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Solution Manual for Econometrics by Example, 2nd Edition

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    SOLUTIONS MANUAL by Inas Kelly

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    CHAPTER 1 EXERCISES 1.1. Consider the regression results given in Table 1.2. a . Suppose you want to test the hypothesis that the true or population regression coefficient of the education variable is 1. How would you test this hypothesis? Show the necessary calculations. The equation we are looking at is: wage i = b 1 + b 2 *(female i ) + b 3 *(nonwhite i ) + b 4 *(union i ) + b 5 *(education i ) + b 6 *(exper i ) + e i Here we are testing: H 0 : β 5 = 1 H 1 : β 5 ≠ 1 From Table 1.2, we have: t = (1.370301 - 1)/0.065904 = 5.618794. From the t table, the critical t statistic for α = 1% is 2.576 (df = 1289 – 6 = 1283, so we can use df = ∞ ). Since 5.619 > 2.576, we can easily reject the null hypothesis at the 1% level. b . Would you reject or not reject the hypothesis that the true union regression coefficient is 1? Here we are testing: H 0 : β 4 = 1 H 1 : β 4 ≠ 1 From Table 1.2, we have: t = (1.095976 - 1)/0.506078 = 0.189647. From the t table, the critical t statistic for α = 10% is 1.645 (using df = ∞). Since 0.190 < 1.645, we cannot even reject the null hypothesis at the 10% level. (Note that from the output, if we were testing H 0 : β 4 = 0 vs. H 1 : β 4 ≠ 0, we could reject the null hypothesis at the 5% level.) c . Can you take the logs of the nominal variables, such as gender, race and union status? Why or why not? No, because these are categorical variables that often take values of 0 or 1. The natural log of 1 is 0, and the natural log of 0 is undefined. Moreover, taking the natural log would not be helpful as the values of the nominal variables to not have a specific meaning. d . What other variables are missing from the model? We could have included control variables for region, marital status, and number of children on the right-hand side. Instead of including a continuous variable for education, we could have controlled for degrees (high school graduate, college graduate, etc). An indicator for the business cycle (such as the unemployment rate) may be helpful. Moreover, we could include state-level policies on the minimum wage and right-to-work laws. e . Would you run separate wage regressions for white and nonwhite workers, male and female workers, and union and non-union workers? And how would you compare them? We would if we felt the two groups were systematically different from one another. We can run the models separately and conduct an F test to see if the two regressions are significantly different. If they are, we should run them separately. The F statistic may be obtained by running the two together – the restricted model – then running the two separately – jointly, the unrestricted model.

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    We then obtain the residual sum of squares for the restricted model (RSS R ) and the residual sum of squares for the unrestricted model (RSS UR , equal to RSS 1 + RSS 2 from two separate models). F = [(RSS R – RSS UR )/k] / [RSS UR /(n-2k)] ~ F k,n-2k . I would then see which model was a better predictor of the outcome variable, wage . f. Some states have right-to-work laws (i.e., union membership is not mandatory) and some do not have such laws (i.e, union membership is permitted). Is it worth adding a dummy variable taking the value of 1 if the right-to-work laws are present and 0 otherwise? A priori, what would you expect if this variable is added to the model? Since we would expect these laws to have an effect on wage, it may be worth adding this variable. A priori, we would expect this variable to have a negative effect on wage, as union wages a re generally higher than nonunion wages. h . Would you add the age of the worker as an explanatory variable to the model? Why or why not? No, we would not add this variable to the model. This is because the variable Exper is defined as (age – education – 6), so it would be perfectly collinear and not add any new information to the model. 1.2. Table 1.5 (available on the companion website) gives data on 654 youths, aged 3 to 19, in the areas of East Boston in the later 1970 ’ s on the following variables: fev = continuous measure (in liters) smoke = smoker coded as 1, non-smoker coded as 0 age = in years ht = height in inches s ex = coded 1 for male and 0 for female fev stands for forced expiratory volume , the volume of air that can be forced out taking a deep breath, an important measure of pulmonary function. The objective of this exercise is to find out the impact of age, height, sex and smoking habits on fev . a. Develop a suitable regression model for this purpose. Fevi = b1 + b2age + b3ht + b4sex + b5smoke + ei Where i denotes the youth. An alternative functional form may be used as well, in which quadratic terms are included for age and height. b. A priori , what is the effect of each regressor on fev ? Do the regression results support your prior expectations? Age : Negative. One would expect that as age increases, pulmonary function decreases. However, since we are analyzing a group of 3 to 19 year olds, this will likely be positive. The result came out positive . Height : Positive. Pulmonary function biologically may be more effective for taller individuals. The result came out positive .

