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Suppose the scientist postulates a model $$ Y_{i}=a+\beta x_{i}+\epsilon_{i}, \quad \dot{i}^{-}-\mid, 2, \ldots, n $$ and \(\alpha\) is a known value, not necessarily zero. (a) What is the appropriate least squares estimator of \(\beta ?\) Justify your answer. (b) What is the variance of the slope estimator?

Short Answer

Expert verified
The least squares estimator of \( \beta \) is \( \beta = \frac{\sum_{i=1}^{n} x_i (Y_i - \alpha)}{\sum_{i=1}^{n} x_i^2} \) and the variance of the slope estimator is \( \sigma^2 / \sum_{i=1}^{n} (x_i - \bar{x})^2 \).

Step by step solution

01

- Formulating the Estimator Function

The sum of squared residuals which is to be minimized in ordinary least square is \( S = \sum_{i=1}^{n} (Y_i - a - \beta x_i )^2 \). However, in the given exercise, 'a' is some known value, which means this can be simplified to \( S = \sum_{i=1}^{n} (Y_i - \alpha - \beta x_i )^2 \). Hence, the criteria for minimization of \( S \) is \( \frac{\partial S}{\partial \beta} = 0 \).
02

- Derivation of the Least Square Estimator

Differentiating w.r.t \( \beta \), we get \( \frac {\partial S}{\partial \beta} = -2 \sum_{i=1}^{n} x_i (Y_i - \alpha - \beta x_i ) \). Setting the derivative equal to zero and simplifying should give the least squares estimator for \( \beta \).
03

- Solving for Beta

Setting the equation to zero, we get \( \sum_{i=1}^{n} x_i Y_i - \alpha \sum_{i=1}^{n} x_i - \beta \sum_{i=1}^{n} x_i^2 = 0 \). This gives us, \( \beta = \frac{\sum_{i=1}^{n} x_i (Y_i - \alpha)}{\sum_{i=1}^{n} x_i^2} \). So, this is the least squares estimator for \( \beta \).
04

- Finding the Variance of the Slope Estimator

Knowing that the variance of the slope estimator \( \beta \) in a simple linear regression model with non-zero intercept is \( \sigma^2 / \sum_{i=1}^{n} (x_i - \bar{x})^2 \), where \( \sigma^2 \) is the variance of the error term and \( \bar{x} \) is the mean of all \( x_i \), but in this case, 'a' (intercept) is a known value (not necessarily zero). As there is no change in the estimating equation of \( \beta \), the variance of the \( \beta \) estimator will be the same as in the case with zero intercept. Hence it remains \( \sigma^2 / \sum_{i=1}^{n} (x_i - \bar{x})^2 \).

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Key Concepts

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

Least Squares Estimator
In the context of linear regression, the least squares estimator is a method used to find the best-fitting line through a set of data points. It works by minimizing the sum of the squares of the residuals, where a residual is the difference between the observed value and the value predicted by the line.

When applied to an equation of the form \( Y_i = \alpha + \beta x_i + \epsilon_i \), where \( \alpha \) is a known constant, the goal is to determine the value of \( \beta \) that minimizes these squared differences. This involves setting the partial derivative of the sum of squared residuals with respect to \( \beta \) to zero as shown by the equation, \( \frac{\partial S}{\partial \beta} = -2 \sum_{i=1}^{n} x_i (Y_i - \alpha - \beta x_i ) \).

Solving for \( \beta \) involves gathering the terms to find: \( \beta = \frac{\sum_{i=1}^{n} x_i (Y_i - \alpha)}{\sum_{i=1}^{n} x_i^2} \). This equation gives the value of the least squares estimator for \( \beta \), ensuring that the calculated line is as close as possible to the actual data points.
Variance of Slope Estimator
Assessing the precision of the slope estimator is essential for understanding the reliability of the regression line. The variance of the slope estimator \( \beta \) provides a measure of this precision. In essence, it tells us how much variability we might expect from sample to sample in the estimated \( \beta \) when using ordinary least squares regression.

In a linear regression model with a known intercept, the variance of the slope estimator \( \beta \) remains the same as in the intercept is zero. The formula for this variance is given by \( \frac{\sigma^2}{\sum_{i=1}^{n} (x_i - \bar{x})^2} \), where \( \sigma^2 \) is the variance of the error term \( \epsilon \), and \( \bar{x} \) is the mean of the \( x_i \) values.

A smaller variance indicates a more reliable slope estimate, as it suggests that the estimator would yield similar values in different samples.
Linear Regression Model
Linear regression is a foundational tool in statistics used to model the relationship between a dependent variable and one or more independent variables. In this exercise, the model is given by \( Y_i = \alpha + \beta x_i + \epsilon_i \), where \( Y_i \) is the dependent variable, \( \alpha \) is a known constant added to the model, \( \beta \) is the slope, \( x_i \) is the independent variable, and \( \epsilon_i \) represents the error term.

