 # Multivariate Regression Model y =    x1 +  x2 +  x3 +… +  The OLS estimates b 0,b 1,b 2, b 3.. …. are sample statistics used to estimate 

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Multivariate Regression Model y =    x1 +  x2 +  x3 +… +  The OLS estimates b 0,b 1,b 2, b 3.. …. are sample statistics used to estimate          respectively y is the DEPENDENT variable Each of the x j is an INDEPENDENT variable

Conditions: Each explanatory variable Xj is assumed (1A) to be deterministic or non-random (1B) : to come from a ‘fixed’ population (1C) : to have a variance V(xj) which is not ‘too large’ The above assumptions are best suited to a situation of a controlled experiment

Assumptions concerning the random term    (IIA) E(  i ) = 0 for all i (IIB) Var(  i ) =    constant for all i (IIC) Covariance (    k ) =  for any i and k (IID) Each of the  i has a normal distribution

Properties of b 0, b 1, b 2, b 3 1. Each of these statistics is a linear functions of the Y values. 2. Therefore, they all have normal distributions 3. Each is an unbiased estimator. That is, E(b k ) =   

4. Each b k is the most efficient estimator of all unbiased estimators.

Best Linear Unbiased Estimator of the respective parameter Thus, each of b 0, b 1, b 2 ….is

Conclusion Each estimator b i has a normal distribution with mean =    and variance =  bi 2 where  bi 2 is unknown.

Income (£ per week) of an individual is regressed on a constant, education (in years), age (in years) and wealth inheritance (in £), using EViews. Number of observations is 20 and the regression output is given below:

Variable Coefficient Std.Error t-Stats Prob. C -1001.87 520.71 -1.92 0.0654 AGE 8.85 5.45 1.62 0.1168 EDUCATION 95.17 38.54 2.46 0.0252 WEALTH 1.51 0.46 3.26 0.0031

Significance Level (  The Maximum Type 1 Error = Significance Level

The smaller the p-value the more significant is the test p-value

The proposed regression model is: Income = ß 0 + ß 1 (Age) +ß 2 (Education) + ß 3 (Wealth Inheritance) ….. (A) We are proposing that Income is the variable dependent on three independent variables: Age, Education and Wealth.

 0 is a constant. It measures the effect of other deterministic factors on Income not included in the model.  1,  2,  3 measure the effect of a marginal change in Age, Education and Wealth, respectively.

However, we recognise that there may be other random factors affecting the dependent variable Income. So we add a random variable  to the model which now becomes: Income = ß 0 + ß 1 (Age) +ß 2 (Education) + ß 3 (Wealth Inherited) +  ….. (B)

We use the least squares technique to estimate the model B. Therefore, our estimation of the proposed model B is Y e = -1001.87 + 8.85*AGE + 95.17*EDUCATION + 1.51*WEALTH INHERITANCE Here Y e is the estimated value of income

-1001.87 is the estimate of ß 0, 8.85 is the estimate of ß 1,; 95.17 is the estimate of ß 2 and 1.51 is the estimate of ß 3 The least-squares estimates of the ß- values are denoted by b-values. Thus, b 1 is the estimate of ß 1 and b 2 is the estimate of ß 2. In our case, b 1 = 8.85 and b 2 = 95.17.

We next make the following assumptions on the specification of model B so that the least-squares method produces ‘good’ estimators.

i.  is normally distributed with mean 0 and an unknown variance  2 . In the context of the model B,  can be thought of as a luck factor which can be good (positive values) or bad (negative values), If the positive and negative values cancel out on average, we can say that mean value is 0.

The  values are uncorrelated across the population (Whether or not you are lucky does not influence my being lucky/unlucky) i. The  values have the same variance (  2  ) across it. (Every individual is exposed to the same extent/chance of good or bad luck)

The  values are uncorrelated with the independent variables Age, Education and Wealth Inheritance. (For example, an old person is as likely to be lucky as a young one; or a university graduate is as likely to be unlucky as someone with no A-levels).

