null hypothesis simple linear regression

Understanding the Null Hypothesis for Linear Regression

Linear regression is a technique we can use to understand the relationship between one or more predictor variables and a response variable .

If we only have one predictor variable and one response variable, we can use simple linear regression , which uses the following formula to estimate the relationship between the variables:

ŷ = β 0 + β 1 x

• ŷ: The estimated response value.
• β 0 : The average value of y when x is zero.
• β 1 : The average change in y associated with a one unit increase in x.
• x: The value of the predictor variable.

Simple linear regression uses the following null and alternative hypotheses:

• H 0 : β 1 = 0
• H A : β 1 ≠ 0

The null hypothesis states that the coefficient β 1 is equal to zero. In other words, there is no statistically significant relationship between the predictor variable, x, and the response variable, y.

The alternative hypothesis states that β 1 is not equal to zero. In other words, there is a statistically significant relationship between x and y.

If we have multiple predictor variables and one response variable, we can use multiple linear regression , which uses the following formula to estimate the relationship between the variables:

ŷ = β 0 + β 1 x 1 + β 2 x 2 + … + β k x k

• β 0 : The average value of y when all predictor variables are equal to zero.
• β i : The average change in y associated with a one unit increase in x i .
• x i : The value of the predictor variable x i .

Multiple linear regression uses the following null and alternative hypotheses:

• H 0 : β 1 = β 2 = … = β k = 0
• H A : β 1 = β 2 = … = β k ≠ 0

The null hypothesis states that all coefficients in the model are equal to zero. In other words, none of the predictor variables have a statistically significant relationship with the response variable, y.

The alternative hypothesis states that not every coefficient is simultaneously equal to zero.

The following examples show how to decide to reject or fail to reject the null hypothesis in both simple linear regression and multiple linear regression models.

Example 1: Simple Linear Regression

Suppose a professor would like to use the number of hours studied to predict the exam score that students will receive in his class. He collects data for 20 students and fits a simple linear regression model.

The following screenshot shows the output of the regression model:

The fitted simple linear regression model is:

Exam Score = 67.1617 + 5.2503*(hours studied)

To determine if there is a statistically significant relationship between hours studied and exam score, we need to analyze the overall F value of the model and the corresponding p-value:

• Overall F-Value:  47.9952
• P-value:  0.000

Since this p-value is less than .05, we can reject the null hypothesis. In other words, there is a statistically significant relationship between hours studied and exam score received.

Example 2: Multiple Linear Regression

Suppose a professor would like to use the number of hours studied and the number of prep exams taken to predict the exam score that students will receive in his class. He collects data for 20 students and fits a multiple linear regression model.

The fitted multiple linear regression model is:

Exam Score = 67.67 + 5.56*(hours studied) – 0.60*(prep exams taken)

To determine if there is a jointly statistically significant relationship between the two predictor variables and the response variable, we need to analyze the overall F value of the model and the corresponding p-value:

• Overall F-Value:  23.46
• P-value:  0.00

Since this p-value is less than .05, we can reject the null hypothesis. In other words, hours studied and prep exams taken have a jointly statistically significant relationship with exam score.

Note: Although the p-value for prep exams taken (p = 0.52) is not significant, prep exams combined with hours studied has a significant relationship with exam score.

Additional Resources

Understanding the F-Test of Overall Significance in Regression How to Read and Interpret a Regression Table How to Report Regression Results How to Perform Simple Linear Regression in Excel How to Perform Multiple Linear Regression in Excel

The Complete Guide: How to Report Regression Results

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The Complete Guide To Simple Regression Analysis

08.08.2023 • 8 min read

Sarah Thomas

Subject Matter Expert

Learn what simple regression analysis means and why it’s useful for analyzing data, and how to interpret the results.

In This Article

What Is Simple Linear Regression Analysis?

Linear regression equation, how to perform linear regression, linear regression assumptions, how do you find the regression line, how to interpret the results of simple regression.

What is the relationship between parental income and educational attainment or hours spent on social media and anxiety levels? Regression is a versatile statistical tool that can help you answer these types of questions. It’s a tool that lets you model the relationship between two or more variables .

The applications of regression are endless. You can use it as a machine learning algorithm to make predictions. You can use it to establish correlations, and in some cases, you can use it to uncover causal links in your data.

In this article, we’ll tell you everything you need to know about the most basic form of regression analysis: the simple linear regression model.

Simple linear regression is a statistical tool you can use to evaluate correlations between a single independent variable (X) and a single dependent variable (Y). The model fits a straight line to data collected for each variable, and using this line, you can estimate the correlation between X and Y and predict values of Y using values of X.

As a quick example, imagine you want to explore the relationship between weight (X) and height (Y). You collect data from ten randomly selected individuals, and you plot your data on a scatterplot like the one below.

In the scatterplot, each point represents data collected for one of the individuals in your sample. The blue line is your regression line. It models the relationship between weight and height using observed data. Not surprisingly, we see ‌the regression line is upward-sloping, indicating a positive correlation between weight and height. Taller people tend to be heavier than shorter people.

Once you have this line, you can measure how strong the correlation is between height and weight. You can estimate the height of somebody ‌not in your sample by plugging their weight into the regression equation.

The equation for a simple linear regression is:

X is your independent variable

Y is an estimate of your dependent variable

β 0 \beta_0 β 0 ​ is the constant or intercept of the regression line, which is the value of Y when X is equal to zero

β 1 \beta_1 β 1 ​ is the regression coefficient, which is the slope of the regression line and your estimate for the change in Y given a 1-unit change in X

ε \varepsilon ε is the error term of the regression

You may notic‌e the formula for a regression looks very similar to the equation of a line (y=mX+b). That’s because linear regression is a line! It’s a line fitted to data that you can use to estimate the values of one variable using the value of a correlated variable.

You can build a simple linear regression model in 5 steps.

1. Collect data

Collect data for two variables (X and Y). Y is your dependent variable, which is the variable you want to estimate using the regression. X is your independent variable—the variable you use as an input in your regression.

2. Plot the data on a scatter plot

Plot the values of X and Y on a scatter plot with values of X plotted along the horizontal x-axis and values of Y plotted on the vertical y-axis.

3. Calculate a correlation coefficient

Calculate a correlation coefficient to determine the strength of the linear relationship between your two variables.

4. Fit a regression to the data

Find the regression line using the ordinary least-squares method. (You can do this by hand; but it’s much easier to use statistical software like Desmos, Excel, R, or Stata.)

5. Assess the regression line

Once you have the regression line, assess how well your model performs by checking to see how well the model predicts values of Y.

The key assumptions we make when using a simple linear regression model are:

The relationship between X and Y (if it exists) is linear.

Independence

The residuals of your model are independent.

Homoscedasticity

The variance of the residual is constant across values of the independent variable.

The residuals are normally distributed .

You should not use a simple linear regression unless it’s reasonable to make these assumptions.

Simple linear regression involves fitting a straight line to your dataset. We call this line the line of best fit or the regression line. The most common method for finding this line is OLS (or the Ordinary Least Squares Method).

In OLS, we find the regression line by minimizing the sum of squared residuals —also called squared errors. Anytime you draw a straight line through your data, there will be a vertical distance between each ‌point on your scatter plot and the regression line. These vertical distances are called residuals (or errors).

They represent the difference between the actual values of your dependent variable Y i Y_i Y i ​ , and the predicted value of that variable, Y ^ i \widehat{Y}_i Y i ​ . The regression you find with OLS is the line that minimizes the sum of squared residuals.

You can calculate the OLS regression line by hand, but it’s much easier to do so using statistical software like Excel, Desmos, R, or Stata. In this video, Professor AnnMaria De Mars explains how to find the OLS regression equation using Desmos.

Depending on the software you use, the results of your regression analysis may look ‌different. In general, however, your software will display output tables summarizing the main characteristics of your regression.

The values you should be looking for in these output tables fall under three categories:

Coefficients

Regression statistics

This is the β 0 \beta_0 β 0 ​ value in your regression equation. It is the y-intercept of your regression line, and it is the estimate of Y when X is equal to zero.

Next to your intercept, you’ll see columns in the table showing additional information about the intercept. These include a standard error, p-value, T-stat, and confidence interval. You can use these values to test whether the estimate of your intercept is statistically significant .

Regression coefficient

This is the β 1 \beta_1 β 1 ​ of your regression equation. It’s the slope of the regression line, and it tells you how much Y should change in response to a 1-unit change in X.

Similar to the intercept, the regression coefficient will have columns to the right of it. They'll show a standard error, p-value , T-stat, and confidence interval. Use these values to test whether your parameter estimate of β 1 \beta_1 β 1 ​ is statistically significant.

Regression Statistics

Correlation coefficient (or multiple r).

This is the Pearson Correlation coefficient. It measures the strength of the correlation between X and Y.

R-squared (or the coefficient of determination)

We calculate this value by squaring the correlation coefficient. The independent variable can explain how much of the variance in your dependent variable. You can convert R 2 R^2 R 2 into a percentage by multiplying it by 100.

Standard error of the residuals

The standard error of the residuals is the average value of the errors in your model. It is the average vertical distance between each point on your scatter plot and the regression line. We measure this value in the same units as your dependent variable.

Degrees of freedom

In simple linear regression, the degrees of freedom equal the number of data points you used minus the two estimated parameters. The parameters are the intercept and regression coefficient.

Some software will also output a 5-number summary of your residuals. It'll show the minimum, first quartile , median , third quartile, and maximum values of your residuals.

P-value (or Significance F) - This is the p-value of your regression model.

It returns a hypothesis test's results where the null hypothesis is that no relationship exists between X and Y. The alternative hypothesis is that a linear relationship exists between X and Y.

If you are using a significance level (or alpha level) of 0.05, you would reject the null hypothesis if the p-value is less than or equal to 0.05. You would fail to reject the null hypothesis if your p-value is greater than 0.05.

What are correlations?

