Bivariate Linear Regression

Research Scenario

Mary conducts a nonexperimental study to evaluate what she refers to as the strength-injury hypothesis. It states that overall body strength in elderly women determines the number and severity of accidents that cause bodily injury. If the results of her prediction study support her strength-injury hypothesis, she plans to conduct an experimental study to assess whether weight training reduces injuries in elderly women.

In the prediction study, Mary collects data from 100 women who range in age from 60 to 75 years old at the time the study begins. The women initially are evaluated on a number of measures that assess lower- and upper-body strength. These measures are summarized using an overall index of body strength. Over the next five years, the women record each time they have an accident that results in a bodily injury and describe fully the extent of the injury. On the basis of these data, Mary calculates an overall injury index for each woman.

Question

Mary is interested in conducting a regression analysis with the overall index of body strength as the predictor (independent) variable and the overall injury index as the criterion (dependent) variable. Specifically, Mary would like to:

  1. develop a linear equation that predicts the extent of physical injury from body strength for elderly women
  2. evaluate how accurately her equation predicts the extent of physical injuries

Instrument & Scoring

Mary’s data contains 100 observation scores on six variables, five individual strength measures (quads, gluts, abdoms, grip, and arms) that are to be combined to yield an overall index of body strength and the criterion variable, the overall injury index.


Note

Mary’s data set includes the dependent variable for the regression analysis (injury) but does not include the independent variable, an overall index of body strength. Fortunately, the injury index can be calculated by creating z scores (standard scores) for the five strength measures and adding them together to create a total strength score which will be called ‘ztotstr’. An example for how the ztotstr variable is computed is shown below using the scores provided for participant one from Mary’s data. The computed ztotstr for participant one is -5.81.

Developing Predictor Scores
Data Type Quads Gluts Abdoms Arms Grip Ztotstr
Unstandarized 35.00 27.00 30.00 11.00 0.00
Standarized -1.25 -0.71 0.15 -2.27 -1.73 -5.81
Note:
The data used above is for participant one in the set of data.


Research Hypothesis

The overall body strength in elderly women determines the number and severity of accidents that cause bodily injury.

  • \(H_{O}\): Overall body strength has no effect on bodily injury
  • \(H_{A}\): Overall body strength has an effect on bodily injury

Assumptions

  1. The X and Y variables are bivariately normally distributed in the population.

  2. The cases represent a random sample from the population, and the scores on each variable are independent of other scores on the same variable.

Assumption Check

Of the two major assumptions for Bivariate Linear Regression, only the first assumption can be evaluated using data alone. The second assumption is something that is best addressed through implementing proper research/sampling practices—simply put, if this assumption is found to be violated in the data analysis phase there is little the researcher can do to address the issue.

If the variables are bivariately normally distributed, each variable is normally distributed ignoring the other variable and each variable is normally distributed at all levels of the other variable. If the bivariate normality assumptions met, the only type of relationship that can exist between two variables is a linear relationship. However, if the assumption is violated, a nonlinear relationship may exist. It is important to determine if a nonlinear relationship exists between two the X and Y variables before describing the results with the Bivariate Linear Regression, otherwise, the significance test results for regression analysis will yield inaccurate p values.


Note

If a nonlinear relationship did exist in Mary’s set of data, she would need to transform her data so that the nonlinear relationship is no longer present. Possible transformations that could be applied to induce a linear relationship include:

  • Exponential
  • Quadratic
  • Reciprocal
  • Logarithmic
  • Power

However, the linear relationship could come at the cost of interpretability. Mary’s transformed data would now be in the the units of his selected process, e.g. a logarithmic transformation would result in a self-concept scores in log units. For an illustrative example, consider reviewing the Pearson Product-Moment Correlation Coefficient walkthrough

To evaluate whether her X and Y variables are bivariately normally distributed, Mary would need to develop a scatterplot for overall index of body strength scores (predictor) and overall injury index scores (criterion). As a reminder, Mary’s predictor (overall index of body strength) is a computed variable comprised of five separate measures. Mary’s scatterplot is presented below.