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    Sex : Ambiguous. No clear expectation for differences in pulmonary function between males and females, although males may have stronger lungs, and thus, the coefficient may be positive. The result came out positive . Smoke : Negative. Smoking adversely affects pulmonary function. The result came out negative . Results in Stata are: . reg fev age ht sex smoke Source | SS df MS Number of obs = 654 -------------+------------------------------ F( 4, 649) = 560.02 Model | 380.64028 4 95.1600701 Prob > F = 0.0000 Residual | 110.279553 649 .16992227 R-squared = 0.7754 -------------+------------------------------ Adj R-squared = 0.7740 Total | 490.919833 653 .751791475 Root MSE = .41222 ------------------------------------------------------------------------------ fev | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- age | .0655093 .0094886 6.90 0.000 .0468774 .0841413 ht | .1041994 .0047577 21.90 0.000 .0948571 .1135418 sex | .1571029 .0332071 4.73 0.000 .0918967 .2223092 smoke | -.0872464 .0592535 -1.47 0.141 -.2035981 .0291054 _cons | -4.456974 .2228392 -20.00 0.000 -4.894547 -4.019401 ------------------------------------------------------------------------------ c . Which of the explanatory variables, or regressors, are individually statistically significant, say, at the 5% level? What are the estimated p values? Age, height, and sex are all statistically significant at the 5% level, which p-values of zero. d . If the estimated p values are greater than the 5% value, does that mean the relevant regressor is not of practical importance? No. In fact, the p-value for smoke is 0.141, suggesting that this explanatory variable is insignificant. However, we would expect smoking to have an effect on pulmonary function; thus, smoke theoretically belongs in the equation and should not be excluded. Excluding a relevant variable because it is not significant may also bias other coefficients in the model. e . Would you expect age and height to be correlated? If so, would you expect that your model suffers from multicollinearity? Do you have any idea what you could do about this problem? Show the necessary calculations. If you do not have the answer, do not be discouraged because we will discuss multicollinearity in some depth in Ch.4. Yes, I would expect age and height to be strongly correlated, especially for youths aged 3 to 19. This is because they are still growing, and the older they are, the taller they are. In fact, we find that the correlation coefficient in this sample is 0.7919. However, one of the suggested indicators of multicollinearity is individual insignificance but joint significance. This is not a problem here, since both age and height are separately very significant. More detailed tests, such as looking at the variance inflation factor (VIF), will be introduced later. f . Would you reject the hypothesis that the (slope) coefficients of all the regressors are statistically insignificant? Which test do you use? Show the necessary calculations. Yes, I would reject this hypothesis. The appropriate test is an F test, and the null and alternative hypotheses are:

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    H 0 : R 2 = 0 H 1 : R 2 ≠ 0 The Stata output reveals that the actual F value, with 4 df in the numerator and 649 df in the denominator, is 560.02. The probability associated with this value is 0, suggesting that we can reject the null hypothesis at all significance levels. g . Set up the analysis of variance (AOV) table. What does this table tell you? This is given in Stata: Source | SS df MS Number of obs = 654 -------------+------------------------------ F( 4, 649) = 560.02 Model | 380.64028 4 95.1600701 Prob > F = 0.0000 Residual | 110.279553 649 .16992227 R-squared = 0.7754 -------------+------------------------------ Adj R-squared = 0.7740 Total | 490.919833 653 .751791475 Root MSE = .41222 Since the formula for the F test is F = [ (ESS/df) / (RSS/df) ], where ESS is the explained sum of squares, RSS is the residual sum of squares, and df are degrees of freedom, the information above tells us that we can compute the F statistic as follows: F = (380.64028/4) / (110.279553/649) = 95.1600701 / .16992227 = 560.02. These values are all provided in the ANOVA table provided by Stata, and can give us information about the joint significance of the explanatory variables. h . What is the R 2 value of your regression model? How would interpret this value? As seen in the output above, the R 2 value is 0.7754. This can be computed by taking the explained sum of squares (ESS) divided by the total sum of squares (TSS). This value tells us that 77.54% of the variation in fev can be explained by the variations in the explanatory variables: age, height, sex, and smoke. i . Compute the adjusted- R 2 value? How does this value compare with the computed R 2 value? The adjusted R 2 value is computed using the following formula: Adjusted R 2 = 1 – (1 – R 2 )*((n-1)/(n-k)) = 1 – (1-0.7754)*(653/649) = 0.7740. This takes degrees of freedom into account and is slightly lower than the value of R 2 . j . Would you conclude from this example that smoking is bad for fev? Explain. There is not sufficient empirical evidence in this example to show that smoking is bad for fev. Although the relationship between the two variables is negative, it is insignificant. This could be due to the age range being analyzed; the smokers in the sample likely have not been smoking for long, and the effects on pulmonary function have not yet been realized. 1.3. Consider the bivariate regression model: 1 2 i i i Y B B X u   