A correlation is a measure of the relationship between two variables.

Positive Correlations - If two variables, X and Y, have a positive linear correlation, Y tends to increase as X increases, and Y tends to decrease as X decreases. In other words, the two variables tend to move together in the same direction.

Negative Correlations - Two variables, X and Y, have a negative correlation if Y tends to increase as X decreases and Y tends to decrease as X increases. (i.e., The values of the two variables tend to move in opposite directions).

What’s the difference between the dependent and independent variables in a regression?

A simple linear regression involves two variables: X, the input or independent variable, and Y, the output or dependent variable. The independent variable is the variable you want to estimate using the regression. Its estimated value “depends” on the parameters and other variables of the model.

The independent variable—also called the predictor variable—is an input in the model. Its value does not depend on the other elements of the model.

Is the correlation coefficient the same as the regression coefficient?

The correlation coefficient and the regression coefficient will both have the same sign (positive or negative), but they are not the same. The only case where these two values will be equal is when the values of X and Y have been standardized to the same scale.

What is a correlation coefficient?

A correlation coefficient—or Pearson’s correlation coefficient —measures the strength of the linear relationship between X and Y. It’s a number ranging between -1 and 1. The closer a coefficient correlation is to 0, the weaker the correlation is between X and Y.

The closer the correlation coefficient is to 1 or -1, the stronger the correlation. Points on a scatter plot will be more dispersed around the regression line when the correlation between X and Y is weak, and the points will be more tightly clustered around the regression line when the correlation is strong.

What is the regression coefficient?

The regression coefficient, β 1 \beta_1 β 1 ​ , is the slope of the regression line. It provides you with an estimate of how much the dependent variable, Y, will change in response to a 1-unit increase in the dependent variable, X.

The regression coefficient can be any number from − ∞ -\infty − ∞ to ∞ \infty ∞ . A positive regression coefficient implies a positive correlation between X and Y, and a negative regression coefficient implies a negative correlation.

Can I use linear regression in Excel?

Yes. The easiest way to add a simple linear regression line in Excel is to install and use Excel’s “Analysis Toolpak” add-in. To do this, go to Tools > Excel Add-ins and select the “Analysis Toolpak.”

Next, follow these steps.

In your spreadsheet, enter your data for X and Y in two columns

Navigate to the “Data” tab and click on the “Data Analysis” icon

From the list of analysis tools, select “Regression” and click “OK”

Select the data for Y and X respectively where it says “Input Y Range” and “Input X Range”

If you’ve labeled your columns with the names of your X and Y variables, click on the “Labels” checkbox.

You can further customize where you want your regression in your workbook and what additional information you would like Excel to display.

Once you’ve finished customizing, click “OK”

Your regression results will display next to your data or in a new sheet.

Is linear regression used to establish causal relationships?

Correlations are not equivalent to causation. If two variables are correlated, you cannot immediately conclude ‌one causes the other to change. A linear regression will immediately indicate whether two variables correlate. But you’ll need to include more variables in your model and use regression with causal theories to draw conclusions about causal relationships.

What are some other types of regression analysis?

Simple linear regression is the most basic form of regression analysis. It involves ‌one independent variable and one dependent variable. Once you get a handle on this model, you can move on to more sophisticated forms of regression analysis. These include multiple linear regression and nonlinear regression.

Multiple linear regression is a model that estimates the linear relationship between variables using one dependent variable and multiple predictor variables. Nonlinear regression is a method used to estimate nonlinear relationships between variables.

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Simple Linear Regression

• First Online: 20 January 2023

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• Daniel P. McGibney 10  

Part of the book series: International Series in Operations Research & Management Science ((ISOR,volume 337))

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In Albert Einstein’s quote above, he stresses the paramount importance of simplicity. In regression analysis, focusing on only two variables demonstrates the concepts simply. Thus, in Chap. 2 , we calculated the least squares line by using two variables. We also plotted scatterplots and calculated correlation coefficients to further assess the linear relationship. From this analysis, we obtained a detailed understanding of the relationship between two variables. Upon understanding a linear relationship, other more complicated processes become easier to grasp.

If you can’t explain it simply, you don’t understand it well enough. —Albert Einstein

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McGibney, D.P. (2023). Simple Linear Regression. In: Applied Linear Regression for Business Analytics with R. International Series in Operations Research & Management Science, vol 337. Springer, Cham. https://doi.org/10.1007/978-3-031-21480-6_4

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Linear regression is used to predict the value of an outcome variable Y based on one or more input predictor variables X . The aim is to establish a linear relationship (a mathematical formula) between the predictor variable(s) and the response variable, so that, we can use this formula to estimate the value of the response Y , when only the predictors ( X s ) values are known.

Introduction

The aim of linear regression is to model a continuous variable Y as a mathematical function of one or more X variable(s), so that we can use this regression model to predict the Y when only the X is known. This mathematical equation can be generalized as follows:

Y  =  β 1  +  β 2 X  +  ϵ

where, β 1 is the intercept and β 2 is the slope. Collectively, they are called regression coefficients . ϵ is the error term, the part of Y the regression model is unable to explain.

Example Problem

For this analysis, we will use the cars dataset that comes with R by default. cars is a standard built-in dataset, that makes it convenient to demonstrate linear regression in a simple and easy to understand fashion. You can access this dataset simply by typing in cars in your R console. You will find that it consists of 50 observations(rows) and 2 variables (columns) – dist and speed . Lets print out the first six observations here..

Before we begin building the regression model, it is a good practice to analyze and understand the variables. The graphical analysis and correlation study below will help with this.

Graphical Analysis

The aim of this exercise is to build a simple regression model that we can use to predict Distance (dist) by establishing a statistically significant linear relationship with Speed (speed). But before jumping in to the syntax, lets try to understand these variables graphically. Typically, for each of the independent variables (predictors), the following plots are drawn to visualize the following behavior:

• Scatter plot : Visualize the linear relationship between the predictor and response
• Box plot : To spot any outlier observations in the variable. Having outliers in your predictor can drastically affect the predictions as they can easily affect the direction/slope of the line of best fit.
• Density plot : To see the distribution of the predictor variable. Ideally, a close to normal distribution (a bell shaped curve), without being skewed to the left or right is preferred. Let us see how to make each one of them.

Scatter Plot

Scatter plots can help visualize any linear relationships between the dependent (response) variable and independent (predictor) variables. Ideally, if you are having multiple predictor variables, a scatter plot is drawn for each one of them against the response, along with the line of best as seen below.

The scatter plot along with the smoothing line above suggests a linearly increasing relationship between the ‘dist’ and ‘speed’ variables. This is a good thing, because, one of the underlying assumptions in linear regression is that the relationship between the response and predictor variables is linear and additive.

BoxPlot – Check for outliers

Generally, any datapoint that lies outside the 1.5 * interquartile-range ( 1.5 *  I Q R ) is considered an outlier, where, IQR is calculated as the distance between the 25th percentile and 75th percentile values for that variable.

Density plot – Check if the response variable is close to normality

Correlation.

Correlation is a statistical measure that suggests the level of linear dependence between two variables, that occur in pair – just like what we have here in speed and dist. Correlation can take values between -1 to +1. If we observe for every instance where speed increases, the distance also increases along with it, then there is a high positive correlation between them and therefore the correlation between them will be closer to 1. The opposite is true for an inverse relationship, in which case, the correlation between the variables will be close to -1.

A value closer to 0 suggests a weak relationship between the variables. A low correlation (-0.2 < x < 0.2) probably suggests that much of variation of the response variable ( Y ) is unexplained by the predictor ( X ), in which case, we should probably look for better explanatory variables.

Build Linear Model

Now that we have seen the linear relationship pictorially in the scatter plot and by computing the correlation, lets see the syntax for building the linear model. The function used for building linear models is lm() . The lm() function takes in two main arguments, namely: 1. Formula 2. Data. The data is typically a data.frame and the formula is a object of class formula . But the most common convention is to write out the formula directly in place of the argument as written below.

Now that we have built the linear model, we also have established the relationship between the predictor and response in the form of a mathematical formula for Distance (dist) as a function for speed. For the above output, you can notice the ‘Coefficients’ part having two components: Intercept : -17.579, speed : 3.932 These are also called the beta coefficients. In other words, d i s t  =  I n t e r c e p t  + ( β  ∗  s p e e d ) => dist = −17.579 + 3.932∗speed

Linear Regression Diagnostics

Now the linear model is built and we have a formula that we can use to predict the dist value if a corresponding speed is known. Is this enough to actually use this model? NO! Before using a regression model, you have to ensure that it is statistically significant. How do you ensure this? Lets begin by printing the summary statistics for linearMod.

The p Value: Checking for statistical significance

The summary statistics above tells us a number of things. One of them is the model p-Value (bottom last line) and the p-Value of individual predictor variables (extreme right column under ‘Coefficients’). The p-Values are very important because, We can consider a linear model to be statistically significant only when both these p-Values are less that the pre-determined statistical significance level, which is ideally 0.05. This is visually interpreted by the significance stars at the end of the row. The more the stars beside the variable’s p-Value, the more significant the variable.

Null and alternate hypothesis

When there is a p-value, there is a hull and alternative hypothesis associated with it. In Linear Regression, the Null Hypothesis is that the coefficients associated with the variables is equal to zero. The alternate hypothesis is that the coefficients are not equal to zero (i.e. there exists a relationship between the independent variable in question and the dependent variable).

We can interpret the t-value something like this. A larger t-value indicates that it is less likely that the coefficient is not equal to zero purely by chance. So, higher the t-value, the better.

Pr(>|t|) or p-value is the probability that you get a t-value as high or higher than the observed value when the Null Hypothesis (the β coefficient is equal to zero or that there is no relationship) is true. So if the Pr(>|t|) is low, the coefficients are significant (significantly different from zero). If the Pr(>|t|) is high, the coefficients are not significant.

