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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

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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

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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

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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 ) =

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

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

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

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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:

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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

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Significance Level ( The Maximum Type 1 Error = Significance Level

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

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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.

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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.

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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)

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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

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-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.

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

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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.

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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)

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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).

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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

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

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

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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.

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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

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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.

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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.

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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

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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.

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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.

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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)

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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

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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

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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.

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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

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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:

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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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