The idea is to use available data to understand or predict the outcome \( Y_i \) based on the inputs \( x_i \). The regression aims to find the line that minimizes the distance (errors) between the predicted \( Y_i \) from the actual \( Y_i \). Ordinary least squares is one common method for finding this best-fitting line by minimizing the squared differences between observed and predicted values. This model is exceptionally useful because it simplifies complex relationships into straightforward calculations and visualizations, making it easier to interpret and draw conclusions from data.

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Most popular questions from this chapter

The thrust of an engine \((y)\) is a function of exhaust temperature \((x)\) in \({ }^{\circ} \mathrm{F}\) when other important variables arc held constant. Consider the following data. $$ \begin{array}{cc|cc} y & X & y & X \\ \hline 4300 & 1760 & 4010 & 1665 \\ 4650 & 1652 & 3810 & 1550 \\ 3200 & 1485 & 4500 & 1700 \\ 3150 & 1390 & 3008 & 1270 \\ 4950 & 1820 & & \end{array} $$ (a) Plot the data. (b) Fit a simple linear regression to the data and plot the line through the data.

The following data were obtained in a study of the relationship between the weight and chest size of infants at birth: $$ \begin{array}{cc} \text { Weight (kg) } & \text { Chest Size (cm) } \\ \hline 2.75 & 29.5 \\ 2.15 & 26.3 \\ 4.41 & 32.2 \\ 5.52 & 36.5 \\ 3.21 & 27.2 \\ 4.32 & 27.7 \\ 2.31 & 28.3 \\ 4.30 & 30.3 \\ 3.71 & 28.7 \end{array} $$ (a) Calculate \(r\). (b) Test the null hypothesis that \(p=0\) against the alternative that \(p>0\) at the 0.0 i level of significance. (c) What percentage of the variation in the infant chest sizes is explained by difference in weight?

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The following data represent the chemistry grades for a random sample of 12 freshmen at a certain college along with their scores on an intelligence test administered while they were still seniors in high school: $$ \begin{array}{ccc} & \text { Test } & \text { Chemistry } \\ \text { Student } & \text { Score, } \boldsymbol{x} & \text { Grade, } \boldsymbol{y} \\ \hline \mathbf{1} & 65 & 85 \\ \mathbf{2} & 50 & 74 \\ \mathbf{3} & 55 & 76 \\ 4 & 65 & 90 \\ 5 & 55 & 85 \\ 6 & 70 & 87 \\ 7 & 65 & 94 \\ 8 & 70 & 98 \\ 9 & 55 & 81 \\ 10 & 70 & 91 \\ 11 & 50 & 76 \\ 12 & 55 & 74 \end{array} $$ (a) Compute and interpret the sample correlation coefficient. (b) State necessary assumptions on random variables. (c) Test the hypothesis that \(p=0.5\) against the alternative that \(p>0.5 .\) Use a P-value in the conclusion.

Transistor gain in an integrated circuit device between emitter and collector (hFE) is related to two variables [Myers and Montgomery (2002)] that can be controlled at the deposition process, emitter drivein time \(\left(x_{1},\right.\) in minutes \(),\) and emitter dose \(\left(x_{2},\right.\) in ions \(x \quad 10^{14}\) ). Fourteen samples were observed following deposition, and the resulting data shown in the table below. We will consider linear regression models using gain as the response and emitter drive-in time or emitter dose as the regressor variables. $$ \begin{array}{rccc} & x_{1},(\text { drive-in } & x_{2},(\text { dose }, & y, \text { (gain } \\ \text { Obs. } & \text { time, min) } & \text { ions } \times 10^{14} & \text { or hFE }) \\ \hline 1 & 195 & 4.00 & 1004 \\ 2 & 255 & 4.00 & 1636 \\ 3 & 195 & 4.60 & 852 \\ 4 & 255 & 4.60 & 1506 \\ 5 & 255 & 4.20 & 1272 \\ 6 & 255 & 4.10 & 1270 \\ 7 & 255 & 4.60 & 1269 \\ 8 & 195 & 4.30 & 903 \\ 9 & 255 & 4.30 & 1555 \\ 10 & 255 & 4.00 & 1260 \\ 11 & 255 & 4.70 & 114 C \\ 12 & 255 & 4.30 & 1276 \\ 13 & 255 & 4.72 & 1225 \\ 14 & 340 & 4.30 & 1321 \end{array} $$ (a) Determine if emitter drive-in time influences gain in a linear relationship. That is, test \(H_{0}: 3 \mid=0\), where \(\beta_{1}\), is the slope of the regressor variable. (b) Do a lack-of-fit test to determine if the linear relationship is adequate. Draw conclusions. (c) Determine if emitter dose influences gain in a linear relationship. Which regressor variable is the better predictor of gain?

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