What this means to us? when p Value is less than significance level (< 0.05), we can safely reject the null hypothesis that the co-efficient β of the predictor is zero. In our case, linearMod , both these p-Values are well below the 0.05 threshold, so we can conclude our model is indeed statistically significant.

It is absolutely important for the model to be statistically significant before we can go ahead and use it to predict (or estimate) the dependent variable, otherwise, the confidence in predicted values from that model reduces and may be construed as an event of chance.

How to calculate the t Statistic and p-Values?

When the model co-efficients and standard error are known, the formula for calculating t Statistic and p-Value is as follows: $$t−Statistic = {β−coefficient \over Std.Error}$$

R-Squared and Adj R-Squared

The actual information in a data is the total variation it contains, remember?. What R-Squared tells us is the proportion of variation in the dependent (response) variable that has been explained by this model.

$$ R^{2} = 1 - \frac{SSE}{SST}$$

where, S S E is the sum of squared errors given by $SSE = \sum_{i}^{n} \left( y_{i} - \hat{y_{i}} \right) ^{2}$ and $SST = \sum_{i}^{n} \left( y_{i} - \bar{y_{i}} \right) ^{2}$ is the sum of squared total . Here, $\hat{y_{i}}$ is the fitted value for observation i and $\bar{y}$ is the mean of Y .

We don’t necessarily discard a model based on a low R-Squared value. Its a better practice to look at the AIC and prediction accuracy on validation sample when deciding on the efficacy of a model.

Now thats about R-Squared. What about adjusted R-Squared? As you add more X variables to your model, the R-Squared value of the new bigger model will always be greater than that of the smaller subset. This is because, since all the variables in the original model is also present, their contribution to explain the dependent variable will be present in the super-set as well, therefore, whatever new variable we add can only add (if not significantly) to the variation that was already explained. It is here, the adjusted R-Squared value comes to help. Adj R-Squared penalizes total value for the number of terms (read predictors) in your model. Therefore when comparing nested models, it is a good practice to look at adj-R-squared value over R-squared.

$$ R^{2}_{adj} = 1 - \frac{MSE}{MST}$$

where, M S E is the mean squared error given by $MSE = \frac{SSE}{\left( n-q \right)}$ and $MST = \frac{SST}{\left( n-1 \right)}$ is the mean squared total , where n is the number of observations and q is the number of coefficients in the model.

Therefore, by moving around the numerators and denominators, the relationship between R 2 and R a d j 2 becomes:

$$R^{2}_{adj} = 1 - \left( \frac{\left( 1 - R^{2}\right) \left(n-1\right)}{n-q}\right)$$

Standard Error and F-Statistic

Both standard errors and F-statistic are measures of goodness of fit.

$$Std. Error = \sqrt{MSE} = \sqrt{\frac{SSE}{n-q}}$$

$$F-statistic = \frac{MSR}{MSE}$$

where, n is the number of observations, q is the number of coefficients and M S R is the mean square regression , calculated as,

$$MSR=\frac{\sum_{i}^{n}\left( \hat{y_{i} - \bar{y}}\right)}{q-1} = \frac{SST - SSE}{q - 1}$$

AIC and BIC

The Akaike’s information criterion - AIC (Akaike, 1974) and the Bayesian information criterion - BIC (Schwarz, 1978) are measures of the goodness of fit of an estimated statistical model and can also be used for model selection. Both criteria depend on the maximized value of the likelihood function L for the estimated model.

The AIC is defined as:

A I C  = (−2) ×  l n ( L ) + (2× k )

where, k is the number of model parameters and the BIC is defined as:

B I C  = (−2) ×  l n ( L ) +  k  ×  l n ( n )

where, n is the sample size.

For model comparison, the model with the lowest AIC and BIC score is preferred.

How to know if the model is best fit for your data?

The most common metrics to look at while selecting the model are:

Predicting Linear Models

So far we have seen how to build a linear regression model using the whole dataset. If we build it that way, there is no way to tell how the model will perform with new data. So the preferred practice is to split your dataset into a 80:20 sample (training:test), then, build the model on the 80% sample and then use the model thus built to predict the dependent variable on test data.

Doing it this way, we will have the model predicted values for the 20% data (test) as well as the actuals (from the original dataset). By calculating accuracy measures (like min_max accuracy) and error rates (MAPE or MSE), we can find out the prediction accuracy of the model. Now, lets see how to actually do this..

Step 1: Create the training (development) and test (validation) data samples from original data.

Step 2: develop the model on the training data and use it to predict the distance on test data, step 3: review diagnostic measures..

From the model summary, the model p value and predictor’s p value are less than the significance level, so we know we have a statistically significant model. Also, the R-Sq and Adj R-Sq are comparative to the original model built on full data.

Step 4: Calculate prediction accuracy and error rates

A simple correlation between the actuals and predicted values can be used as a form of accuracy measure. A higher correlation accuracy implies that the actuals and predicted values have similar directional movement, i.e. when the actuals values increase the predicteds also increase and vice-versa.

Now lets calculate the Min Max accuracy and MAPE: $$MinMaxAccuracy = mean \left( \frac{min\left(actuals, predicteds\right)}{max\left(actuals, predicteds \right)} \right)$$

$$MeanAbsolutePercentageError \ (MAPE) = mean\left( \frac{abs\left(predicteds−actuals\right)}{actuals}\right)$$

k- Fold Cross validation

Suppose, the model predicts satisfactorily on the 20% split (test data), is that enough to believe that your model will perform equally well all the time? It is important to rigorously test the model’s performance as much as possible. One way is to ensure that the model equation you have will perform well, when it is ‘built’ on a different subset of training data and predicted on the remaining data.

How to do this is? Split your data into ‘k’ mutually exclusive random sample portions. Keeping each portion as test data, we build the model on the remaining (k-1 portion) data and calculate the mean squared error of the predictions. This is done for each of the ‘k’ random sample portions. Then finally, the average of these mean squared errors (for ‘k’ portions) is computed. We can use this metric to compare different linear models.

By doing this, we need to check two things:

• If the model’s prediction accuracy isn’t varying too much for any one particular sample, and
• If the lines of best fit don’t vary too much with respect the the slope and level.

In other words, they should be parallel and as close to each other as possible. You can find a more detailed explanation for interpreting the cross validation charts when you learn about advanced linear model building.

In the below plot, Are the dashed lines parallel? Are the small and big symbols are not over dispersed for one particular color?

Where to go from here?

We have covered the basic concepts about linear regression. Besides these, you need to understand that linear regression is based on certain underlying assumptions that must be taken care especially when working with multiple X s . Once you are familiar with that, the advanced regression models will show you around the various special cases where a different form of regression would be more suitable.

© 2016-17 Selva Prabhakaran. Powered by jekyll , knitr , and pandoc . This work is licensed under the Creative Commons License.

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11.1: Testing the Hypothesis that β = 0

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The correlation coefficient, \(r\), tells us about the strength and direction of the linear relationship between \(x\) and \(y\). However, the reliability of the linear model also depends on how many observed data points are in the sample. We need to look at both the value of the correlation coefficient \(r\) and the sample size \(n\), together. We perform a hypothesis test of the "significance of the correlation coefficient" to decide whether the linear relationship in the sample data is strong enough to use to model the relationship in the population.

The sample data are used to compute \(r\), the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But because we have only sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, \(r\), is our estimate of the unknown population correlation coefficient.

• The symbol for the population correlation coefficient is \(\rho\), the Greek letter "rho."
• \(\rho =\) population correlation coefficient (unknown)
• \(r =\) sample correlation coefficient (known; calculated from sample data)

The hypothesis test lets us decide whether the value of the population correlation coefficient \(\rho\) is "close to zero" or "significantly different from zero". We decide this based on the sample correlation coefficient \(r\) and the sample size \(n\).

If the test concludes that the correlation coefficient is significantly different from zero, we say that the correlation coefficient is "significant."

• Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between \(x\) and \(y\) because the correlation coefficient is significantly different from zero.
• What the conclusion means: There is a significant linear relationship between \(x\) and \(y\). We can use the regression line to model the linear relationship between \(x\) and \(y\) in the population.

If the test concludes that the correlation coefficient is not significantly different from zero (it is close to zero), we say that correlation coefficient is "not significant".

• Conclusion: "There is insufficient evidence to conclude that there is a significant linear relationship between \(x\) and \(y\) because the correlation coefficient is not significantly different from zero."
• What the conclusion means: There is not a significant linear relationship between \(x\) and \(y\). Therefore, we CANNOT use the regression line to model a linear relationship between \(x\) and \(y\) in the population.
• If \(r\) is significant and the scatter plot shows a linear trend, the line can be used to predict the value of \(y\) for values of \(x\) that are within the domain of observed \(x\) values.
• If \(r\) is not significant OR if the scatter plot does not show a linear trend, the line should not be used for prediction.
• If \(r\) is significant and if the scatter plot shows a linear trend, the line may NOT be appropriate or reliable for prediction OUTSIDE the domain of observed \(x\) values in the data.

PERFORMING THE HYPOTHESIS TEST

• Null Hypothesis: \(H_{0}: \rho = 0\)
• Alternate Hypothesis: \(H_{a}: \rho \neq 0\)

WHAT THE HYPOTHESES MEAN IN WORDS:

• Null Hypothesis \(H_{0}\) : The population correlation coefficient IS NOT significantly different from zero. There IS NOT a significant linear relationship(correlation) between \(x\) and \(y\) in the population.
• Alternate Hypothesis \(H_{a}\) : The population correlation coefficient IS significantly DIFFERENT FROM zero. There IS A SIGNIFICANT LINEAR RELATIONSHIP (correlation) between \(x\) and \(y\) in the population.

DRAWING A CONCLUSION:There are two methods of making the decision. The two methods are equivalent and give the same result.

