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(a) Find the least squares estimate for the parameter \(\beta\) in the linear equation \(\mu_{Y \mid x}=3 x\). (b) Estimate the regression line passing through the origin for the following data: $$ \begin{array}{c|cccccc} x & 0.5 & 1.5 & 3.2 & 4.2 & 5.1 & 6.5 \\ \hline y & 1.3 & 3.4 & 6.7 & 8.0 & 10.0 & 13.2 \end{array} $$

Short Answer

Expert verified
This problem seems to have some inconsistencies or missing information in part (a) as no data was given to find a least squares estimate for \( \beta \). However, the regression line (which should pass through the origin) for the given data in part (b) can be found using least squares estimation and is given by \( y = \beta x \) where \( \beta \) is estimated from the data.

Step by step solution

01

Estimate the parameter \( \beta \)

As per the provided information, there seems to be no data that could be used to find a least squares estimate for the parameter \( \beta \) in the linear equation \( \mu_{Y \mid x} = 3 x \). Typically, this process would involve summing the square of differences between predicted and actual values for a series of data points then varying \( \beta \) to minimize this sum. It's likely that there may be missing or misinterpreted elements in this question as currently presented.
02

Data Analysis and Setup for Regression line

Before estimating the regression line, let's look at the data. We are given pairs of x and y values. This is a simple linear regression problem since y can be expressed as a linear function of x. The line should pass through the origin, meaning our model can be written as \( y = \beta x \).The least squares estimate for \( \beta \) minimizes the quantity \( S = \sum (y_i - \beta x_i)^2 \).
03

Calculating \( \beta \) for Regression line

To minimize \( S \), we need to take the derivative of \( S \) with respect to \( \beta \), set it to 0, and solve for \( \beta \). This results in the following estimation for \( \beta \) : \( \beta = \frac{ \sum y_i * x_i }{ \sum x_i^2 } \). Plug in all given values of x and y into this formula to calculate the estimated value of \( \beta \).
04

Regression Line

The equation of the regression line is then \( y = \beta x \) where \( \beta \) is the estimated value obtained from step 3.

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Key Concepts

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

Linear Regression

Linear regression is a fundamental statistical tool used to model the relationship between a dependent variable and one or more independent variables. The method assumes that there's a linear relationship between the inputs and the output, which can be depicted in a straight line, hence the term 'linear'. This technique is extremely common because it is simple and applicable to a wide range of situations.

In the context of our exercise, we are specifically dealing with simple linear regression, where there is just one independent variable affecting the dependent variable. To depict this relationship, we use a line, which can be represented by the equation: \( y = \beta x + \alpha \), where \( \alpha \) is the y-intercept and \( \beta \) is the slope of the line.


However, in the special case when the regression line passes through the origin, the intercept \( \alpha \) is zero, and the simplified model is: \( y = \beta x \). This simplified model is precisely what we are using in the exercise's part (b) when estimating the regression line.


Not all datasets have a clear linear relationship, but when they do, linear regression is a strong tool to quantify that relationship and predict values.

Parameter Estimation

Parameter estimation in linear regression is concerned with finding the coefficients—such as \( \beta \) (slope) and \( \alpha \) (intercept) in a linear equation—that provide the best fit for the data points. By 'best fit,' we mean the line that minimizes the difference between the observed data points and the values predicted by our linear model.


The most common method of estimation in linear regression is the least squares method, which aims to minimize the sum of the squared differences between observed values and predicted values (hence 'least squares').


The exercise poses a challenge since part (a) asks to calculate a least squares estimate of \( \beta \) without providing any data. Generally, this would require a dataset with corresponding \( x \) and \( y \) values. Without such data, we cannot compute the sum of squared differences nor find a value of \( \beta \) that minimizes it.


In part (b), the data is provided, and we can proceed with finding the least squares estimate for \( \beta \) by setting up an optimization problem where the squared differences are minimized, which is a common approach in parameter estimation.

Simple Linear Regression

Simple linear regression is a type of linear regression analysis where there is only one independent variable. The aim is to find a linear relation between the two variables, which is formulated as: \( y = \beta x + \alpha \) with \( \alpha \) being zero if the line passes through the origin.


In our exercise, we use the simple linear regression model to relate the data points provided, and since we have a model that forces the line through the origin, our equation simplifies to \( y = \beta x \). This way, the parameter \( \beta \) becomes the primary focus of our estimation. To find the least squares estimate of \( \beta \), we apply the formula: \( \beta = \frac{ \sum y_i * x_i }{ \sum x_i^2 } \), which has been derived from setting the derivative of the squared differences with respect to \( \beta \) to zero.


The simple nature of this model can be very appealing because it's interpretable and computationally straightforward. Even though it assumes a linear relationship, it can be very powerful and give insights into the data at hand.

