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Hypothesis Test for Regression Slope

This lesson describes how to conduct a hypothesis test to determine whether there is a significant linear relationship between an independent variable X and a dependent variable Y .

The test focuses on the slope of the regression line

Y = Β 0 + Β 1 X

where Β 0 is a constant, Β 1 is the slope (also called the regression coefficient), X is the value of the independent variable, and Y is the value of the dependent variable.

If we find that the slope of the regression line is significantly different from zero, we will conclude that there is a significant relationship between the independent and dependent variables.

Test Requirements

The approach described in this lesson is valid whenever the standard requirements for simple linear regression are met.

• The dependent variable Y has a linear relationship to the independent variable X .
• For each value of X, the probability distribution of Y has the same standard deviation σ.
• The Y values are independent.
• The Y values are roughly normally distributed (i.e., symmetric and unimodal ). A little skewness is ok if the sample size is large.

The test procedure consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results.

State the Hypotheses

If there is a significant linear relationship between the independent variable X and the dependent variable Y , the slope will not equal zero.

H o : Β 1 = 0

H a : Β 1 ≠ 0

The null hypothesis states that the slope is equal to zero, and the alternative hypothesis states that the slope is not equal to zero.

Formulate an Analysis Plan

The analysis plan describes how to use sample data to accept or reject the null hypothesis. The plan should specify the following elements.

• Significance level. Often, researchers choose significance levels equal to 0.01, 0.05, or 0.10; but any value between 0 and 1 can be used.
• Test method. Use a linear regression t-test (described in the next section) to determine whether the slope of the regression line differs significantly from zero.

Analyze Sample Data

Using sample data, find the standard error of the slope, the slope of the regression line, the degrees of freedom, the test statistic, and the P-value associated with the test statistic. The approach described in this section is illustrated in the sample problem at the end of this lesson.

SE = s b 1 = sqrt [ Σ(y i - ŷ i ) 2 / (n - 2) ] / sqrt [ Σ(x i - x ) 2 ]

• Slope. Like the standard error, the slope of the regression line will be provided by most statistics software packages. In the hypothetical output above, the slope is equal to 35.

t = b 1 / SE

• P-value. The P-value is the probability of observing a sample statistic as extreme as the test statistic. Since the test statistic is a t statistic, use the t Distribution Calculator to assess the probability associated with the test statistic. Use the degrees of freedom computed above.

Interpret Results

If the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level.

Test Your Understanding

The local utility company surveys 101 randomly selected customers. For each survey participant, the company collects the following: annual electric bill (in dollars) and home size (in square feet). Output from a regression analysis appears below.

Is there a significant linear relationship between annual bill and home size? Use a 0.05 level of significance.

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

H o : The slope of the regression line is equal to zero.

H a : The slope of the regression line is not equal to zero.

• Formulate an analysis plan . For this analysis, the significance level is 0.05. Using sample data, we will conduct a linear regression t-test to determine whether the slope of the regression line differs significantly from zero.

We get the slope (b 1 ) and the standard error (SE) from the regression output.

b 1 = 0.55       SE = 0.24

We compute the degrees of freedom and the t statistic, using the following equations.

DF = n - 2 = 101 - 2 = 99

t = b 1 /SE = 0.55/0.24 = 2.29

where DF is the degrees of freedom, n is the number of observations in the sample, b 1 is the slope of the regression line, and SE is the standard error of the slope.

• Interpret results . Since the P-value (0.0242) is less than the significance level (0.05), we cannot accept the null hypothesis.

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12.2.1: Hypothesis Test for Linear Regression

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• Page ID 34850

• Rachel Webb
• Portland State University

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To test to see if the slope is significant we will be doing a two-tailed test with hypotheses. The population least squares regression line would be \(y = \beta_{0} + \beta_{1} + \varepsilon\) where \(\beta_{0}\) (pronounced “beta-naught”) is the population \(y\)-intercept, \(\beta_{1}\) (pronounced “beta-one”) is the population slope and \(\varepsilon\) is called the error term.

If the slope were horizontal (equal to zero), the regression line would give the same \(y\)-value for every input of \(x\) and would be of no use. If there is a statistically significant linear relationship then the slope needs to be different from zero. We will only do the two-tailed test, but the same rules for hypothesis testing apply for a one-tailed test.