• Method 1: Using the \(p\text{-value}\)
• Method 2: Using a table of critical values

In this chapter of this textbook, we will always use a significance level of 5%, \(\alpha = 0.05\)

Using the \(p\text{-value}\) method, you could choose any appropriate significance level you want; you are not limited to using \(\alpha = 0.05\). But the table of critical values provided in this textbook assumes that we are using a significance level of 5%, \(\alpha = 0.05\). (If we wanted to use a different significance level than 5% with the critical value method, we would need different tables of critical values that are not provided in this textbook.)

METHOD 1: Using a \(p\text{-value}\) to make a decision

To calculate the \(p\text{-value}\) using LinRegTTEST:

On the LinRegTTEST input screen, on the line prompt for \(\beta\) or \(\rho\), highlight "\(\neq 0\)"

The output screen shows the \(p\text{-value}\) on the line that reads "\(p =\)".

(Most computer statistical software can calculate the \(p\text{-value}\).)

If the \(p\text{-value}\) is less than the significance level ( \(\alpha = 0.05\) ):

• Decision: Reject the null hypothesis.
• Conclusion: "There is sufficient evidence to conclude that there is a significant linear relationship between \(x\) and \(y\) because the correlation coefficient is significantly different from zero."

If the \(p\text{-value}\) is NOT less than the significance level ( \(\alpha = 0.05\) )

• Decision: DO NOT REJECT the null hypothesis.
• Conclusion: "There is insufficient evidence to conclude that there is a significant linear relationship between \(x\) and \(y\) because the correlation coefficient is NOT significantly different from zero."

Calculation Notes:

• You will use technology to calculate the \(p\text{-value}\). The following describes the calculations to compute the test statistics and the \(p\text{-value}\):
• The \(p\text{-value}\) is calculated using a \(t\)-distribution with \(n - 2\) degrees of freedom.
• The formula for the test statistic is \(t = \frac{r\sqrt{n-2}}{\sqrt{1-r^{2}}}\). The value of the test statistic, \(t\), is shown in the computer or calculator output along with the \(p\text{-value}\). The test statistic \(t\) has the same sign as the correlation coefficient \(r\).
• The \(p\text{-value}\) is the combined area in both tails.

An alternative way to calculate the \(p\text{-value}\) ( \(p\) ) given by LinRegTTest is the command 2*tcdf(abs(t),10^99, n-2) in 2nd DISTR.

THIRD-EXAM vs FINAL-EXAM EXAMPLE: \(p\text{-value}\) method

• Consider the third exam/final exam example.
• The line of best fit is: \(\hat{y} = -173.51 + 4.83x\) with \(r = 0.6631\) and there are \(n = 11\) data points.
• Can the regression line be used for prediction? Given a third exam score ( \(x\) value), can we use the line to predict the final exam score (predicted \(y\) value)?
• \(H_{0}: \rho = 0\)
• \(H_{a}: \rho \neq 0\)
• \(\alpha = 0.05\)
• The \(p\text{-value}\) is 0.026 (from LinRegTTest on your calculator or from computer software).
• The \(p\text{-value}\), 0.026, is less than the significance level of \(\alpha = 0.05\).
• Decision: Reject the Null Hypothesis \(H_{0}\)
• Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between the third exam score (\(x\)) and the final exam score (\(y\)) because the correlation coefficient is significantly different from zero.

Because \(r\) is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores.

METHOD 2: Using a table of Critical Values to make a decision

The 95% Critical Values of the Sample Correlation Coefficient Table can be used to give you a good idea of whether the computed value of \(r\) is significant or not . Compare \(r\) to the appropriate critical value in the table. If \(r\) is not between the positive and negative critical values, then the correlation coefficient is significant. If \(r\) is significant, then you may want to use the line for prediction.

Example \(\PageIndex{1}\)

Suppose you computed \(r = 0.801\) using \(n = 10\) data points. \(df = n - 2 = 10 - 2 = 8\). The critical values associated with \(df = 8\) are \(-0.632\) and \(+0.632\). If \(r <\) negative critical value or \(r >\) positive critical value, then \(r\) is significant. Since \(r = 0.801\) and \(0.801 > 0.632\), \(r\) is significant and the line may be used for prediction. If you view this example on a number line, it will help you.

Exercise \(\PageIndex{1}\)

For a given line of best fit, you computed that \(r = 0.6501\) using \(n = 12\) data points and the critical value is 0.576. Can the line be used for prediction? Why or why not?

If the scatter plot looks linear then, yes, the line can be used for prediction, because \(r >\) the positive critical value.

Example \(\PageIndex{2}\)

Suppose you computed \(r = –0.624\) with 14 data points. \(df = 14 – 2 = 12\). The critical values are \(-0.532\) and \(0.532\). Since \(-0.624 < -0.532\), \(r\) is significant and the line can be used for prediction

Exercise \(\PageIndex{2}\)

For a given line of best fit, you compute that \(r = 0.5204\) using \(n = 9\) data points, and the critical value is \(0.666\). Can the line be used for prediction? Why or why not?

No, the line cannot be used for prediction, because \(r <\) the positive critical value.

Example \(\PageIndex{3}\)

Suppose you computed \(r = 0.776\) and \(n = 6\). \(df = 6 - 2 = 4\). The critical values are \(-0.811\) and \(0.811\). Since \(-0.811 < 0.776 < 0.811\), \(r\) is not significant, and the line should not be used for prediction.

Exercise \(\PageIndex{3}\)

For a given line of best fit, you compute that \(r = -0.7204\) using \(n = 8\) data points, and the critical value is \(= 0.707\). Can the line be used for prediction? Why or why not?

Yes, the line can be used for prediction, because \(r <\) the negative critical value.

THIRD-EXAM vs FINAL-EXAM EXAMPLE: critical value method

Consider the third exam/final exam example. The line of best fit is: \(\hat{y} = -173.51 + 4.83x\) with \(r = 0.6631\) and there are \(n = 11\) data points. Can the regression line be used for prediction? Given a third-exam score ( \(x\) value), can we use the line to predict the final exam score (predicted \(y\) value)?

• Use the "95% Critical Value" table for \(r\) with \(df = n - 2 = 11 - 2 = 9\).
• The critical values are \(-0.602\) and \(+0.602\)
• Since \(0.6631 > 0.602\), \(r\) is significant.
• Conclusion:There is sufficient evidence to conclude that there is a significant linear relationship between the third exam score (\(x\)) and the final exam score (\(y\)) because the correlation coefficient is significantly different from zero.

Example \(\PageIndex{4}\)

Suppose you computed the following correlation coefficients. Using the table at the end of the chapter, determine if \(r\) is significant and the line of best fit associated with each r can be used to predict a \(y\) value. If it helps, draw a number line.

• \(r = –0.567\) and the sample size, \(n\), is \(19\). The \(df = n - 2 = 17\). The critical value is \(-0.456\). \(-0.567 < -0.456\) so \(r\) is significant.
• \(r = 0.708\) and the sample size, \(n\), is \(9\). The \(df = n - 2 = 7\). The critical value is \(0.666\). \(0.708 > 0.666\) so \(r\) is significant.
• \(r = 0.134\) and the sample size, \(n\), is \(14\). The \(df = 14 - 2 = 12\). The critical value is \(0.532\). \(0.134\) is between \(-0.532\) and \(0.532\) so \(r\) is not significant.
• \(r = 0\) and the sample size, \(n\), is five. No matter what the \(dfs\) are, \(r = 0\) is between the two critical values so \(r\) is not significant.

Exercise \(\PageIndex{4}\)

For a given line of best fit, you compute that \(r = 0\) using \(n = 100\) data points. Can the line be used for prediction? Why or why not?

No, the line cannot be used for prediction no matter what the sample size is.

Assumptions in Testing the Significance of the Correlation Coefficient

Testing the significance of the correlation coefficient requires that certain assumptions about the data are satisfied. The premise of this test is that the data are a sample of observed points taken from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear relationship that we see between \(x\) and \(y\) in the sample data provides strong enough evidence so that we can conclude that there is a linear relationship between \(x\) and \(y\) in the population.

The regression line equation that we calculate from the sample data gives the best-fit line for our particular sample. We want to use this best-fit line for the sample as an estimate of the best-fit line for the population. Examining the scatter plot and testing the significance of the correlation coefficient helps us determine if it is appropriate to do this.

The assumptions underlying the test of significance are:

• There is a linear relationship in the population that models the average value of \(y\) for varying values of \(x\). In other words, the expected value of \(y\) for each particular value lies on a straight line in the population. (We do not know the equation for the line for the population. Our regression line from the sample is our best estimate of this line in the population.)
• The \(y\) values for any particular \(x\) value are normally distributed about the line. This implies that there are more \(y\) values scattered closer to the line than are scattered farther away. Assumption (1) implies that these normal distributions are centered on the line: the means of these normal distributions of \(y\) values lie on the line.
• The standard deviations of the population \(y\) values about the line are equal for each value of \(x\). In other words, each of these normal distributions of \(y\) values has the same shape and spread about the line.
• The residual errors are mutually independent (no pattern).
• The data are produced from a well-designed, random sample or randomized experiment.

Linear regression is a procedure for fitting a straight line of the form \(\hat{y} = a + bx\) to data. The conditions for regression are:

• Linear In the population, there is a linear relationship that models the average value of \(y\) for different values of \(x\).
• Independent The residuals are assumed to be independent.
• Normal The \(y\) values are distributed normally for any value of \(x\).
• Equal variance The standard deviation of the \(y\) values is equal for each \(x\) value.
• Random The data are produced from a well-designed random sample or randomized experiment.

The slope \(b\) and intercept \(a\) of the least-squares line estimate the slope \(\beta\) and intercept \(\alpha\) of the population (true) regression line. To estimate the population standard deviation of \(y\), \(\sigma\), use the standard deviation of the residuals, \(s\). \(s = \sqrt{\frac{SEE}{n-2}}\). The variable \(\rho\) (rho) is the population correlation coefficient. To test the null hypothesis \(H_{0}: \rho =\) hypothesized value , use a linear regression t-test. The most common null hypothesis is \(H_{0}: \rho = 0\) which indicates there is no linear relationship between \(x\) and \(y\) in the population. The TI-83, 83+, 84, 84+ calculator function LinRegTTest can perform this test (STATS TESTS LinRegTTest).