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    Verify that the OLS estimators for this model are as follows: 2 2 i i i x y b x    1 2 b Y b X     2 2 2 i e n      where 1 2 ( ), ( ) , ( ) i i i i i i i x X X y Y Y e Y b b X         

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    The sample variance is of the estimate, sigma-hat squared, is simply equal to the residual sum of squares (RSS) divided by degrees of freedom, equal to n-k. Since we have only two parameters in this bivariate regression model, k=2. 1.4. Consider the following regression model: 1 2 i i i y B B x u   

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    where x i and y i are as defined in Exercise 1.3. Show that in this model b 1 = 0. What is the advantage of this model over the model in Exercise 1.3? Since this model takes deviations from the mean for all variables, the calculations are simpler. The slope remains the same, while the y-intercept is simply zero (the origin). Note that, from Exercise 1.3, we can see that the y-intercept is equal to 1 2 b Y b X     . Since we are taking deviations from the mean, the mean of y is now zero. Similarly, the mean of x is zero. Substituting, we can see that this means that b1 is equal to zero. 1.5. Interaction among regressors. Consider the wage regression model given in Table 1.3. Suppose you decide to add the variable education.experience, the product of the two regressors, to the model. What is the logic behind introducing such a variable, called an interaction variable , to the model? Reestimate the model in Table 1.3 with this added variable and interpret your results. The logic behind introducing such a variable is to account for the possibility that education ’ s effect on wages relies in part on experience. In other words, the coefficient on education is incomplete on its own; likewise, the partial slope on experience is incomplete. In this example, we may believe that there is something about both having more experience and a higher education that increases wages. When we run the regression in Stata, it gives us the following results: . reg wage female nonwhite union education exper education_exper Source | SS df MS Number of obs = 1289 -------------+------------------------------ F( 6, 1282) = 102.44 Model | 26026.2103 6 4337.70172 Prob > F = 0.0000 Residual | 54283.6144 1282 42.3429129 R-squared = 0.3241 -------------+------------------------------ Adj R-squared = 0.3209 Total | 80309.8247 1288 62.3523484 Root MSE = 6.5071 ------------------------------------------------------------------------------ wage | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- female | -3.089394 .3647682 -8.47 0.000 -3.805002 -2.373786 nonwhite | -1.55922 .509136 -3.06 0.002 -2.558051 -.5603885 union | 1.090656 .5060209 2.16 0.031 .0979362 2.083376 education | 1.501845 .1295197 11.60 0.000 1.247751 1.755939 exper | .2437558 .0673361 3.62 0.000 .1116547 .3758569 education_~r | - .0061015 .005172 - 1.18 0.238 - .0162481 .004045 _cons | -8.883978 1.763414 -5.04 0.000 -12.34347 -5.424483 ------------------------------------------------------------------------------ Interestingly, the coefficient on the interaction term (education.experience) is negative and insignificant.

