 indicates a perfect negative linear relationship. - An \( r \) of 0 suggests no linear relationship.With \( r = 0.596 \), we see a tendency but not certainty that a positive increase in January usually leads to a positive yearly increase. This moderate correlation is crucial to understanding how much January's performance could be used as a predictor for the entire year.

Explanatory Variable

Explanatory variables are used to explain variations in a response variable, essentially acting as predictors. In our exercise about stock market predictions, the explanatory variable \( x \) represents the percent change in a stock market index in January.The main role of the explanatory variable is to use its value to predict or explain the behavior of another variable. It is important to remember:- The explanatory variable is independent, meaning its value is not dependent on other variables in the study.- It provides the basis for creating a prediction model, like the least-squares regression line.Understanding how to identify the explanatory variable helps in analyzing data sets effectively, aiding in predictions and conclusions about the relationships between variables.

Response Variable

The response variable is the outcome or the variable that one aims to predict using an explanatory variable. In the stock market exercise, the response variable \( y \) is the percent change for the entire year.A few key points regarding response variables:- Dependent in nature, its value changes in response to the explanatory variable.- In regression analysis, it is the output we are interested in assessing based on predictors.When performing any regression analysis, pinpointing the response variable is crucial as it enables us to define what output we are aiming to predict or analyze.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. In our stock market exercise, standard deviations for January (\( s_x = 5.36\% \)) and the entire year (\( s_y = 15.35\% \)) are provided.- A small standard deviation suggests that the data points are close to the mean.- A larger standard deviation indicates a wider spread of data around the mean.By having the standard deviation, we can better understand the volatility of January's changes or the overall year's changes. This helps in assessing the reliability of predictions made using models like the least-squares regression line. It gives insight into the confidence we might have in the mean as a representation of the data.