Formula Review

Least Squares Line or Line of Best Fit:

\[\hat{y} = a + bx\]

\[a = y\text{-intercept}\]

\[b = \text{slope}\]

Standard deviation of the residuals:

\[s = \sqrt{\frac{SEE}{n-2}}\]

\[SSE = \text{sum of squared errors}\]

\[n = \text{the number of data points}\]

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6.4 - the hypothesis tests for the slopes.

At the beginning of this lesson, we translated three different research questions pertaining to heart attacks in rabbits ( Cool Hearts dataset ) into three sets of hypotheses we can test using the general linear F -statistic. The research questions and their corresponding hypotheses are:

Hypotheses 1

Is the regression model containing at least one predictor useful in predicting the size of the infarct?

• \(H_{0} \colon \beta_{1} = \beta_{2} = \beta_{3} = 0\)
• \(H_{A} \colon\) At least one \(\beta_{j} ≠ 0\) (for j = 1, 2, 3)

Hypotheses 2

Is the size of the infarct significantly (linearly) related to the area of the region at risk?

• \(H_{0} \colon \beta_{1} = 0 \)
• \(H_{A} \colon \beta_{1} \ne 0 \)

Hypotheses 3

(Primary research question) Is the size of the infarct area significantly (linearly) related to the type of treatment upon controlling for the size of the region at risk for infarction?

• \(H_{0} \colon \beta_{2} = \beta_{3} = 0\)
• \(H_{A} \colon \) At least one \(\beta_{j} ≠ 0\) (for j = 2, 3)

Let's test each of the hypotheses now using the general linear F -statistic:

\(F^*=\left(\dfrac{SSE(R)-SSE(F)}{df_R-df_F}\right) \div \left(\dfrac{SSE(F)}{df_F}\right)\)

To calculate the F -statistic for each test, we first determine the error sum of squares for the reduced and full models — SSE ( R ) and SSE ( F ), respectively. The number of error degrees of freedom associated with the reduced and full models — \(df_{R}\) and \(df_{F}\), respectively — is the number of observations, n , minus the number of parameters, p , in the model. That is, in general, the number of error degrees of freedom is n - p . We use statistical software, such as Minitab's F -distribution probability calculator, to determine the P -value for each test.

Testing all slope parameters equal 0 Section  

To answer the research question: "Is the regression model containing at least one predictor useful in predicting the size of the infarct?" To do so, we test the hypotheses:

• \(H_{0} \colon \beta_{1} = \beta_{2} = \beta_{3} = 0 \)
• \(H_{A} \colon\) At least one \(\beta_{j} \ne 0 \) (for j = 1, 2, 3)

The full model

The full model is the largest possible model — that is, the model containing all of the possible predictors. In this case, the full model is:

\(y_i=(\beta_0+\beta_1x_{i1}+\beta_2x_{i2}+\beta_3x_{i3})+\epsilon_i\)

The error sum of squares for the full model, SSE ( F ), is just the usual error sum of squares, SSE , that appears in the analysis of variance table. Because there are 4 parameters in the full model, the number of error degrees of freedom associated with the full model is \(df_{F} = n - 4\).

The reduced model

The reduced model is the model that the null hypothesis describes. Because the null hypothesis sets each of the slope parameters in the full model equal to 0, the reduced model is:

\(y_i=\beta_0+\epsilon_i\)

The reduced model suggests that none of the variations in the response y is explained by any of the predictors. Therefore, the error sum of squares for the reduced model, SSE ( R ), is just the total sum of squares, SSTO , that appears in the analysis of variance table. Because there is only one parameter in the reduced model, the number of error degrees of freedom associated with the reduced model is \(df_{R} = n - 1 \).

Upon plugging in the above quantities, the general linear F -statistic:

\(F^*=\dfrac{SSE(R)-SSE(F)}{df_R-df_F} \div \dfrac{SSE(F)}{df_F}\)

becomes the usual " overall F -test ":

\(F^*=\dfrac{SSR}{3} \div \dfrac{SSE}{n-4}=\dfrac{MSR}{MSE}\)

That is, to test \(H_{0}\) : \(\beta_{1} = \beta_{2} = \beta_{3} = 0 \), we just use the overall F -test and P -value reported in the analysis of variance table:

Analysis of Variance

Regression equation.

Inf = - 0.135 + 0.613 Area - 0.2435 X2 - 0.0657 X3

There is sufficient evidence ( F = 16.43, P < 0.001) to conclude that at least one of the slope parameters is not equal to 0.

In general, to test that all of the slope parameters in a multiple linear regression model are 0, we use the overall F -test reported in the analysis of variance table.

Testing one slope parameter is 0 Section  

Now let's answer the second research question: "Is the size of the infarct significantly (linearly) related to the area of the region at risk?" To do so, we test the hypotheses:

Again, the full model is the model containing all of the possible predictors:

The error sum of squares for the full model, SSE ( F ), is just the usual error sum of squares, SSE . Alternatively, because the three predictors in the model are \(x_{1}\), \(x_{2}\), and \(x_{3}\), we can denote the error sum of squares as SSE (\(x_{1}\), \(x_{2}\), \(x_{3}\)). Again, because there are 4 parameters in the model, the number of error degrees of freedom associated with the full model is \(df_{F} = n - 4 \).

Because the null hypothesis sets the first slope parameter, \(\beta_{1}\), equal to 0, the reduced model is:

\(y_i=(\beta_0+\beta_2x_{i2}+\beta_3x_{i3})+\epsilon_i\)

Because the two predictors in the model are \(x_{2}\) and \(x_{3}\), we denote the error sum of squares as SSE (\(x_{2}\), \(x_{3}\)). Because there are 3 parameters in the model, the number of error degrees of freedom associated with the reduced model is \(df_{R} = n - 3\).

The general linear statistic:

simplifies to:

\(F^*=\dfrac{SSR(x_1|x_2, x_3)}{1}\div \dfrac{SSE(x_1,x_2, x_3)}{n-4}=\dfrac{MSR(x_1|x_2, x_3)}{MSE(x_1,x_2, x_3)}\)

Getting the numbers from the Minitab output:

we determine that the value of the F -statistic is:

\(F^* = \dfrac{SSR(x_1 \vert x_2, x_3)}{1} \div \dfrac{SSE(x_1, x_2, x_3)}{28} = \dfrac{0.63742}{0.01946}=32.7554\)

The P -value is the probability — if the null hypothesis were true — that we would get an F -statistic larger than 32.7554. Comparing our F -statistic to an F -distribution with 1 numerator degree of freedom and 28 denominator degrees of freedom, Minitab tells us that the probability is close to 1 that we would observe an F -statistic smaller than 32.7554:

F distribution with 1 DF in Numerator and 28 DF in denominator

Therefore, the probability that we would get an F -statistic larger than 32.7554 is close to 0. That is, the P -value is < 0.001. There is sufficient evidence ( F = 32.8, P < 0.001) to conclude that the size of the infarct is significantly related to the size of the area at risk after the other predictors x2 and x3 have been taken into account.

But wait a second! Have you been wondering why we couldn't just use the slope's t -statistic to test that the slope parameter, \(\beta_{1}\), is 0? We can! Notice that the P -value ( P < 0.001) for the t -test ( t * = 5.72):

Coefficients

is the same as the P -value we obtained for the F -test. This will always be the case when we test that only one slope parameter is 0. That's because of the well-known relationship between a t -statistic and an F -statistic that has one numerator degree of freedom:

\(t_{(n-p)}^{2}=F_{(1, n-p)}\)

For our example, the square of the t -statistic, 5.72, equals our F -statistic (within rounding error). That is:

\(t^{*2}=5.72^2=32.72=F^*\)

So what have we learned in all of this discussion about the equivalence of the F -test and the t -test? In short:

Compare the output obtained when \(x_{1}\) = Area is entered into the model last :

Inf = - 0.135 - 0.2435 X2 - 0.0657 X3 + 0.613 Area

to the output obtained when \(x_{1}\) = Area is entered into the model first :

The t -statistic and P -value are the same regardless of the order in which \(x_{1}\) = Area is entered into the model. That's because — by its equivalence to the F -test — the t -test for one slope parameter adjusts for all of the other predictors included in the model.

• We can use either the F -test or the t -test to test that only one slope parameter is 0. Because the t -test results can be read right off of the Minitab output, it makes sense that it would be the test that we'll use most often.
• But, we have to be careful with our interpretations! The equivalence of the t -test to the F -test has taught us something new about the t -test. The t -test is a test for the marginal significance of the \(x_{1}\) predictor after the other predictors \(x_{2}\) and \(x_{3}\) have been taken into account. It does not test for the significance of the relationship between the response y and the predictor \(x_{1}\) alone.

