This chapter deals with the general linear hypothesis. In a wide variety of problems the experimenter is interested in making inferences about a vector parameter. For example, he may wish to estimate the mean of a multivariate normal or to test some hypotheses concerning the mean vector. The problem of estimation can be solved, for example, by resorting to the method of maximum likelihood estimation, discussed in Section 8.7. In this chapter we restrict ourselves to the so-called linear model problems and concern ourselves mainly with problems of hypotheses testing.
In Section 12.2 we formally describe the general model and derive a test in complete generality. In the next four sections we demonstrate the power of this test by solving four important testing problems. We will need a considerable amount of linear algebra in Section 12.2.
A wide variety of problems of hypotheses testing can be treated under a general setup. In this section we state the general problem and derive the test statistic and its distribution. Consider the following examples.
where t is a mathematical variable, β 0, β1, β2 are unknown parameters, and ε is a nonobservable RV. The experimenter takes observations Y1,Y2,…,Yn at predetermined values t1,t2,…,tn, respectively, and is interested in testing the hypothesis that the relation is in fact linear, that is, .
Examples of the type discussed above and their much more complicated variants can all be treated under a general setup. To fix ideas, let us first make the following definition.
In what follows we will assume that ε1,2,…,εn are independent, normal RVs with common variance σ2 and , . In view of (2), it follows that Y1,Y2,…,Yn are independent normal RVs with
We will assume that H is a matrix of full rank r, , and X is a matrix of full rank . Some remarks are in order.
Remark 1. Clearly, Y satisfies a linear model if the vector of means lies in a k-dimensional subspace generated by the linearly independent column vectors x1,x2,…,xk of the matrix X. Indeed, (1) states that EY is a linear combination of the known vectors x1,…,xk. The general linear hypothesis states that the parameters β1,β2,…,βk satisfy r independent homogeneous linear restrictions. It follows that, under H0, EY lies in a -dimensional subspace of the k-space generated by x1,…,xk.
Remark 2. The assumption of normality, which is conventional, is made to compute the likelihood ratio test statistic of H0 and its distribution. If the problem is to estimate β, no such assumption is needed. One can use the principle of least squares and estimate β by minimizing the sum of squares,
The minimizing value is known as a least square estimate of β. This is not a difficult problem, and we will not discuss it here in any detail but will mention only that any solution of the so-called normal equations
is a least square estimator. If the rank of X is , then X' X, which has the same rank as X, is a nonsingular matrix that can be inverted to give a unique least square estimator
If the rank of X is , then X' X is singular and the normal equations do not have a unique solution. One can show, for example, that is unbiased for β, and if the Yi’s are uncorrelated with common variance σ2, the variance-covariance matrix of the ’s is given by
Remark 3. One can similarly compute the so-called restricted least square estimator of β by the usual method of Lagrange multipliers. For example, under one simply minimizes subject to to get the restricted least square estimator . The important point is that, if ε is assumed to be a multivariate normal RV with mean vector 0 and dispersion matrix σ2In, the MLE of β is the same as the least square estimator. In fact, one can show that is the UMVUE of , by the usual methods.
Similarly, we may assume that the regression of Y on x is quadratic:
and we may wish to test that a linear function will be sufficient to describe the relationship, that is, . Here X is the matrix
and H is the matrix (0, 0, 1).
In another example of regression, the Y's can be written as
and we wish to test the hypothesis that . In this case, X is the matrix
and H may be chosen to be the matrix
The model described in this example is frequently referred to as a one-way analysis of variance model. This is a very simple example of an analysis of variance model. Note that the matrix X is of a very special type, namely, the elements of X are either 0 or 1. X is known as a design matrix.
