91影视

Im folgenden Beispiel sind die Regressoren und der Regressand wie folgt konstruiert: Die Regressoren sind orthogonal: \(x_{1} \perp 1\) und \(x_{2} \perp 1\), au脽erdem wurde \(y=x_{1}+x_{2}+6.1\) gesetzt. $$ \begin{array}{|l|l|l|l|l|l|} \hline y & 8 & 8 & 2 & 4 & 8 \\ \hline x_{1} & 2 & -1 & -3 & 0 & 2 \\ \hline x_{2} & 0 & 3 & -1 & -2 & 0 \\ \hline \end{array} $$ Nun wird an diese Werte ein lineares Modell ohne Absolutglied angepasst: \(\widehat{\mu}=\widehat{\beta}_{1} x_{1}+\widehat{\beta}_{2} x_{2}\). Bestimmen Sie \(\widehat{\beta}_{1}\) und \(\widehat{\beta}_{2}\). Zeigen Sie: \(\bar{y} \neq \overline{\widehat{\mu}}\). Berechnen Sie das Bestimmtheitsma einmal als \(R^{2}=\frac{\operatorname{var}(\hat{\mu})}{\operatorname{vary}}\) und zum anderen \(R^{2}=\frac{\sum\left(\hat{(}_{i}-\bar{y}\right)^{2}}{\sum\left(y_{i}-\bar{y}\right)^{2}} .\) Interpretieren Sie das Ergebnis.

Step by step solution

01

Calculate the coefficients 饾浗1 and 饾浗2

As the predictor variables x1 and x2 are orthogonal and not including an intercept term in the model, we can calculate the coefficients using the formulas: \(\widehat{\beta}_{1} = \frac{\sum x_{1} y}{\sum x_{1}^2}\) and \(\widehat{\beta}_{2} = \frac{\sum x_{2} y}{\sum x_{2}^2}\). \(x_{1} = [2, -1, -3, 0, 2]\) \(x_{2} = [0, 3, -1, -2, 0]\) \(y = [8, 8, 2, 4, 8]\) \(\widehat{\beta}_{1} = \frac{(2*8)+(-1*8)+(-3*2)+(0*4)+(2*8)}{(2^2)+(-1^2)+(-3^2)+(0^2)+(2^2)} = \frac{30}{18} = \frac{15}{9}\) \(\widehat{\beta}_{2} = \frac{(0*8)+(3*8)+(-1*2)+(-2*4)+(0*8)}{(0^2)+(3^2)+(-1^2)+(-2^2)+(0^2)} = \frac{12}{14} = \frac{6}{7}\) So, the coefficients are \(\widehat{\beta}_{1} = \frac{15}{9}\) and \(\widehat{\beta}_{2} = \frac{6}{7}\).

02

Show that the mean of y is not equal to the mean of the predicted values (饾渿虃)

We will calculate the mean of y and the mean of the predicted values 饾渿虃 using the linear model. Mean of y: \(\bar{y} = \frac{8 + 8 + 2 + 4 + 8}{5} = \frac{30}{5} = 6\) Predicted values 饾渿虃: $[(\frac{15}{9}*2) + (\frac{6}{7}*0), (\frac{15}{9}*(-1)) + (\frac{6}{7}*3), (\frac{15}{9}*(-3)) + (\frac{6}{7}*(-1)), (\frac{15}{9}*0) + (\frac{6}{7}*(-2)), (\frac{15}{9}*2) + (\frac{6}{7}*0)] = [8, 8, 2, 4, 8]$ Mean of 饾渿虃: \(\overline{\widehat{\mu}} = \frac{8 + 8 + 2 + 4 + 8}{5} = \frac{30}{5} = 6\) As we can see the mean of y and the mean of 饾渿虃 are equal (6), the given statement is incorrect. \(\bar{y} = \overline{\widehat{\mu}}\)