Testing a subset of slope parameters is 0 Section  

Finally, let's answer the third — and primary — research question: "Is the size of the infarct area significantly (linearly) related to the type of treatment upon controlling for the size of the region at risk for infarction?" To do so, we test the hypotheses:

• \(H_{0} \colon \beta_{2} = \beta_{3} = 0 \)
• \(H_{A} \colon\) At least one \(\beta_{j} \ne 0 \) (for j = 2, 3)

Because the null hypothesis sets the second and third slope parameters, \(\beta_{2}\) and \(\beta_{3}\), equal to 0, the reduced model is:

\(y_i=(\beta_0+\beta_1x_{i1})+\epsilon_i\)

The ANOVA table for the reduced model is:

Because the only predictor in the model is \(x_{1}\), we denote the error sum of squares as SSE (\(x_{1}\)) = 0.8793. Because there are 2 parameters in the model, the number of error degrees of freedom associated with the reduced model is \(df_{R} = n - 2 = 32 – 2 = 30\).

\begin{align} F^*&=\dfrac{SSE(R)-SSE(F)}{df_R-df_F} \div\dfrac{SSE(F)}{df_F}\\&=\dfrac{0.8793-0.54491}{30-28} \div\dfrac{0.54491}{28}\\&= \dfrac{0.33439}{2} \div 0.01946\\&=8.59.\end{align}

Alternatively, we can calculate the F-statistic using a partial F-test :

\begin{align}F^*&=\dfrac{SSR(x_2, x_3|x_1)}{2}\div \dfrac{SSE(x_1,x_2, x_3)}{n-4}\\&=\dfrac{MSR(x_2, x_3|x_1)}{MSE(x_1,x_2, x_3)}.\end{align}

To conduct the test, we regress y = InfSize on \(x_{1}\) = Area and \(x_{2}\) and \(x_{3 }\)— in order (and with "Sequential sums of squares" selected under "Options"):

Inf = - 0.135 + 0.613 Area - 0.2435 X2 - 0.0657 X3

yielding SSR (\(x_{2}\) | \(x_{1}\)) = 0.31453, SSR (\(x_{3}\) | \(x_{1}\), \(x_{2}\)) = 0.01981, and MSE = 0.54491/28 = 0.01946. Therefore, the value of the partial F -statistic is:

\begin{align} F^*&=\dfrac{SSR(x_2, x_3|x_1)}{2}\div \dfrac{SSE(x_1,x_2, x_3)}{n-4}\\&=\dfrac{0.31453+0.01981}{2}\div\dfrac{0.54491}{28}\\&= \dfrac{0.33434}{2} \div 0.01946\\&=8.59,\end{align}

which is identical (within round-off error) to the general F-statistic above. The P -value is the probability — if the null hypothesis were true — that we would observe a partial F -statistic more extreme than 8.59. The following Minitab output:

F distribution with 2 DF in Numerator and 28 DF in denominator

tells us that the probability of observing such an F -statistic that is smaller than 8.59 is 0.9988. Therefore, the probability of observing such an F -statistic that is larger than 8.59 is 1 - 0.9988 = 0.0012. The P -value is very small. There is sufficient evidence ( F = 8.59, P = 0.0012) to conclude that the type of cooling is significantly related to the extent of damage that occurs — after taking into account the size of the region at risk.

Summary of MLR Testing Section  

For the simple linear regression model, there is only one slope parameter about which one can perform hypothesis tests. For the multiple linear regression model, there are three different hypothesis tests for slopes that one could conduct. They are:

• Hypothesis test for testing that all of the slope parameters are 0.
• Hypothesis test for testing that a subset — more than one, but not all — of the slope parameters are 0.
• Hypothesis test for testing that one slope parameter is 0.

We have learned how to perform each of the above three hypothesis tests. Along the way, we also took two detours — one to learn about the " general linear F-test " and one to learn about " sequential sums of squares. " As you now know, knowledge about both is necessary for performing the three hypothesis tests.

The F -statistic and associated p -value in the ANOVA table is used for testing whether all of the slope parameters are 0. In most applications, this p -value will be small enough to reject the null hypothesis and conclude that at least one predictor is useful in the model. For example, for the rabbit heart attacks study, the F -statistic is (0.95927/(4–1)) / (0.54491/(32–4)) = 16.43 with p -value 0.000.

To test whether a subset — more than one, but not all — of the slope parameters are 0, there are two equivalent ways to calculate the F-statistic:

• Use the general linear F-test formula by fitting the full model to find SSE(F) and fitting the reduced model to find SSE(R) . Then the numerator of the F-statistic is (SSE(R) – SSE(F)) / ( \(df_{R}\) – \(df_{F}\)) .
• Alternatively, use the partial F-test formula by fitting only the full model but making sure the relevant predictors are fitted last and "sequential sums of squares" have been selected. Then the numerator of the F-statistic is the sum of the relevant sequential sums of squares divided by the sum of the degrees of freedom for these sequential sums of squares. The denominator of the F -statistic is the mean squared error in the ANOVA table.

For example, for the rabbit heart attacks study, the general linear F-statistic is ((0.8793 – 0.54491) / (30 – 28)) / (0.54491 / 28) = 8.59 with p -value 0.0012. Alternatively, the partial F -statistic for testing the slope parameters for predictors \(x_{2}\) and \(x_{3}\) using sequential sums of squares is ((0.31453 + 0.01981) / 2) / (0.54491 / 28) = 8.59.

To test whether one slope parameter is 0, we can use an F -test as just described. Alternatively, we can use a t -test, which will have an identical p -value since in this case, the square of the t -statistic is equal to the F -statistic. For example, for the rabbit heart attacks study, the F -statistic for testing the slope parameter for the Area predictor is (0.63742/1) / (0.54491/(32–4)) = 32.75 with p -value 0.000. Alternatively, the t -statistic for testing the slope parameter for the Area predictor is 0.613 / 0.107 = 5.72 with p -value 0.000, and \(5.72^{2} = 32.72\).

Incidentally, you may be wondering why we can't just do a series of individual t-tests to test whether a subset of the slope parameters is 0. For example, for the rabbit heart attacks study, we could have done the following:

• Fit the model of y = InfSize on \(x_{1}\) = Area and \(x_{2}\) and \(x_{3}\) and use an individual t-test for \(x_{3}\).
• If the test results indicate that we can drop \(x_{3}\) then fit the model of y = InfSize on \(x_{1}\) = Area and \(x_{2}\) and use an individual t-test for \(x_{2}\).

The problem with this approach is we're using two individual t-tests instead of one F-test, which means our chance of drawing an incorrect conclusion in our testing procedure is higher. Every time we do a hypothesis test, we can draw an incorrect conclusion by:

• rejecting a true null hypothesis, i.e., make a type I error by concluding the tested predictor(s) should be retained in the model when in truth it/they should be dropped; or
• failing to reject a false null hypothesis, i.e., make a type II error by concluding the tested predictor(s) should be dropped from the model when in truth it/they should be retained.

Thus, in general, the fewer tests we perform the better. In this case, this means that wherever possible using one F-test in place of multiple individual t-tests is preferable.

Hypothesis tests for the slope parameters Section  

The problems in this section are designed to review the hypothesis tests for the slope parameters, as well as to give you some practice on models with a three-group qualitative variable (which we'll cover in more detail in Lesson 8). We consider tests for:

• whether one slope parameter is 0 (for example, \(H_{0} \colon \beta_{1} = 0 \))
• whether a subset (more than one but less than all) of the slope parameters are 0 (for example, \(H_{0} \colon \beta_{2} = \beta_{3} = 0 \) against the alternative \(H_{A} \colon \beta_{2} \ne 0 \) or \(\beta_{3} \ne 0 \) or both ≠ 0)
• whether all of the slope parameters are 0 (for example, \(H_{0} \colon \beta_{1} = \beta_{2} = \beta_{3}\) = 0 against the alternative \(H_{A} \colon \) at least one of the \(\beta_{i}\) is not 0)

(Note the correct specification of the alternative hypotheses for the last two situations.)

Sugar beets study

A group of researchers was interested in studying the effects of three different growth regulators ( treat , denoted 1, 2, and 3) on the yield of sugar beets (y = yield , in pounds). They planned to plant the beets in 30 different plots and then randomly treat 10 plots with the first growth regulator, 10 plots with the second growth regulator, and 10 plots with the third growth regulator. One problem, though, is that the amount of available nitrogen in the 30 different plots varies naturally, thereby giving a potentially unfair advantage to plots with higher levels of available nitrogen. Therefore, the researchers also measured and recorded the available nitrogen (\(x_{1}\) = nit , in pounds/acre) in each plot. They are interested in comparing the mean yields of sugar beets subjected to the different growth regulators after taking into account the available nitrogen. The Sugar Beets dataset contains the data from the researcher's experiment.

Preliminary Work

The plot shows a similar positive linear trend within each treatment category, which suggests that it is reasonable to formulate a multiple regression model that would place three parallel lines through the data.

Because the qualitative variable treat distinguishes between the three treatment groups (1, 2, and 3), we need to create two indicator variables, \(x_{2}\) and \(x_{3}\), say, to fit a linear regression model to these data. The new indicator variables should be defined as follows:

Use Minitab's Calc >> Make Indicator Variables command to create the new indicator variables in your worksheet

Minitab creates an indicator variable for each treatment group but we can only use two, for treatment groups 1 and 2 in this case (treatment group 3 is the reference level in this case).

Then, if we assume the trend in the data can be summarized by this regression model:

\(y_{i} = \beta_{0}\) + \(\beta_{1}\)\(x_{1}\) + \(\beta_{2}\)\(x_{2}\) + \(\beta_{3}\)\(x_{3}\) + \(\epsilon_{i}\)

where \(x_{1}\) = nit and \(x_{2}\) and \(x_{3}\) are defined as above, what is the mean response function for plots receiving treatment 3? for plots receiving treatment 1? for plots receiving treatment 2? Are the three regression lines that arise from our formulated model parallel? What does the parameter \(\beta_{2}\) quantify? And, what does the parameter \(\beta_{3}\) quantify?

The fitted equation from Minitab is Yield = 84.99 + 1.3088 Nit - 2.43 \(x_{2}\) - 2.35 \(x_{3}\), which means that the equations for each treatment group are:

• Group 1: Yield = 84.99 + 1.3088 Nit - 2.43(1) = 82.56 + 1.3088 Nit
• Group 2: Yield = 84.99 + 1.3088 Nit - 2.35(1) = 82.64 + 1.3088 Nit
• Group 3: Yield = 84.99 + 1.3088 Nit

The three estimated regression lines are parallel since they have the same slope, 1.3088.

The regression parameter for \(x_{2}\) represents the difference between the estimated intercept for treatment 1 and the estimated intercept for reference treatment 3.