Returning to our general model
we wish to test the null hypothesis . We will compute the likelihood ratio test and the distribution of the test statistic. In order to do so, we assume that ε has a multivariate normal distribution with mean vector 0 and variance-covariance matrix σ2in, where σ2 is unknown and In is the identity matrix. This means that Y has an n-variate normal distribution with mean Xβ and dispersion matrix σ2In for some β and some σ2, both unknown. Here the parameter space Θ is the set of -tuples , and the joint PDF of the X's is given by
It remains to determine the distribution of the test statistic. For this purpose it is convenient to reduce the problem to the canonical form. Let Vn be the vector space of the observation vector Y, Vk be the subspace of Vn generated by the column vectors x1, x2,…,xk of X, and be the subspace of Vk in which EY is postulated to lie under H0. We change variables from Y1,Y2,…,Yn to Z1,Z2,…,Zn, where Z1,Z2,…,Zn are independent normal RVs with common variance σ2 and means , . This is done as follows. Let us choose an orthonormal basis of column vectors {ai} for , say . We extend this to an orthonormal basis for Vk, and then extend once again to an orthonormal basis for Vn. This is always possible.
Let z1, z2,…,zn be the coordinates of y relative to the basis {α1, α2, …, αn}. Then and , where P is an orthogonal matrix with ith row . Thus , and . Since Xβ ∈ Vk (Remark 1), it follows that for . Similarly, under H0, , so that for . Let us write . Then , and under . Finally, from Corollary 2 of Theorem 6 it follows that Z1, Z2,…,Zn are independent normal RVs with the same variance σ2 and . We have thus transformed the problem to the following simpler canonical form:
Now
The quantity is minimized if we choose , so that
Under , so that will be minimized if we choose . Thus
It follows that
Now has a distribution, and, under has a χ2(r) distribution. Since and are independent, we see that is distributed as under H0, as asserted. This completes the proof of the theorem.
Remark 4. In practice, one does not need to find a transformation that reduces the problem to the canonical form. As will be done in the following sections, one simply computes the estimators and and then computes the test statistic in any of the equivalent forms (14), (15), or (16) to apply the F-test.
Remark 5. The computation of is greatly facilitated, in view of Remark 3, by using the principle of least squares. Indeed, this was done in the proof of Theorem 1 when we reduced the problem of maximum likelihood estimation to that of minimization of sum of squares .
Remark 6. The distribution of the test statistic under H1 is easily determined. We note that for , so that has a noncentral chi-square distribution with r d.f. and noncentrality parameter . It follows that has a noncentral F-distribution with d.f. and noncentrality parameter δ. Under , so that has a central distribution. Since , it follows from (19) and (20) that if we replace each observation Yi bu its expected value in the numerator of (16), we get σ2δ.
Remark 7. The general linear hypothesis makes use of the assumption of common variance. For instance, in Example 4, . Let us suppose that .. Then we need to test that before we can apply Theorem 1. The case has already been considered in Section 10.3. For the case where one can show that a UMP unbiased test does not exist. A large-sample approximation is described in Lehmann [64, pp. 376–377]. It is beyond the scope of this book to consider the effects of departures from the underlying assumptions. We refer the reader to Scheffé [101, Chapter 10], for a discussion of this topic.
Remark 8. The general linear model (GLM) is widely used in social sciences where Y is often referred to as the response (or dependent) variable and X as the explanatory (or independent) variable. In this language the GLM “predicts” a response variable from a linear combination of one or more explanatory variables. It should be noted that dependent and independent in this context do not have the same meaning as in Chapter 4. Moreover, dependence does not imply causality.
In this section we study regression analysis, which is a tool to investigate the interrelationship between two or more variables. Typically, in its simplest form a response random variable Y is hypothesized to be related to one or more explanatory nonrandom variables xi's. Regression analysis with a single explanatory RV is known as simple regression and if, in addition, the relationship is thought to be linear, it is called simple linear regression (Example 12.2.3). In the case where several explanatory variables xi's are involved the regression is referred to as multiple linear regression. Regression analysis is widely used in forecasting and prediction. Again this is a special case of GLM.
This section is divided into three subsections. The first subsection deals with multiple linear regression where the RV Y is of the continuous type. In the next two subsections we study the case when Y is either Bernoulli or a count variable.