Conclusion

In reviewing the scatterplot, Mary’s X and Y variables appear to be bivariately normally distributed. The first assumption is confirmed.

Computing the Bivariate Linear Regression

A regression analysis results in an equation that is a model interested in understanding the specific relationship between a predictor variable and criterion variable; in Mary’s case her predictor is Ztotstr and her criterion is Injury. Ideally, if a predictor is associated with a criterion variable, the resulting regression model regression equation should allow a researcher to predict his/her outcome with a relatively small amount of error. If Mary’s hypothesis that strength is predictive of injury, her resulting regression model and associated equation should provide her a degree of certainty in identifying elderly individuals at risk of injury. As the steps of the bivariate linear regression are reviewed, readers will leave with an understanding for how the scatterplot and its best fit (presented above) were created as well as understand whether Mary’s hypotheses can be rejected.

As one will see in the sections below, there are multiple steps involved in conducting the linear regression; something that isn’t necessarily appreciated with the statistical package programs and languages such as SPSS, SAS, Stata, R, etc. Therefore, the intent here is not to suggest that someone should not use software to conduct analyses—that would be untenable, and really just downright silly—rather the reader should leave with a better appreciation of the steps that go into such an analysis as well as offer insight into the array of numbers and figures produced by statistical software solutions. It is one thing have output, but it is another to understand and appreciate what it actually means.


Note

The following sections will walk through the steps needed to conduct a bivariate linear regression. It will/should become clear that many of the steps outlined below are similar to ones addressed in other walkthrough materials. As a result, it is assumed that the reader has at least a basic understanding of both one-way ANOVA as well as Pearson Correlation. If these concepts aren’t familiar, it is recommended that those materials be reviewed before proceeding.


Pearson Correlation

The first step in conducting a regression analysis is to conduct a Pearson correlation among the predictor and criterion variables as Mary will need it later to create her regression equation.

Since Mary is only interested in evaluating the effect of a single variable (Ztotstr, the variable created earlier) and its impact on injury severity among elderly women, only a single Pearson correlation will be conducted.

Presented below is the equation that was used to compute the Pearson correlation coefficient in the prior lesson. The same equation will be used here to compute the Pearson correlation for Mary’s study as well.

\[ r = \frac{ \Bigg(\frac{\sum\limits_{i\text{ = 1}}^{n}(X - \overline X)(Y - \overline Y)}{N - 1}\Bigg)}{S_XS_Y} \]

Note

Remember the Pearson correlation is a measure of the strength of a linear association between two continuous variables.

  • A value of 0 indicates that there is no association between the two variables.
  • A value greater than 0 indicates a positive association; that is, as the value of one variable increases, so does the value of the other variable.
  • A value less than 0 indicates a negative association; that is, as the value of one variable increases, the value of the other variable decreases.

For a more detailed description of the Pearson correlation, readers should review the Pearson correlation walkthrough materials.

Simplify the Equation

Simplifying the formula to Mary’s experiment results in the equation below.

\[ r = \frac{ \Bigg(\frac{\text{Sum}(\text{Mean Difference}_{\text{Injury}\huge\cdot}\text{Mean Difference}_{\text{Ztotsr}})}{N - 1}\Bigg)}{\text{SD}_{\text{Injury}\huge{\cdot}}\text{SD}_{\text{Ztotstr}}} \]

Data Needed to Compute Pearson Correlation

To compute a Pearson correlation coefficient for her predictor, (Ztotstr), and criterion, (Injury), Mary needs four pieces of information which are outlined below.