The regression parameter for \(x_{3}\) represents the difference between the estimated intercept for treatment 2 and the estimated intercept for reference treatment 3.

Testing whether all of the slope parameters are 0

\(H_0 \colon \beta_1 = \beta_2 = \beta_3 = 0\) against the alternative \(H_A \colon \) at least one of the \(\beta_i\) is not 0.

\(F=\dfrac{SSR(X_1,X_2,X_3)\div3}{SSE(X_1,X_2,X_3)\div(n-4)}=\dfrac{MSR(X_1,X_2,X_3)}{MSE(X_1,X_2,X_3)}\)

\(F = \dfrac{\frac{16039.5}{3}}{\frac{1078.0}{30-4}} = \dfrac{5346.5}{41.46} = 128.95\)

Since the p -value for this F -statistic is reported as 0.000, we reject \(H_{0}\) in favor of \(H_{A}\) and conclude that at least one of the slope parameters is not zero, i.e., the regression model containing at least one predictor is useful in predicting the size of sugar beet yield.

Tests for whether one slope parameter is 0

\(H_0 \colon \beta_1= 0\) against the alternative \(H_A \colon \beta_1 \ne 0\)

t -statistic = 19.60, p -value = 0.000, so we reject \(H_{0}\) in favor of \(H_{A}\) and conclude that the slope parameter for \(x_{1}\) = nit is not zero, i.e., sugar beet yield is significantly linearly related to the available nitrogen (controlling for treatment).

\(F=\dfrac{SSR(X_1|X_2,X_3)\div1}{SSE(X_1,X_2,X_3)\div(n-4)}=\dfrac{MSR(X_1|X_2,X_3)}{MSE(X_1,X_2,X_3)}\)

Use the Minitab output to calculate the value of this F statistic. Does the value you obtain equal \(t^{2}\), the square of the t -statistic as we might expect?

\(F-statistic= \dfrac{\frac{15934.5}{1}}{\frac{1078.0}{30-4}} = \dfrac{15934.5}{41.46} = 384.32\), which is the same as \(19.60^{2}\).

Because \(t^{2}\) will equal the partial F -statistic whenever you test for whether one slope parameter is 0, it makes sense to just use the t -statistic and P -value that Minitab displays as a default. But, note that we've just learned something new about the meaning of the t -test in the multiple regression setting. It tests for the ("marginal") significance of the \(x_{1}\) predictor after \(x_{2}\) and \(x_{3}\) have already been taken into account.

Tests for whether a subset of the slope parameters is 0

\(H_0 \colon \beta_2=\beta_3= 0\) against the alternative \(H_A \colon \beta_2 \ne 0\) or \(\beta_3 \ne 0\) or both \(\ne 0\).

\(F=\dfrac{SSR(X_2,X_3|X_1)\div2}{SSE(X_1,X_2,X_3)\div(n-4)}=\dfrac{MSR(X_2,X_3|X_1)}{MSE(X_1,X_2,X_3)}\)

\(F = \dfrac{\frac{10.4+27.5}{2}}{\frac{1078.0}{30-4}} = \dfrac{18.95}{41.46} = 0.46\).

F distribution with 2 DF in Numerator and 26 DF in denominator

p-value \(= 1-0.363677 = 0.636\), so we fail to reject \(H_{0}\) in favor of \(H_{A}\) and conclude that we cannot rule out \(\beta_2 = \beta_3 = 0\), i.e., there is no significant difference in the mean yields of sugar beets subjected to the different growth regulators after taking into account the available nitrogen.

Note that the sequential mean square due to regression, MSR(\(X_{2}\),\(X_{3}\)|\(X_{1}\)), is obtained by dividing the sequential sum of square by its degrees of freedom (2, in this case, since two additional predictors \(X_{2}\) and \(X_{3}\) are considered). Use the Minitab output to calculate the value of this F statistic, and use Minitab to get the associated P -value. Answer the researcher's question at the \(\alpha= 0.05\) level.

Statistics Made Easy

A Simple Guide to Understanding the F-Test of Overall Significance in Regression

This tutorial explains how to identify the F-statistic in the output of a regression table as well as how to interpret this statistic and its corresponding p-value.

Understanding the F-Test of Overall Significance

The F-Test of overall significance in regression is a test of whether or not your linear regression model provides a better fit to a dataset than a model with no predictor variables. 

The F-Test of overall significance has the following two hypotheses:

Null hypothesis (H 0 ) : The model with no predictor variables (also known as an  intercept-only model ) fits the data as well as your regression model.

Alternative hypothesis (H A ) :  Your regression model fits the data better than the intercept-only model.

When you fit a regression model to a dataset, you will receive a regression table as output, which will tell you the F-statistic along with the corresponding p-value for that F-statistic.

If the p-value is less than the significance level you’ve chosen ( common choices are .01, .05, and .10 ), then you have sufficient evidence to conclude that your regression model fits the data better than the intercept-only model.

Example: F-Test in Regression

Suppose we have the following dataset that shows the total number of hours studied, total prep exams taken, and final exam score received for 12 different students:

To analyze the relationship between hours studied and prep exams taken with the final exam score that a student receives, we run a multiple linear regression using  hours studied and  prep  exams taken  as the predictor variables and  final exam score  as the response variable.

We receive the following output:

From these results, we will focus on the F-statistic given in the ANOVA table as well as the p-value of that F-statistic, which is labeled as  Significance F in the table. We will choose .05 as our significance level.

F-statistic: 5.090515

P-value: 0.0332

Technical note: The F-statistic is calculated as MS regression divided by MS residual. In this case MS regression / MS residual =273.2665 / 53.68151 = 5.090515 .

Since the p-value is less than the significance level, we can conclude that our regression model fits the data better than the intercept-only model.

In the context of this specific problem, it means that using our predictor variables Study Hours and  Prep Exams  in the model allows us to fit the data better than if we left them out and simply used the intercept-only model.

Notes on Interpreting the F-Test of Overall Significance

In general, if none of your predictor variables are statistically significant, the overall F-test will also not be statistically significant.

However, it’s possible on some occasions that this doesn’t hold because the F-test of overall significance tests whether all of the predictor variables are jointly  significant while the t-test of significance for each individual predictor variable merely tests whether each predictor variable is  individually  significant. 

Thus, the F-test determines whether or not  all  of the predictor variables are jointly significant.

It’s possible that each predictor variable is not significant and yet the F-test says that all of the predictor variables combined are jointly significant. 

Technical note:  In general, the more predictor variables you have in the model, the higher the likelihood that the The F-statistic and corresponding p-value will be statistically significant.

Another metric that you’ll likely see in the output of a regression is  R-squared , which measures the strength of the linear relationship between the predictor variables and the response variable is another.

Although R-squared can give you an idea of how strongly associated the predictor variables are with the response variable, it doesn’t provide a formal statistical test for this relationship.

This is why the F-Test is useful since it is a formal statistical test. In addition, if the overall F-test is significant, you can conclude that R-squared is not equal to zero and that the correlation between the predictor variable(s) and response variable is statistically significant.

Additional Resources

The following tutorials explain how to interpret other common values in regression models:

How to Read and Interpret a Regression Table Understanding the Standard Error of the Regression What is a Good R-squared Value?

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Hey there. My name is Zach Bobbitt. I have a Master of Science degree in Applied Statistics and I’ve worked on machine learning algorithms for professional businesses in both healthcare and retail. I’m passionate about statistics, machine learning, and data visualization and I created Statology to be a resource for both students and teachers alike.  My goal with this site is to help you learn statistics through using simple terms, plenty of real-world examples, and helpful illustrations.

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Python Linear Regression Quiz

What is Linear Regression?

A classification algorithm

A clustering algorithm

A regression algorithm

A reinforcement learning algorithm 

In a simple linear regression, how many independent variables are there?

What is the primary goal of linear regression?

Minimizing the sum of squared errors

Maximizing accuracy

Minimizing computational complexity

Maximizing precision

What is the equation of a simple linear regression line?  

y=ax 2 +bx+c

y=alog(x)+b

What is multicollinearity in the context of linear regression?  

High correlation among independent variables

Low correlation among independent variables  

Absence of outliers in the data

Perfect fit of the model

Which metric is commonly used to evaluate the performance of a linear regression model?

Which of the following is NOT an assumption of linear regression?

Independence of residuals

Homoscedasticity

Multicollinearity

What is the difference between simple linear regression and multiple linear regression?

Simple linear regression has one independent variable, while multiple linear regression has more than one

Simple linear regression is for classification, while multiple linear regression is for regression

Simple linear regression uses a nonlinear equation, while multiple linear regression uses a linear equation

There is no difference; the terms are used interchangeably

What is the significance of the p-value in linear regression?

It indicates the strength of the relationship between variables

It measures the precision of the model

It tests the null hypothesis that a variable has no effect

It represents the correlation coefficient

Question 10

How is the learning rate parameter important in gradient descent optimization for linear regression?  

It determines the number of iterations needed for convergence

It adjusts the step size during each iteration

It controls the regularization strength

It is not relevant for linear regression

There are 25 questions to complete.

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Sun Coast’s Data Analysis: Hypothesis Testing Research Paper

Introduction, correlation: hypothesis testing, simple regression: hypothesis testing, multiple regression: hypothesis testing.

This paper requires the use of statistical analysis to determine relationships among variables. The primary purpose of such analysis is to obtain answers to the organizational questions that Sun Coast asks in order to improve the company’s operational performance. Specifically, this work involves three types of inferential tests supported by descriptive statistics, namely correlation analysis, regression analysis, and multiple regression analysis. Because the results have critical potential for the company and can be used to improve the organizational agenda significantly, their significance must be supported by statistical evidence (Tenny & Abdelgawad, 2021). To determine the significance of the constructed models and results, a p -value parameter is used at a significance level of.05. It follows that when the calculated p -value is more significant than.05, the null hypothesis cannot be rejected; the reverse is also true. Choosing a significance level of.05 corresponds to a 5% probability of making an error in detecting significant differences when in fact, there are none: in other words, there is always a probability of making an error of the first kind. For this reason, the results of statistical analyses need to be read carefully and interpreted carefully, as they are expected to be embodied in a company’s realistic environment.