We now test (at 10% significance) the following hypothesis: Education has a positive effect on income Step 1: Set up the hypotheses H 0 : ß 2 = 0 (Education has no effect) H 1 : ß 2 > 0(Education has a positive effect) one-tailed test

Step 2: Select statistic The estimator b 2 is the test-statistic Step3 : Identify the distribution of b 2

Best Linear in the dependent variable income Unbiased Estimator of  2 Assumptions i-iii above imply that b2 is

Since b 2 is unbiased, E(b 2 ) =  2 b 2 has a normal distribution because it is linear in Income Thus, b 2 ~ N(  2,  2 2 ) where    is unknown.

Therefore, the test statistic is t  (b 2 -  2 ) / (standard error of b 2 ) has a Student’s t-distribution with 20-4 = 16 d.o.f. Step 4: Construct test statistic We use the standard error of b 2 because we do not know what    is

EViews therefore gives us a t-statistic regarding education of 2.46907 As  2 = 0 under the null hypothesis (H0) t = b2 / (standard error of b 2 ) The corresponding probability value is 0.0252.

Select f x /TDIST. For X, enter 2.469607, the t-Statistic value. The degree of freedom is 16. EViews calculates two-tail probability So number of tails is 2. You now get the 2-tail probability of 0.025165 from Excel. Since we are performing a one-tail test, take half the probability value, or 0.0126.

Step 5: Compare with critical value t C t C = 1.336757 for a one-tailed test with significance level (  ) = 0.1 and d.o.f. = 16 t C = 1.336757 < 2.469607

Step 6 : Draw conclusion The test is significant. Reject H0 at 10% and at 5% ( 1.745884 < 2.469607) but not at 1% ( 2.583492 > 2.469607) Step 7: Interpret result The data supports (with at least 98% accuracy) the hypothesis that EDUCATION is an important explanatory variable affecting income.

The probability of a type 1 error is nothing but the area to the right of t- statistic, or 0.0126. In rejecting H 0, we are prone to make a Type 1 Error.

Example 2: Use output 2 to test the hypothesis (at 5% significance) that weightgain is proportional to foodvalue. H 0 : a = 0 (proportionality) H 1 : a  0 (non-proportionality) The estimator a is the test-statistic Step 1: Step 2: The Model :: y =  x +  and add the assumptions (Lec17)

Conditions: The explanatory variable X is assumed (1A) to be deterministic or non-random (1B) : to come from a ‘fixed’ population (1C) : to have a variance V(x) which is not ‘too large’ The above assumptions are best suited to a situation of a controlled experiment

Assumptions concerning the random term    (IIA) E(  i ) = 0 for all i (IIB) Var(  i ) =    constant for all i (IIC) Covariance (    j ) =  for any i and j (IID) Each of the  i has a normal distribution

Step 3: Thus, a~ N( ,   ) where   is unknown. Step 4: Therefore, the test statistic t  (a-  ) / (standard error of a) has a Student’s t-distribution with 10-2 = 8 d.o.f.

The p-value is 0.0169 < 0.05 Foodvalue is not the only variable that affects weightgain Step 6: Draw conclusion Step 5: Compare with critical value t C t C = - 2.31 > -3.005262 t C = - 2.31 for a two-tailed test with significance level (  ) = 0.05 and d.o.f.= 8 The test is significant. Reject H0 at 5% Step 7: Interpret

Use output 3 to test (at 5% significance) the following hypothesis: Exercise has a negative effect on weight gain The proposed regression model is: Weightgain = ß 0 + ß 1 (Foodvalue) +ß 2 (Exercise) +  Example 3:

Step 1: Set up the hypotheses H 0 : ß 2 = 0 (Exercise has no effect) H 1 : ß 2 < 0(Exercise has a negative effect)

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