  • sum of the mean difference products of paired scores (\(\sum\limits_{i\text{ = 1}}^{n}(X - \overline X)(Y - \overline Y)\)) = -5830.79
  • number of observations for Mary’s set of data (\(N\)) = 100
  • standard deviation for for Ztotstr, (\(S_X\)) = 3.47
  • standard deviation for Injury, (\(S_Y\)) = 52.2

Step 1: Input All of the Data

\[ r = \frac{\Bigg(\frac{-5830.79}{100 - 1}\Bigg)}{52.2 \cdot 3.47} \]

Step 2: Simplify the Numerator and Denominator

\[ r = \frac{-58.9}{181.12} \] Step 3: Round the Resulting Pearson Value

\[ r = -0.325 \]

Compute t Value for Pearson Correlation Analysis and Determine Significance

Below is the equation for computing the t statistic for the resulting Pearson coefficient that was discussed in Lesson 31.

\[ t = \frac{\hat{\rho}}{\sqrt{\frac{1-\hat{\rho}^2}{N-2}}} \]

Note

For a more in-depth discussing of conducting a t-test for a Pearson Coefficient, please review the Pearson correlation walkthrough.


Data Needed to Compute t Value Pearson Correlation

  • \({\hat\rho}\) is the Pearson correlation computed above = -0.325
  • \({N}\) is sample size = 100

Step 1: Inpute Data and Simplify Numerator and Denominator

\[ t = \frac{-0.325}{0.096} \]

Step 2: Round the Resulting Pearson Value

\[ t = -3.404 \]

Pearson Correlation Conclusion

Mary’s calculated predictor that she called Z total strength, or Ztotsr, appears to be significant at p < 0.001 since her computed t value of 3.404 is larger than the critical t value of 3.281.

Regression Equation

Compute Slope and Y-Intercept for the Regression Equation

The formula for calculating Mary’s regression equation is provided below along with a description of the terms contained in the equation.

\[ Y_i = b_0 + b_1\cdot X_i + \epsilon_i\]

Regression Equation Terms

  • \(Y_i\) is the criterion variable, in Mary’s case this is Injury score
  • \(X_i\) is the predictor variable, in Mary’s case this is Ztotstr
  • \(b_0\) is the y-intercept for the regression equation. Additionally, this would be the resulting value of the regression equation if the input predictor value were equal to zero. This could be considered a type of baseline or control point. In Mary’s case, in an individual were to have a Ztotstr equal to zero, the resulting predicted value for injury would be equal to the intercept value that will be computed below.
  • \(b_1\) represents the product between the predictor variable and the slope (regression coefficient) that Mary will compute below; the product of this along with the \(b_0\) will produce the predicted outcomes for participants within Mary’s regression model.
  • \(\epsilon_i\) is the uncontrolled or random error present within Mary’s set of data for which she cannot control; for the purpose of this analysis \(\epsilon_i\) will be presumed to be normally distributed with a mean equal to zero which means that \(\epsilon_i\) will not have an impact on Mary’s resulting regression equation.

What is error?

The error, \(\epsilon_i\) in Mary’s represents everything that her model has not taken into account, which is a considerable amount since it is nearly impossible for Mary to control for every possible condition that may interfere with her study’s criterion variable (Injury). Additionally, the error, \(\epsilon_i\), could/would also include measurement discrepancies related to measurement reading or measuring inaccuracies—remember, Ztotstr is comprised of five variables across 100 participants which means there are 500 separate measurements taken along with the 100 observations needed to evaluate injury. For context, a 95% accuracy rate capturing measurements means that 30 of Mary’s measurements, somewhere are incorrect in some fashion.

Why is it assumed to be normally distributed?

Well, because of the central limit theorem. Specifically, it states that the distribution of the sum of a large number of random variables will tend towards a normal distribution. This idea along with the fact that many times ‘real world’ observations of error appear to be distributed this way lends additional support to extrapolating the distribution of the unobservable errors.

Furthermore, since it is assumed that the cases in a regression represent a random sample from the population, and the scores on variables for one case are independent of scores on these variables for other cases another assumption can be made which is that each data point has its own independent associated error, i.e., the errors are independent from one another, which helps provides additional support that error occurs randomly with the data.

And this means, what exactly?

It means a lot… or, more precisely, it means that since the errors are assumed to occur randomly, it can then be expected that each data point has an equal probability of appearing above or below the line of best fit created by the regression (positive error values for the data points with a higher value than the one predicted by the line, and negative error values for the data points with a smaller value predicted by the line).