It is convenient to write GLM in the form
where Y,X,ε, and β are as in Equation (12.2.1), and 1n is the column unit vector (1,1,…, 1). The parameter β0 is usually referred to as the intercept whereas β is known as the slope vector with k parameters. The least estimator (LSE) of β0 and β are easily obtained by minimizing.
resulting in normal equations
or
and
An unbiased estimate of σ2 is given by
Let us now consider the simple linear regression model
The LSEs of (β0, β)′ is given by
and
The covariance matrix is given by
where
Let us now verify these results using the maximum likelihood method.
Clearly, Y1,Y2,…,Yn are independent normal RVs with and , and Y is an n-variate normal random vector with mean Xβ and variance σ2In. The joint PDF of Y is given by
It easily follows that the MLE’s for β0, β1 and σ2 are given by
and
where .
If we wish to test , we take , so that the model is a special case of the general linear hypothesis with , . Under H0 the MLE’s are
and
Thus
From Theorem 12.2.1, the statistic has a central distribution under H0. Since is the square of a , the likelihood ratio test rejects H0 if
where c0 is computed from t-tables for d.f.
For testing , we choose so that the model is again a special case of the general linear hypothesis. In this case
and
It follows that
and since
we can write the numerator of F as
It follows from Theorem 12.2.1 that the statistic
has a central t-distribution with d.f. under . The rejection region is therefore given by
where c0 is determined from the tables of distribution for a given level of significance α.
For testing , we choose , so that the model is again a special case of the general linear hypothesis with . In this case
and
From Theorem 12.2.1, the statistic has a central distribution under . It follows that the α-level rejection region for H0 is given by
where F is given by (26), and c0 is the upper α percent point under the distribution.
Remark 1. It is quite easy to modify the analysis above to obtain tests of null hypotheses , and , where are given real numbers (Problem 4).
Remark 2.. The confidence intervals for β0, β1 are also easily obtained. One can show that -level confidence interval for β0 is given by
and that for β1 is given by
Similarly, one can obtain confidence sets for (β0, β1)′ from the likelihood ratio test of . It can be shown that the collection of sets of points (β0, β1)′ satisfying
is a -level collection of confidence sets (ellipsoids) for (β0, β1)′ centered at .
Remark 3. Sometimes interest lies in constructing a confidence interval on the unknown linear regression function for a given value of x, or on a value of Y, given . We assume that x0 is a value of x distinct from x1,x2,…,xn. Clearly, is the maximum likelihood estimator of . This is also the best linear unbiased estimator. Let us write . Then
which is clearly a linear function of normal RVs Yi. It follows that is also normally distributed with mean and variance
(See Problem 6.) It follows that
is (0,1). But σ is not known, so that we cannot use (32) to construct a confidence interval for . Since is a RV and is independent of (why?), it follows that
has a distribution. Thus, a -level confidence interval for is given by
In a similar manner one can show (Problem 7) that
is a -level confidence interval for , that is, for the estimated value Y0 of Y at x0.
Remark 4. The simple regression model (2) considered above can be generalized in many directions. Thus we may consider EY as a polynomial in x of a degree higher than 1, or we may regard EY as a function of several variables. Some of these generalizations will be taken up in the problems.
Remark 5. Let (X1, Y1), (X2, Y2),…,(Xn, Yn) be a sample from a bivariate normal population with parameters ,,,, and .
In Section 6.6 we computed the PDF of the sample correlation coefficient R and showed (Remark 6.6.4) that the statistic
has a distribution, provided that . If we wish to test , that is, the independence of two jointly normal RVs, we can base a test on the statistic T. Essentially, we are testing that the population covariance is 0, which implies that the population regression coefficients are 0. Thus we are testing, in particular, that . It is therefore not surprising that (36) is identical with (18). We emphasize that we derived (36) for a bivariate normal population, but (18) was derived by taking the X’s as fixed and the distribution of Y’s as normal. Note that for a bivariate normal population is linear, in consistency with our model (1) or (2).
Let us next find a 95 percent confidence interval for . This is given by (34). We have
so that the 95 percent confidence interval is .
(The data were produced from Table ST6, of random numbers with , by letting and so that , which surely lies in the interval.)