We will only be using the two-tailed test for a population slope.

The hypotheses are:

\(H_{0}: \beta_{1} = 0\) \(H_{1}: \beta_{1} \neq 0\)

The null hypothesis of a two-tailed test states that there is not a linear relationship between \(x\) and \(y\). The alternative hypothesis of a two-tailed test states that there is a significant linear relationship between \(x\) and \(y\).

Either a t-test or an F-test may be used to see if the slope is significantly different from zero. The population of the variable \(y\) must be normally distributed.

F-Test for Regression

An F-test can be used instead of a t-test. Both tests will yield the same results, so it is a matter of preference and what technology is available. Figure 12-12 is a template for a regression ANOVA table,

where \(n\) is the number of pairs in the sample and \(p\) is the number of predictor (independent) variables; for now this is just \(p = 1\). Use the F-distribution with degrees of freedom for regression = \(df_{R} = p\), and degrees of freedom for error = \(df_{E} = n - p - 1\). This F-test is always a right-tailed test since ANOVA is testing the variation in the regression model is larger than the variation in the error.

Use an F-test to see if there is a significant relationship between hours studied and grade on the exam. Use \(\alpha\) = 0.05.

T-Test for Regression

If the regression equation has a slope of zero, then every \(x\) value will give the same \(y\) value and the regression equation would be useless for prediction. We should perform a t-test to see if the slope is significantly different from zero before using the regression equation for prediction. The numeric value of t will be the same as the t-test for a correlation. The two test statistic formulas are algebraically equal; however, the formulas are different and we use a different parameter in the hypotheses.

The formula for the t-test statistic is \(t = \frac{b_{1}}{\sqrt{ \left(\frac{MSE}{SS_{xx}}\right) }}\)

Use the t-distribution with degrees of freedom equal to \(n - p - 1\).

The t-test for slope has the same hypotheses as the F-test:

Use a t-test to see if there is a significant relationship between hours studied and grade on the exam, use \(\alpha\) = 0.05.

Understanding the t-Test in Linear Regression

Linear regression is used to quantify the relationship between a predictor variable and a response variable.

Whenever we perform linear regression, we want to know if there is a statistically significant relationship between the predictor variable and the response variable.

We test for significance by performing a t-test for the regression slope. We use the following null and alternative hypothesis for this t-test:

• H 0 : β 1 = 0 (the slope is equal to zero)
• H A : β 1 ≠ 0 (the slope is not equal to zero)

We then calculate the test statistic as follows:

• t = b  / SE b
• b : coefficient estimate
• SE b : standard error of the coefficient estimate

If the p-value that corresponds to t is less than some threshold (e.g. α = .05) then we reject the null hypothesis and conclude that there is a statistically significant relationship between the predictor variable and the response variable.

The following example shows how to perform a t-test for a linear regression model in practice.

Example: Performing a t-Test for Linear Regression

Suppose a professor wants to analyze the relationship between hours studied and exam score received for 40 of his students.

He performs simple linear regression using hours studied as the predictor variable and exam score received as the response variable.

The following table shows the results of the regression model:

To determine if hours studied has a statistically significant relationship with final exam score, we can perform a t-test.

We use the following null and alternative hypothesis for this t-test:

• H 0 : β 1 = 0 (the slope for hours studied is equal to zero)
• H A : β 1 ≠ 0 (the slope for hours studied is not equal to zero)
• t = 1.117 / 1.025

The p-value that corresponds to t = 1.089 with df = n-2 = 40 – 2 = 38 is 0.283 .

Note that we can also use the T Score to P Value Calculator to calculate this p-value:

Since this p-value is not less than .05, we fail to reject the null hypothesis.

This means that hours studied does not have a statistically significant relationship between final exam score.

Additional Resources

The following tutorials offer additional information about linear regression:

Introduction to Simple Linear Regression Introduction to Multiple Linear Regression How to Interpret Regression Coefficients How to Interpret the F-Test of Overall Significance in Regression

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One Reply to “Understanding the t-Test in Linear Regression”

if I t-test a regression vs another regression instead vs zero, ie a slope vs another slope, do I need to compare against a t crit of n – 4?

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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

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

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

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

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

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.