And this is important, why?

Well, if a researcher were to sum up every error within his/her regression analysis, the resulting error term would very close to zero. This is why the error term in regression is presumed to be equal to zero.

Fine Print

Presuming that error term in regression is equal to zero implies that outside of the covariates (predictor/criterion relationship) included in a model, the rest of the variance is completely independent and normally distributed across the observations. However there are certainly times in the ‘real world’ that this isn’t true, so anyone looking to conduct a regression should use proper due diligence to ensure they have adequately confirmed that the error assumptions can be maintained.


Knowing that she doesn’t need to worry about error present within her data, Mary updates the regression formula to exclude her error term, \(\epsilon_i\) which is presented below.

\[ Y_i = b_0 + b_1\cdot X_i \] which can be re-written as:

\[ Y = mX + b \] The equation above is a more commonly recognizable form of the linear regression equation that is likely familiar to anyone who recalls their grade- and high-school math classes.

Where:

  • \(Y\) is the criterion variable, in Mary’s case this is Injury
  • \(X\) is the predictor variable, in Mary’s case this is Ztotstr
  • \(b\) is the y-intercept for the regression equation
  • \(m\) is the slope for Mary’s regression equation
Calculating for Slope for Mary’s Equation

The slope for Mary’s regression equation can be determined using the formula below. After determining her slope, Mary can then determine the y-intercept for her regression equation. It’s important to note that Mary needs to know her equation’s slope so that she can determine her y-intercept. Without this information, Mary cannot determine her equation’s y-intercept. Therefore, Mary will first solve for her equation’s slope and then will solve for her equation’s y-intercept.

Slope for Regression Equation

\[ m = \hat{\rho} \cdot \frac{S_Y}{S_X}\]

Data Needed to Compute Slope of the Regression Equation

The three items below were already presented as part of Mary’s Pearson correlation analysis and were:

  • Pearson correlation, \(\hat{\rho}\) = -0.325
  • standard deviation for for Ztotstr, (\(S_X\)) = 3.47
  • standard deviation for Injury, (\(S_Y\)) = 52.2

Step 1: Input Data

\[ m = -0.325\cdot \frac{52.2}{3.47} \]

Step 2: Simplify the Equation

\[ m = -0.325\cdot 15.04 \]

Step 3: Round the Resulting Slope Value

\[ m = -4.89\]

Calculating for Y-Intercept for Mary’s Equation

After finding her regression equation’s slope to be -4.89, Mary can now compute her equation’s y-intercept.

Y-Intercept for Regression Equation

Intercept for Regression Equation \[ b = \overline Y - (m \cdot \overline X)\]

Data Needed to Compute Y-Intercept of the Regression Equation

  • slope, \(m\), for Mary’s equation = -4.89
  • mean for for Ztotstr, (\(\overline X\)) = 0
  • mean for Injury, (\(\overline Y\)) = 145.8

Step 1: Input Data

\[ b = 145.8 - (-4.89 \cdot 0 )\]

Step 2: Simplify the Equation

\[ b = 145.8\]

Step 3: Round the Resulting Y-Intercept Value

\[ b = 145.8\]

Regression Equation Conclusion

Taking Mary’s slope of 145.8 and y-intercept of 145.8 she can now create her regression equation, which is presented below:

\[ \text{Predicted Overall Injury} = -4.89_{\text{Overall Strength}} + 145.8 \]

Regression ANOVA

ANOVA and Regression, Different but the Same

Regression and ANOVA Are Different

On cursory review, regression and ANOVA appear to be meaningfully different analyses. A researcher uses regression when he/she wants to predict a continuous outcome on the basis of one or more continuous predictor variables. In contrast, a researcher uses ANOVA when he/she wants to predict a continuous outcome on the basis of one or more categorical predictor variables.

Before continuing, yes, if a researcher had a single categorical variable, and it only had two levels (in other words, a binary category), he/she could use a two-sample t-test, though for the purposes of this review, the reader can assume that if there is a single categorical predictor it has three or more levels or alternatively, there are two or more categorical predictor variables in which the t-test is out leaving only an ANOVA model.