In the regression model considered above Y is a continuous type RV. However, in a wide variety of problems Y is either binary or a count variable. Thus in a medical study Y may be the presence or absence of a disease such as diabetes. How do we modify linear regression model to apply in this case? The idea here is to choose a function of E(Y) so that in Section 12.3.1
This can be accomplished by choosing the function f to be the logarithm of the odds ratio
where so that . It follows that
so that logistic regression models the logarithm of odds ratio as a linear function of RVs Xi. The term logistic regression derives from the fact that the function is known as the logistic function.
For simplicity we will only consider the simple linear regression model case so that
Choosing the logistic distribution as
let Y1, Y2,…,Yn be iid binary RVs taking values 0 or 1. Then the joint PMF of Y1, Y2,…,Yn is given by
and the log likelihood function by
It is easy to see that
Since the likelihood equations are nonlinear in the parameters, the MLEs of β0 and β are obtained numerically by using Newton-Raphson method.
Let and be the MLE of β0 and β, respectively. From section 8.7 we note that the variance of is given by
so that the standard error (SE) of is its square root. For large n, the so-called Wald statistic has an approximate N(0, 1) distribution under . Thus we reject H0 at level α if . One can use as a (1–α) -level confidence interval for β.
Yet another choice for testing H0 is to use the LRT statistics –2logλ (see Theorem 10.2.3). Under H0, has a chi-square distribution with 1 d.f. Here
In (40) we chose the DF of a logistic RV. We could have chosen some other DF such as ϕ(x), the DF of a N(0, 1) RV. In that case we have . The resulting model is called probit regression.
We finally consider the case when the RV Y is a count of rare events and has Poisson distribution with parameter λ. Clearly, the GLM is not directly applicable. Again we only consider the linear regression model case. Let , be independent P(λi)RVs where , so that
The log likelihood function is given by
In order to find the MLEs of β0 and β1 we need to solve the likelihood equations
which are nonlinear in β0 and β1. The most common method of obtaining the MLEs is to apply the iteratively weighted least squares algorithm.
Once the MLEs of β0 and β1 are computed, one can compute the SEs of the estimates by using methods of Section 8.7. Using the SE , for example, one can test hypothesis concerning β1 or construct -level confidence interval for β1.
For a detailed discussion of Geometric and Poisson regression we refer Agresti [1] . A wide variety of software is available, which can be used to carry out the computations required.
where β0, β1, β2 are unknown parameters, x1, x2,…,xn are known values of x, and ε1, ε2,…,εn are unobservable RVs that are assumed to be independently normally distributed with common mean 0 and common variance σ2 (see Example 12.2.3). Assume that the coefficient vectors , are linearly independent. Write the normal equations for estimating the β's and derive the generalized likelihood ratio test of .
where xi1, xi2, xi3 are three mathematical variables, and εi are iid (0,1) RVs. Assuming that the matrix X (see Example 3) is of full rank, write the normal equations and derive the likelihood ratio test of the null hypothesis .
In this section we return to the problem of one-way analysis of variance considered in Examples 12.2.1 and 12.2.4. Consider the model
as described in Example 12.2.4. In matrix notation we write
Where
As in Example 12.2.4, y is a vector of n-observations , whose components Yij are subject to random error is a vector of k unknown parameters, and x is a design matrix. We wish to find a test of against all alternatives. We may write H0 in the form , where h is a matrix of rank (k–1), which can be chosen to be
Let us write under H0. The joint PDF of Y is given by
and, under H0,by
It is easy to check that the MLEs are
and
By Theorem 12.2.1, the likelihood ratio test is to reject H0 if
where F0 is the upper α percent point in the distribution. Since
we may rewrite (9) as
It is usual to call the sum of squares in the numerator of (11) the between sum of squares (BSS) and the sum of squares in the denominator of (11) the within sum of squares (WSS). The results are conveniently displayed in a so-called analysis of variance table in the following form.