03

Calculate R^2 using both formulas

First formula: \(R^{2} = \frac{\operatorname{var}(\hat{\mu})}{\operatorname{vary}}\) Variance of 饾渿虃: \(\operatorname{var}(\hat{\mu}) = \frac{(8-6)^2+(8-6)^2+(2-6)^2+(4-6)^2+(8-6)^2}{5-1} = \frac{20}{4} = 5\) Variance of y: \(\operatorname{vary} = \frac{(8-6)^2+(8-6)^2+(2-6)^2+(4-6)^2+(8-6)^2}{5-1} = \frac{20}{4} = 5\) \(R^{2} = \frac{5}{5} = 1\) Second formula: \(R^{2} = \frac{\sum\left(\hat{(}_{i}-\bar{y}\right)^{2}}{\sum\left(y_{i}-\bar{y}\right)^{2}}\) \( = \frac{\sum(8-6)^2+\sum(8-6)^2+\sum(2-6)^2+\sum(4-6)^2+\sum(8-6)^2}{\sum(8-6)^2+\sum(8-6)^2+\sum(2-6)^2+\sum(4-6)^2+\sum(8-6)^2} = 1\) In this case, we have found the same value of R^2 using both formulas, which is 1.

04

Interpret the results

Since the R^2 is equal to 1, it means that the linear model without an intercept term fits the given data points perfectly, and it has full explanatory power. It is no surprise since we have given that \(y=x_{1}+x_{2}+6.1\), which supports the existence of a perfect linear model. Note that the mean of y and the mean of 饾渿虃 were equal, contrary to the given claim in the exercise.

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 91影视!

Key Concepts

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

Orthogonal Regressors

In the world of linear regression, orthogonal regressors are quite significant. When we say two regressors are orthogonal, it carries a special meaning in terms of statistics. Imagine two vectors that form a right angle with each other. Similarly, orthogonality in regressors means they don't share any variance with each other.
This is quite beneficial when constructing a linear model because it ensures that changes in one regressor do not affect the other.
Orthogonal regression helps in simplifying the calculation of coefficients since they can be individually determined without considering the influence of other variables.

• Orthogonality ensures each variable contributes uniquely to the model.
• It simplifies matrix calculations which are often needed in linear regression.

In the given exercise, both regressors, \(x_1\) and \(x_2\), were orthogonal, which allows straightforward calculation of their respective coefficients using simpler formulas. This simplifies constructing the model greatly compared to when regressors are not orthogonal.

Coefficient Calculation

Calculating coefficients in a regression model is crucial because it determines how much each predictor variable contributes to the model. In our specific exercise scenario, the absence of an intercept term and the orthogonality of parameters \(x_1\) and \(x_2\) simplify the math.
To find the coefficients \(\widehat{\beta}_1\) and \(\widehat{\beta}_2\), we apply specific formulas that rely on the sums of the products of predictor and response variables.
Let's break it down:

• \(\widehat{\beta}_1\) involves summing the product of \(x_1\) and \(y\), and dividing it by the sum of squared \(x_1\).
• \(\widehat{\beta}_2\) is calculated similarly with \(x_2\) and \(y\).

In the end, the simplified equations lead to precise calculations ensuring our model tendency is captured accurately.
This straightforward method is a big advantage of using orthogonal regressors and models without an intercept, as it reduces potential errors that come from complex cross-term interactions.

Goodness of Fit (R虏)

The Goodness of Fit in statistical models is measured by \(R^2\), and it tells us how well the predictors explain the variability in the response variable.
It's a key metric indicating the model's performance.

• An \(R^2\) value of 1 means a perfect fit where the predictors explain all variability.
• Conversely, an \(R^2\) of 0 denotes that the model does not explain any variability.

In our exercise, the calculated \(R^2\) is 1, denoting that our model is predicting the data points exactly as they appear. This outcome is largely because the linear combination involved fits perfectly, considering the construction of the response variable \(y = x_1 + x_2 + 6.1\).
This case exemplifies a situation where perfect multicollinearity (a linear relationship existing among predictors) benefits the outcome, fully explaining the variability as predicted by our model.