When to Use Regression and ANOVA

Imagine a researcher interested in predicting the resale values for used cars in the US. He/she could use an array of variables to try and develop his/her resale estimates, and depending upon the variables used, his/her approach would vary. For example he/she could use:

  • total number of miles driven (continuous variable) - regression
  • type of car (categorical variable) - ANOVA
  • total number of service visits (continuous variable) - regression
  • car color (categorical) - ANOVA

Now, what if the researcher had two predictor variables, one continuous and one categorical? Suppose, for example, that a researcher wanted to predict resale values of cars in the US using both total number of miles driven (continuous) and the color of the car (categorical)? Is it a regression? Is it an ANOVA? Or is it something entirely different?

Regression and ANOVA Aren’t Different

Both regression and ANOVA are applicable only when a researcher has a continuous outcome variable. Revisiting the example from earlier, if the researcher wasn’t interested in predicting resale value but was instead interested in understanding a car buyer’s willingness to purchase a used car in a given setting then the outcome would no longer be continuous, but instead A categorical outcome variable which would rule out both regression and ANOVA.

ANOVA partitions total variation (Total SS) into variation between groups (Treatment SS) and variation within groups (Error SS). The same sort of partitioning approach is used for a regression, with total variation (total SS) partitioned into variation due to the model (Regression SS) and variation unexplained by the model (Error SS).

Many of the diagnostic procedures used to examine the underlying assumptions for both an ANOVA and regression are the same. In particular, a researcher can compute residuals in both models and the plots involving those residuals are often very helpful.

So What About that Car Resale Example that Had Both Continuous and Categorical Variables?

Well, believe it or not, categorical predictors can be incorporated into a regression model by using indicator variables. Specifically, an indicator variable is a variable that that takes the value of 0 or 1 to indicate the absence or presence of a categorical effect that may be expected change relative to the criterion of interest in a regression model. For example, if the researcher from earlier had categorized car color as black, white, red and all others, he/she would have a four-level categorical variable that could be split into three indicators variables. For simplicity of presentation, four observations for car color and the associated indicator variables are presented in the table below.

Indicator Variables for a Car Color Categorical Variable
Car Color Black Indicator White Indicator Red Indicator
All Other 0 0 0
Black 1 0 0
White 0 0 1
Red 0 1 0
Note:
Note: the last indicator, in this instance the color grouping All Other is not included since it is redunant.

The resulting regression output is effectively the same results as an ANOVA model—the difference being interpretation.


As mentioned above, similar to the one-way ANOVA analysis, regression uses an ANOVA-like approach to partition the total variation of data, \(SS_{\text{Total}}\), into variance due to the regression model, \(SS_{\text{Regression}}\) and variance due to error, \(SS_{\text{Error}}\). The equation for \(SS_{\text{Total}}\) within regression is very similar to that of the equation used within one-way ANOVA, and is presented below with each of its constituent parts labeled.

\[ \begin{array}{ccccc} & & \text{Regression Sum of Squares Partitioned}\\ & & & & \\ SS_{\text{Total}} & = & SS_{\text{Regression}} & + & SS_\text{Error} \\ & & & & \\ \sum\limits_{i\text{ = 1}}^{n}(Y_i - \overline{Y})^2 & = & \sum\limits_{i\text{ = 1}}^{n}(\hat{Y}_i - \overline{Y})^2 & + & \sum\limits_{i\text{ = 1}}^{n}(Y_i - \hat{Y}_i)^2 & & & & & & \\ \end{array} \]

Equation Components

Moving from left to right, the equation components are outlined below.

  • \(n\) is the number of observations within Mary’s set of data.

  • \(Y_{i}\) are the individual scores within a set of data; Mary’s total set of data includes 100 separate scores.

  • \(\overline Y\) is the mean for Injury scores within Mary’s set of data, which is 145.8.

  • \(\hat{Y}_i\) are the individual predicted injury scores within a set of data; Mary would compute these values using her regression equation from earlier. An example for computing a predicted injury score for participant one in Mary’s data is presented below.