One-Way Analysis of Variance | |||||
Source Variation |
Sum of Squares | Degrees of Freedom | Mean Sum of Squares | F-Ratio | |
Between | BSS/(k–1) | ||||
Within | |||||
Mean | 1 | ||||
Total | n |
The third row, designated “Mean,” has been included to make the total of the second column add up to the total sum of squares (TSS),
From the data , and
Also, the grand mean is
Thus
Analysis of Variance | ||||
Source | SS | d.f. | MSS | F-Ratio |
Between | 1140 | 2 | 570 | |
Within | 1600 | 12 | 133.33 |
Choosing , we see that . Thus we reject at level .
From the data , , . Also, the grand mean is
Thus
Analysis of Variance | ||||
Source | SS | d.f. | MSS | F-Ratio |
Between | 303.41 | 2 | 151.70 | 151.70/489.67 |
Within | 11,752.00 | 24 | 489.67 |
We therefore cannot reject the null hypothesis that the average grades given by the three instructors are the same.
Test whether there is a significant difference between the yields of the varieties.
Distance Traveled (Miles) per Gallon of Gasoline | ||||||
Brand | ||||||
Car | D | F | G | P | V | T |
1 | 42 | 38 | 28 | 32 | 30 | 25 |
2 | 35 | 33 | 32 | 36 | 35 | 32 |
3 | 37 | 28 | 35 | 27 | 25 | 24 |
4 | 37 | 37 | 26 | 30 | ||
5 | 28 | 30 | ||||
6 | 19 |
Should the consumer accept his friend's advice?
University | ||
A | B | C |
17 | 16 | 21 |
19 | 16 | 23 |
20 | 19 | 22 |
21 | 20 | |
18 | 19 |
Test the hypothesis that the average ages of entering freshman at these universities are the same.
Brand | Tar Lavel (mg) |
A | 4.2, 4.8, 4.6, 4.0, 4.4 |
B | 4.9, 4.8, 4.7, 5.0, 4.9, 5.2 |
C | 5.4, 5.3, 5.4, 5.2, 5.5 |
D | 5.8, 5.6, 5.5, 5.4, 5.6, 5.8 |
E | 5.9, 6.2, 6.2, 6.8, 6.4, 6.3 |
Can the differences among the sample means be attributed to chance?
Location | Quantity of Dissolved Oxygen (%) |
A | 7.8, 6.4, 8.2, 6.9 |
B | 6.7, 6.8, 7.1, 6.9, 7.3 |
C | 7.2, 7.4, 6.9, 6.4, 6.5 |
D | 6.0, 7.4, 6.5, 6.9, 7.2, 6.8 |
Do the data indicate a significant difference in the average amount of dissolved oxygen for the four locations?
In many practical problems one is interested in investigating the effects of two factors that influence an outcome. For example, the variety of grain and the type of fertilizer used both affect the yield of a plot or the score on a standard examination is influenced by the size of the class and the instructor.
Let us suppose that two factors affect the outcome of an experiment. Suppose also that one observation is available at each of a number of levels of these two factors. Let be the observation when the first factor is at the ith level, and the second factor at the jth level. Assume that
where αi is the effect of the ith level of the first factor, βj is the effect of the jth level of the second factor, and εij is the random error, which is assumed to be normally distributed with mean 0 and variance σ2. We will assume that the εij’s are independent. It follows that Yij are independent normal RVs with means and variance σ2. There is no loss of generality in assuming that for, if , we can write
and Here we have written and for the means of ’s and ’s, respectively. Thus Yij may denote the yield from the use of the ith variety of some grain and the jth type of some fertilizer. The two hypotheses of interest are
The first of these, for example, says that the first factor has no effect on the outcome of the experiment.
In view of the fact that and and we can write our model in matrix notation as
where
and
The vector of unknown parameters β is and the matrix X is (b blocks of a rows each). We leave the reader to check that X is of full rank, . The hypothesis or can easily be put into the form . For example, for Hβ we can choose H to be the matrix of full rank , given by
Clearly, the model described above is a special case of the general linear hypothesis, and we can use Theorem 12.2.1 to test Hβ.
To apply Theorem 12.2.1 we need the estimators and . It is easily checked that
and
where Also, under Hβ, for example,
In the notation of Theorem 12.2.1, , so that , and
Since
we may write
It follows that, under has a central distribution.