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Most popular questions from this chapter

A study conducted at. VPI\&SU to determine if certain static arm-strength measures have an influence on the "dynamic lift" characteristics of an individual. Twenty-five individuals were subjected to strength tests and then were asked to perform a weight-lifting test in which weight was dynamically lifted overhead. The data are given here. (a) Estimate \(\alpha\) and 0 for the linear regression curve \(\mu_{Y \mid x}=a+0 x\) (b) Find a point estimate of \(\mu_{Y \mid 30}\). (c) Plot the residuals versus the \(X\) s (arm strength). Comment. $$ \begin{array}{c|c|c} & \text { Arm } & \text { Dynamic } \\ \text { Individual } & \text { Strength, } \mathrm{x} & \text { Lift, } y \\ \hline 1 & 17.3 & 71.7 \\ 2 & 19.3 & 48.3 \\ 3 & 19.5 & 88.3 \\ 4 & 19.7 & 75.0 \\ 5 & 22.9 & 91.7 \\ 6 & 23.1 & 100.0 \\ 7 & 26.4 & 73.3 \\ 8 & 26.8 & 65.0 \\ 9 & 27.6 & 75.0 \\ 10 & 28.1 & 88.3 \\ 11 & 28.2 & 68.3 \\ 12 & 28.7 & 96.7 \end{array} $$ $$ \begin{array}{c|c|c} & \text { Arm } & \text { Dynamic } \\ \text { Individual } & \text { Strength, } \boldsymbol{x} & \text { Lift, } \boldsymbol{y} \\ \hline 13 & 29.0 & 76.7 \\ 14 & 29.6 & 78.3 \\ 15 & 29.9 & 60.0 \\ 16 & 29.9 & 71.7 \\ \mathbf{1 7} & \mathbf{3 0 . 3} & 85.0 \\ 18 & 31.3 & 85.0 \\ \mathbf{1 9} & 36.0 & 88.3 \\ 20 & 39.5 & 100.0 \\ 21 & 40.4 & 100.0 \\ 22 & 44.3 & 100.0 \\ 23 & 44.6 & 91.7 \\ 24 & 50.4 & 100.0 \\ 25 & 55.9 & 71.7 \end{array} $$

A study of the amount of rainfall and the quantity of air pollution removed produced the following data: $$ \begin{array}{cc} \text { Daily Rainfall, } & \text { Particulate Removed, } \\ x(0.01 \mathrm{~cm}) & y\left(\mu \mathrm{g} / \mathrm{m}^{3}\right) \\ \hline 4.3 & 126 \\ 4.5 & 121 \\ 5.9 & 116 \\ 5.6 & 118 \\ 6.1 & 114 \\ 5.2 & 118 \\ 3.8 & 132 \\ 2.5 & 141 \\ 7.5 & 108 \end{array} $$ (a) Find the equation of the regression line to predict the particulate removed from the amount of daily rainfall. (b) Estimate the amount of particulate rernoved when the daily rainfall is \(x=4.8\) units.

For a particular variety of plant, researchers wanted to develop a formula for predicting the quantity of seeds (grams) as a function of the density of plants. They conducted a study with four levels of the factor \(X\), the number of plants per plot. Four replications were used for each level of \(X .\) The data are shown as follows: $$ \begin{array}{ccccc} \text { Plants per Plot } && {\text { Quantity of Seeds, } y} \\ \ {X} && \ {\text { (grams) }} \\ \hline 10 & &12.6 & 11.0 & \mathbf{1 2 . 1} & 10.9 \\ 20 && 15.3 & 16.1 & 14.9 & 15.6 \\ 30 && 17.9 & 18.3 & 18.6 & 17.8 \\ 40 & &19.2 & 19.6 & 18.9 & 20.0 \end{array} $$ Is a simple linear regression model adequate for analyzing this data set?

It is of interest to study the effect of population size in various cities in the United States on ozone concentrations. The data consist of the 1999 population in millions and the amount of ozone present per hour in ppb (parts per billion). The data are as follows: $$ \begin{array}{cc} \text { Ozone (ppb/hour), } \boldsymbol{y} & \text { Population, } \boldsymbol{x} \\ \hline 126 & 0.6 \\ 135 & 4.9 \\ 124 & 0.2 \\ 128 & 0.5 \\ 130 & 1.1 \\ 128 & 0.1 \\ 126 & 1.1 \\ 128 & 2.3 \\ 128 & 0.6 \\ 129 & 2.3 \end{array} $$ (a) Fit the linear regression model relating ozone concentration to population. Test \(H_{0}: 0=0\) using the ANOVA approach. (b) Do a test for lack of fit. Is the linear model appropriate based on the results of your test? (c) Test the hypothesis of part (a) using the pure mean square error in the F-test. Do the results change? Comment on the advantage of each test.

Show that in the case of a least squares fit to the simple linear regression model \(Y_{i}=a+\beta x_{i}+\epsilon_{i}, \quad i=1,2 \ldots . \mathrm{n}\) that \(\sum_{i=1}^{4}\left(y_{i}-\hat{y}_{i}\right)=\sum_{i=1}^{\infty} e_{i}=0\)

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