Note

Computing a Predicted Injury Score for Participant One in Mary’s Study

\[ \text{Predicted Overall Injury} = (-4.89 \;{\huge\cdot} -5.81) + 145.8 \]

\[ \text{Predicted Overall Injury} = 174.23 \]


Given that the steps for computing sum of squared deviations for regression are very similar to the steps discussed, in the one-way ANOVA, the intermediate steps for computing each Sum of Squares are not included here. However, for a more in-depth discussion for how to compute sum of squared deviations for an ANOVA, the reader should review the Lesson 25’s walkthrough of the One-Way ANOVA. Therefore, the resulting sum of squares values for Mary’s regression are presented below.

\(SS_{\text{Error}}\). The equation for \(SS_{\text{Total}}\) within regression is very similar to that of the equation used within one-way ANOVA and is presented below with each of its constituent parts labeled.

\[\begin{array}{ccccc} & & \text{Regression Sum of Squares Partitioned}\\ & & & & \\ SS_{\text{Total}} & = & SS_{\text{Regression}} & + & SS_\text{Error} \\ & & & & \\ \sum\limits_{i\text{ = 1}}^{n}(Y_i - \overline{Y})^2 & = & \sum\limits_{i\text{ = 1}}^{n}(\hat{Y}_i - \overline{Y})^2 & + & \sum\limits_{i\text{ = 1}}^{n}(Y_i - \hat{Y}_i)^2 & &\\ & & & & \\ 269714& = & 28520.19 & + & 241193.81 & & & & \\ \end{array} \]

A slightly adjusted ANOVA equation for Mary’s regression is:

\[\text{Regression ANOVA}\\~\\F =\frac{\text{Mean Squares Regression}_{(MSR)}}{\text{Mean Squares Error}_{(MSE)}} \]

\[\text{Mean Squares Regression}_{(MSR)} = \text{SS}_{\text{Regression}} \]

\[\text{Mean Squares Treatment}_{(MST)} = 28520.19\]

\[\text{Mean Squares Error}_{(MSE)} = \frac{\text{SS}_{\text{Error}}}{N - 2} \]

  • \(N\) is the total number of participants in Mary’s study, which is 100.

\[\text{Mean Squares Error}_{(MSE)} = 2461.16\]
Plugging the Mean Squares information into the ANOVA equation results in:

\[ F = \frac{\text{Mean Squares Regression}_{(MSR)}}{\text{Mean Squares Error}_{(MSE)}} \]
\[ F = \frac{28520.19}{2461.16} \]

\[ F = 11.588 \]

ANOVA Conclusion

Mary’s regression model appears to be significant at p < .01 since her computed F of 11.588 was greater than the critical value of 6.901.

ANOVA
Model Sum Sq Df Mean Sq F value Significance
Regression 28520.19 1 28520.191 11.588 0.001
Residual 241193.81 98 2461.161
Total 269714.00 99 30981.353
a Criterion: INJURY
b Predictors: (Intercept), ztotstr


Model Summary
R R Square Adjusted R Square Standard Error of the Estimate
-0.325 0.106 0.097 49.61
Note:
Predictors: (Intercept), ztotstr

The table above presents a summary of analyses for Mary’s regression. Each of the items is described below.

  • \(R\) is the Pearson correlation of Mary’s predictor, Ztotstr, and criterion, Injury; the value is reported as the absolute value of the ‘true’ correlation, therefore negative correlations have the negative sign dropped.
  • \(R^2\) provides the relative measure of the percentage of the dependent variable variance that the model explains that can range from 0 to 100%.
  • \(\text{Adjusted }R^2\) sample \(R^2\) typically overestimates the population \(R^2\) and needs to be adjusted downward, this is what \(\text{Adjusted }R^2\) attempts to do. Though this is typically only meaningful for regression analyses that contain more than a single predictor. Therefore, further explanation of \(\text{Adjusted }R^2\) will be pushed to the Multiple Regression lesson, for the time being.
  • Standard Error of the Estimate provides an absolute measure of the typical distance that data points fall from the regression line, i.e. how far data points are from Mary’s regression line on average.
Coefficients
B Standard Error t Significance 95% CI Lower Bound 95% CI Upper Bound
145.800 4.961 29.389 0.001 135.955 155.645
-4.891 1.437 -3.404 0.001 -7.743 -2.040
Note:
Criterion: INJURY