The numerator of F in (8) measures the variability between the means and the denominator measures the variability that exists once the effects due to the two factors have been subtracted.
If Hα is the null hypothesis to be tested, one can show that under Hα the MLEs are
As before, , but . Also,
which may be rewritten as
It follows that, under F has a central distribution. The numerator of F in (11) measures the variability between the means .
If the data are put into the following form:
so that the rows represent various levels of factor 1, and the columns, the levels of factor 2, one can write
Similarly,
It is usual to write error or residual sum of squares (SSE) for the denominator of (8) or (11). These results are conveniently presented in an analysis of variance table as follows.
Two-Way Analysis of Variance Table with One Observation per Cell | ||||
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F-Ratio |
Rows | SS1 | MS1/MSE | ||
Columns | SS2 | MS2/MSE | ||
Error | SSE | |||
Mean | 1 | |||
Total | ab |
In our notation, , , , , , , , , , .
Also,
The results are shown in the following table:
Analysis of Variance | ||||
Source | SS | d.f. | MS | F-Ratio |
Variety of wheat | 34.67 | 2 | 17.33 | 14.2 |
Fertilizer | 4.67 | 3 | 1.56 | 1.28 |
Error | 7.33 | 6 | 1.22 | |
Mean | 481.33 | 1 | 481.33 | |
Total | 528.00 | 12 | 44.00 |
Now and . Since , we reject Hβ, that there is equality in the average yield of the three varieties; but, since , we accept Hα, that the four fertilizers are equally effective.
Machinist | ||||
Machine | A1 | A2 | A3 | A4 |
B1 | 15 | 14 | 19 | 18 |
B2 | 17 | 12 | 20 | 16 |
B3 | 16 | 18 | 16 | 17 |
B4 | 16 | 16 | 15 | 15 |
Test whether the differences in the performances of the machinists are significant and also whether the differences in the performances of the four brands of machines are significant. Use .
Teaching Method | |||
Ability Group | A | B | C |
1 | 15 | 19 | 14 |
2 | 18 | 17 | 12 |
3 | 22 | 25 | 17 |
4 | 17 | 21 | 19 |
Test the hypothesis that the teaching methods yield the same results. That is, that the teaching methods are equally effective.
Variety of Wheat | ||||
Fertilizer | A | B | C | D |
α | 8 | 3 | 6 | 7 |
β | 10 | 4 | 5 | 8 |
γ | 8 | 4 | 6 | 7 |
Test the hypotheses that the four varieties of wheat yield the same average yield and that the three fertilizers are equally effective.
The model described in Section 12.5 assumes that the two factors act independently, that is, are additive. In practice this is an assumption that needs testing. In this section we allow for the possibility that the two factors might jointly affect the outcome, that is, there might be so-called interactions. More precisely, if Yij is the observation in the (i,j)th cell, we will consider the model
where represent row effects (or effects due to factor 1), represent column effects (or effects due to factor 2), and γij represent interactions or joint effects. We will assume that εij are independently (0,σ2). We will further assume that
The hypothesis of interest is
One may also be interested in testing that all α’s are 0 or that all β’s are 0 in the presence of interactions γij.
We first note that (2) is not restrictive since we can write
where , and do not satisfy (2), as
and then (2) is satisfied by choosing
Here
Next note that, unless we replicate, that is, take more than one observation per cell, there are no degrees of freedom left to estimate the error SS (see Remark 1).