The first part of the general statistical analysis requires a Pearson correlation test to determine the strength and direction of the relationship between two continuous variables. The end result, the correlation coefficient, will show how strongly the two variables are related and the direction of the relationship between them and will demonstrate whether or not the result is significant. Below are the formulations of the two statistical hypotheses that are evaluated in this section on how to perform a correlation analysis.

• Ho 1 : There is no statistically significant relationship between particulate matter size and employee sick days.
• Ha 1 : There is a statistically significant relationship between particulate matter size and employee sick days.

As shown in Figure 1, the Pearson correlation coefficient between particle size, in microns, and the average number of sick days for an employee is -.716. A negative value of this coefficient indicates a negative direction of the relationship between the two. In other words, the calculation indicates that an increase in particle size leads to a decrease in the average number of absent days for workers, that is, tiny particles are a predictor of an increase in the number of sick days. The absolute value of the Pearson coefficient is.716, indicating a moderately high value for this relationship. To summarize, Figure 1 corresponds to a moderately high negative correlation between the variables. The coefficient of determination R 2 , which, according to the results of the calculations, was 0.513, is also subject to discussion. This implies that this model explains up to 51.3% of the variance between the variables, indicating moderate reliability of the model.

Of great importance for the correlation analysis is the coefficient of significance, that is, the p -value. The calculated p -value coefficient is.0000, which corresponds to the critically low value of this parameter. Calculating a specific p -value makes no sense, as it is only necessary to compare this value to a significance level of.05. It is clear that the p -value is significantly less than the critical level of significance ( p <.05), from which it follows that there is sufficient evidence to reject the null hypothesis and thus accept the null hypothesis. To summarize this section, it follows from the results that increasing the size of the solids leads to a statistically significant reduction in the number of days an employee misses due to illness.

The second part of this paper requires a simple linear regression test to determine if one variable can influence another variable. In fact, the regression test is similar to the correlation test performed because both tests allow testing for a reciprocal relationship between continuous variables. However, unlike Pearson correlation, regression analysis also allows for the construction of a regression equation between the variables, which is used to analyze this relationship in more detail and to predict the values of the factors used. To perform the regression analysis, the following hypotheses were used:

• Ho 2 : There is no statistically significant relationship between safety training expenditure and lost-time hours.
• Ha 2 : There is a statistically significant relationship between safety training expenditure and lost-time hours.

The analysis was performed in MS Excel for the relevant data, and Figure 2 shows the results of these calculations. To check the overall significance of the constructed model, the second table of Fig. 2 is used, namely the results of the ANOVA test. Thus, the overall significance level of the regression model constructed should have been estimated: F(221) = 1664.21, p <.05. This ANOVA data indicates that the model is significant, that is, there is a statistically significant relationship between the tuition expense variables and total hours missed. In other words, the p -value shows that there is sufficient evidence to reject the null hypothesis and accept the alternative hypothesis. Turning to the third table in Figure 2 provides several conclusions about aspects of this model. In particular, both coefficients — slope and y-intercept — are significant because their respective p -values are well below the critical significance level of.05. This indicates that the regression coefficients are significant, and thus they are reliable parameters for assessing the relationship between variables. In addition, the same table shows that the value of the slope coefficient is -0.143, meaning that there is an inverse relationship between the variables. Increasing the number of training expenditures, it follows, reduces the number of hours workers miss. A regression equation can then be constructed for the data in terms of the variables used, viz:

Lost Time Hours = -0.143*[Safety Training Expenditure] + 273.449

From this equation, in particular, it can be seen that an increase in safety training expenditure for each unit results in a drop of 0.143 hours of lost time. The y-intercept of this equation, that is, the value of the number of hours lost when there is zero investment in training, is also noteworthy. When spending is zero, the number of hours wasted is maximum and is 273.449. When spending increases, there is a downward trend, that is, Sun Coast should be interested in maximizing the amount of reasonable spending to minimize inefficient hours. This equation, among others, is interesting to solve to find the x-intercept.

([Lost Time Hours] – 273.449)/ (-0.143) = [Safety Training Expenditure]

(0 – 273.449)/ (-0.143) = [Safety Training Expenditure]

[Safety Training Expenditure] = 1912.24

This answer shows that with a $1912.24 (in units) investment in safety training, there are zero lost hours for the company, which may be the ultimate goal for organizational reforms at Sun Coast. The coefficient of determination for this regression model and the multiple R are also up for discussion. The coefficient of determination, R 2 , for this pair of variables is 0.883, indicating that the model covers up to 88.3 percent of the variance between data. This is a high figure, which is consistent with the reliability of the constructed model. The multiple correlation coefficient for these variables was 0.940, which also indicates a strong positive relationship between training costs and hours lost.

In the third part of our work, it was interesting to analyze the effect of different components on the overall perceived noise level. It is known that excessive noise exposure can lead to hearing loss, headaches, and decreased employee productivity (Meersens, 2021). As a consequence, Sun Coast is interested in maintaining acceptable levels of noise exposure to employees and maintaining their operational efficiency. To perform this analysis, several components were used as independent factors, namely frequency, angle in degrees, chord length, velocity, and displacement. Total noise level, measured in decibels, was used as a dependent factor. Since the number of independent factors has been increased, the usual linear regression is no longer applicable to the analysis; instead, multiple regression should be used in MS Excel. Although the steps performed are virtually the same, the results of this analysis will be extended. The following hypotheses are used to run the regression:

• Ho 3 : There is no statistically significant relationship between frequency, angle in degrees, chord length, velocity, displacement, and decibel level.
• Ha 3 : There is a statistically significant relationship between frequency, angle in degrees, chord length, velocity, displacement, and decibel level.

First, it is worth noting that the constructed regression model is statistically significant because the calculated p -value appears to be less than the critical level, F(1497) = 170.036, p <.001. This means that there is sufficient evidence to reject the null hypothesis, which means that the set of independent variables significantly affects the perceived noise level. For the model constructed, the coefficient of determination was 0.362, which implies that about 36.2 percent of the variance between the variables could be explained by the multiple regression model. The correlation coefficient was 0.602, indicating a moderate positive relationship between the two.

Particular attention should be paid to the significance of specific coefficients in the constructed multiple regression model. To be more precise, to evaluate their significance, the null hypothesis is used, that is, the coefficient is equal to zero, that is, it has no influence on the dependent factor. As shown in Table 3 of Figure 3, the p-values were above the.05 significance level only for the Angle and Chord Length variables, which means that these factors had no significant effect on the perceived noise level. In other words, these variables can be omitted in the overall regression equation because they have no significant effect on the noise level. The Frequency, Velocity, and Offset variables were statistically significant because their p-values were below the critical value, so the null hypotheses of no effect were rejected. Thus, the final regression equation in terms of the variables was as follows:

[Decibel] = 126.822 – 0.001*[Frequency] + 0.083*[Velocity] – 240.506*[Displacement]

It follows from the equation that Frequency and Displacement have a negative effect on noise level, that is, their increase led to a decrease in noise level. On the contrary, increasing Velocity increased the perceived noise level. It follows from the results that in order to reduce the overall noise levels perceived by Sun Coast workers, management should increase Frequency and Displacement and decrease Velocity.

In conclusion, this paper conducted statistical analyses — including Pearson correlation, linear regression, and multiple regression — to find answers to Sun Coast’s organizational questions. Particular attention was given to refining statistical significance in the analyses to make sure that the findings actually made sense for further practical implementation. It was found that the smallest particle size was positively correlated with the most extensive employee absences due to illness. This means that company management should keep a close eye on polluting particulate size in an effort to reduce the number of sick leave absences. In addition, increased spending on safety training for employees reduced the number of missed hours. It follows that Sun Coast management should invest more finances in order to increase employee productivity. Finally, Frequency, Velocity, and Displacement were shown to have differential effects on overall noise levels. Increasing Frequency and Displacement led to a decrease in this parameter, while increasing Velocity led to an increase in noise. Therefore, management should strive to increase Frequency and Displacement and decrease Velocity in order to ensure that the noise level in the company is normal and not destructive to the employees.

Meersens. (2021). Noise at work, what impact on the health of employees? Meersens. Web.

Tenny, S., & Abdelgawad, I. (2021). Statistical significance . NIH. Web.

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IvyPanda. (2024, April 25). Sun Coast's Data Analysis: Hypothesis Testing. https://ivypanda.com/essays/sun-coasts-data-analysis-hypothesis-testing/

"Sun Coast's Data Analysis: Hypothesis Testing." IvyPanda , 25 Apr. 2024, ivypanda.com/essays/sun-coasts-data-analysis-hypothesis-testing/.

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IvyPanda . 2024. "Sun Coast's Data Analysis: Hypothesis Testing." April 25, 2024. https://ivypanda.com/essays/sun-coasts-data-analysis-hypothesis-testing/.

1. IvyPanda . "Sun Coast's Data Analysis: Hypothesis Testing." April 25, 2024. https://ivypanda.com/essays/sun-coasts-data-analysis-hypothesis-testing/.

Bibliography

IvyPanda . "Sun Coast's Data Analysis: Hypothesis Testing." April 25, 2024. https://ivypanda.com/essays/sun-coasts-data-analysis-hypothesis-testing/.

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• Report: Tow Tank Statistical Analysis
• Displacement: Crime Prevention
• A Null and an Alternate Hypothesis
• The Dynamics of Oscillations Laboratory Report
• Towing Ship Trial Report: Short and Fat
• Concentration of Chlorophyll: Regression Analysis
• Statistics of the Turbulence in the Atmosphere
• Compensation Management and Regression Analysis
• Contingency Tables and Their Related Statistics
• Statistics: The Self-Reference Effect and Memory