Overall Conclusion

The 95% confidence interval for the slope, —7.74 to —2.04 does not contain the value of zero, and therefore overall strength is significantly related to the overall injury index. As Mary hypothesized, elderly women who are stronger tended to have lower overall injury scores. Accuracy in predicting the overall injury index was moderate. The correlation between the strength index and the injury index was -.32. Approximately 11% of the variance of the injury index was accounted for by its linear relationship with the strength index.



Deeper Dive into Mary’s Analyses

Standard Error of the Estimate

While working through Mary’s regression analysis, several analyses have been conducted along the way. While most were covered during the process, the standard error of the estimate and coefficients, the t value for the intercept and the 95% confidence intervals were only reported and not covered in detail due to brevity. However, for completeness the analyses are presented below.

\[ s_{\text{est}} = \sqrt{\frac{\sum(Y_{i} - \hat{Y_i})^2}{N - 2}} = \sqrt{\frac{\text{Mean Squares Error}_{{(MSE)}}}{N - 2}} = \sqrt{\frac{2461.1613135}{100}} = 49.61\]
Standard Error of the Coefficient

The standard error of the coefficient similar measure the precision of the estimate of the coefficient. The smaller the standard error, the more precise the estimate.

ztotsr

\[ s_{\text{error}B_1} = s_{\text{est}} \cdot \sqrt{\frac{N}{N\cdot \sum(X^2) - (\sum X)^2}} = 49.61 \cdot \sqrt{\frac{100}{100 \cdot 1192.0708778 - 0}} \]

\[ s_{\text{error}B_1} = 49.61 \cdot 0.029 = 1.437 \]

Y-Intercept

\[ s_{\text{error}B_0} = s_{\text{est}} \cdot \sqrt{\frac{\sum X^2}{N\cdot \sum(X^2) - (\sum X)^2}} = 49.61 \cdot \sqrt{\frac{1192.0708778}{100 \cdot 1192.0708778 - 0}}\]

\[ s_{\text{error}B_0} = 49.61 \cdot 0.1 = 4.961\]

t Value for the Y-Intercept

\[ t = \frac{\beta_0}{\beta_{\text{Standard Error}}} \]

\[ t = \frac{145.8}{4.961} \]

\[ t = 29.389 \]

95% Confidence Intervals for Coefficients

ztotstr:

\[ \text{CI}_{0.95}^{\beta_1} = \left[ \hat{\beta}_1 - \text{critical value for }t \times SE(\hat{\beta}_1) \, , \, \hat{\beta}_1 + \text{critical value for }t \times SE(\hat{\beta}_1) \right] \]

\[ \text{CI}_{0.95}^{\beta_1} = \left[ -4.891 - 1.984 \times 1.437 \, , \, -4.891 + 1.984 \times 1.437 \right] \]

\[ \text{CI}_{0.95}^{\beta_1} = \left[ -7.743 \, , \, -2.04 \right] \]

Injury:

\[ \text{CI}_{0.95}^{\beta_0} = \left[ \hat{\beta}_0 - \text{critical value for }t \times SE(\hat{\beta}_0) \, , \, \hat{\beta}_0 + \text{critical value for }t \times SE(\hat{\beta}_0) \right] \]

\[ \text{CI}_{0.95}^{\beta_0} = \left[ 145.8 - 1.984 \times 4.961 \, , \, 145.8 + 1.984 \times 4.961 \right] \]
\[ \text{CI}_{0.95}^{\beta_0} = \left[ 135.955 \, , \, 155.645 \right] \]

 

A work by Alex Aguilar

aaguilar@thechicagoschool.edu