Let Yijs be the sth observation when the first factor is at the ith level, and the second factor at the jth level, . Then the model becomes as follows:
Levels of Factor 1 | ||||
Levels of Factor 2 | 1 | 2 | … | b |
1 | y111 | y121 | … | y1b1 |
∙ | ∙ | … | ∙ | |
∙ | ∙ | … | ∙ | |
∙ | ∙ | … | ∙ | |
y11m | y12m | … | y1bm | |
2 | y211 | y221 | … | y2b1 |
∙ | ∙ | … | ∙ | |
∙ | ∙ | … | ∙ | |
∙ | ∙ | … | ∙ | |
y21m | y22m | … | y2bm | |
∙ | ∙ | … | ∙ | |
∙ | ∙ | … | ∙ | |
∙ | ∙ | … | ∙ | |
a | ya11 | ya21 | … | yab1 |
∙ | ∙ | … | ∙ | |
∙ | ∙ | … | ∙ | |
∙ | ∙ | … | ∙ | |
ya1m | ya2m | … | yabm |
, and , where εijs’s are independent (0,σ2). We assume that . Suppose that we wish to test . We leave the reader to check that model (4) is then a special case of the general linear hypothesis with , and .
Let us write
Then it can be easily checked that
It follows from Theorem 12.2.1 that
Since
we can write (7) as
Under Hα the statistic has the central distribution, so that the likelihood ratio test rejects Hα if
A similar analysis holds for testing .
Next consider the test of hypothesis for all i, j, that is, that the two factors are independent and the effects are additive. In this case , and . It can be shown that
Thus
Now
so that we may write
Under Hγ, the statistic has the distribution. The likelihood ratio test rejects Hγ if
Let us write
and
Then we may summarize the above results in the following table.
Two-Way Analysis of Variance Table with Interaction | ||||
Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F-Ratio |
Rows | SS1 | MS1/MSE | ||
Columns | SS2 | MS2/MSE | ||
Interaction | SSI | MSI/MSE | ||
Error | SSE | |||
Mean | 1 | |||
Total | abm |
Remark 1. Note that, if , there are no d.f. associated with the SSE. Indeed, SSE if Hence, we cannot make tests of hypotheses when , and for this reason we assume .
Instructor | |||
Teaching Method | I | II | III |
1 | 95 | 60 | 86 |
85 | 90 | 77 | |
74 | 80 | 75 | |
74 | 70 | 70 | |
2 | 90 | 89 | 83 |
80 | 90 | 70 | |
92 | 91 | 75 | |
82 | 86 | 72 | |
3 | 70 | 68 | 74 |
80 | 73 | 86 | |
85 | 78 | 91 | |
85 | 93 | 89 |
From the data the table of means is as follows:
82 | 75 | 77 | 78.0 | |
86 | 89 | 75 | 83.3 | |
80 | 78 | 85 | 81.0 | |
82.7 | 80.7 | 79.0 |
Then
Analysis of Variance | ||||
Source | SS | d.f. | MSS | F-Ratio |
Methods | 169.56 | 2 | 84.78 | 1.25 |
Instructors | 82.32 | 2 | 41.16 | 0.61 |
Interactions | 561.80 | 4 | 140.45 | 2.07 |
Error | 1830.00 | 27 | 67.78 |
With , we see from the tables that and , so that we cannot reject any of the three hypotheses that the three methods are equally effective, that the three instructors are equally effective, and that the interactions are all 0.
Machinist | |||
Machine | A | B | C |
B1 | 15, 15, 17 | 19, 19, 16 | 16, 18, 21 |
B2 | 17, 17, 17 | 15, 15, 15 | 19, 22, 22 |
B3 | 15, 17, 16 | 18, 17, 16 | 18, 18, 18 |
B4 | 18, 20, 22 | 15, 16, 17 | 17, 17, 17 |
Using a 0.05 level of significance, test whether (a) the differences among the machinists are significant, (b) the differences among the machines are significant, and (c) the interactions are significant.
Automobile Make | ||||
Cities | A | B | C | D |
Cleveland | 92.3, 104.1 | 90.4, 103.8 | 110.2, 115.0 | 120.0, 125.4 |
Detroit | 96.2, 98.6 | 91.8, 100.4 | 112.3, 111.7 | 124.1, 121.1 |
San Francisco | 90.8, 96.2 | 90.3, 89.1 | 107.2, 103.8 | 118.4, 115.6 |
Denver | 98.5, 97.3 | 96.8, 98.8 | 115.2, 110.2 | 126.2, 120.4 |
Construct the analysis of variance table. Test the hypothesis of no automobile effect, no city effect, and no interactions. Use .