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    CHAPTER 2 EXERCISES 2.1. Consider the following production function, known in the literature as the transcendental production function (TPF). 3 4 5 2 1 i i B B L B K B i i i Q B L K e   where Q , L and K represent output, labor and capital, respectively. ( a ) How would you linearize this function? (Hint: logarithms.) Taking the natural log of both sides, the transcendental production function above can be written in linear form as: i i i i i i u K B L B K B L B B Q       5 4 3 2 1 ln ln ln ln ( b ) What is the interpretation of the various coefficients in the TPF? The coefficients may be interpreted as follows: ln B 1 is the y-intercept, which may not have any viable economic interpretation, although B 1 may be interpreted as a technology constant in the Cobb-Douglas production function. The elasticity of output with respect to labor may be interpreted as (B 2 + B 4 *L). This is because L B B L B B L Q i i 4 2 4 2 1 ln ln       . Recall that   i i i i L L Q L Q      1 ln ln ln . Similarly, the elasticity of output with respect to capital can be expressed as (B 3 + B 5 *K). ( c ) Given the data in Table 2.1, estimate the parameters of the TPF. The parameters of the transcendental production function are given in the following Stata output: . reg lnoutput lnlabor lncapital labor capital Source | SS df MS Number of obs = 51 -------------+------------------------------ F( 4, 46) = 312.65 Model | 91.95773 4 22.9894325 Prob > F = 0.0000 Residual | 3.38240102 46 .073530457 R-squared = 0.9645 -------------+------------------------------ Adj R-squared = 0.9614 Total | 95.340131 50 1.90680262 Root MSE = .27116 ------------------------------------------------------------------------------ lnoutput | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- lnlabor | .5208141 .1347469 3.87 0.000 .2495826 .7920456 lncapital | .4717828 .1231899 3.83 0.000 .2238144 .7197511 labor | -2.52e-07 4.20e-07 -0.60 0.552 -1.10e-06 5.94e-07 capital | 3.55e - 08 5.30e - 08 0.67 0.506 - 7.11e - 08 1.42e - 07 _cons | 3.949841 .5660371 6.98 0.000 2.810468 5.089215 ------------------------------------------------------------------------------ B 1 = e 3.949841 = 51.9271. B 2 = 0.5208141 B 3 = 0.4717828 B 4 = -2.52e-07

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    B 5 = 3.55e-08 Evaluated at the mean value of labor (373,914.5), the elasticity of output with respect to labor is 0.4266. Evaluated at the mean value of capital (2,516,181), the elasticity of output with respect to capital is 0.5612. ( d ) Suppose you want to test the hypothesis that B 4 = B 5 = 0. How would you test these hypotheses? Show the necessary calculations. (Hint: restricted least squares.) I would conduct an F test for the coefficients on labor and capital. The output in Stata for this test gives the following: . test labor capital ( 1) labor = 0 ( 2) capital = 0 F( 2, 46) = 0.23 Prob > F = 0.7992 This shows that the null hypothesis of B 4 = B 5 = 0 cannot be rejected in favor of the alternative hypothesis of B 4 ≠ B 5 ≠ 0. We may thus question the choice of using a transcendental production function over a standard Cobb-Douglas production function. We can also use restricted least squares and perform this calculation “by hand” by conducting an F test as follows: 46 , 2 ~ ) /( ) 2 /( ) ( F k n RSS k n k n RSS RSS F UR UR R        The restricted regression is: i i i i u K B L B B Q     ln ln ln ln 3 2 1 , which gives the following Stata output: . reg lnoutput lnlabor lncapital; Source | SS df MS Number of obs = 51 -------------+------------------------------ F( 2, 48) = 645.93 Model | 91.9246133 2 45.9623067 Prob > F = 0.0000 Residual | 3.41551772 48 .071156619 R-squared = 0.9642 -------------+------------------------------ Adj R-squared = 0.9627 Total | 95.340131 50 1.90680262 Root MSE = .26675 ------------------------------------------------------------------------------ lnoutput | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- lnlabor | .4683318 .0989259 4.73 0.000 .269428 .6672357 lncapital | .5212795 .096887 5.38 0.000 .326475 .7160839 _cons | 3.887599 .3962281 9.81 0.000 3.090929 4.684269 ------------------------------------------------------------------------------ The unrestricted regression is the original one shown in 2(c). This gives the following: 46 , 2 ~ .22519 0 ) 5 51 /( 3.382401 ) 5 51 2 5 51 /( ) 3.382401 3.4155177 ( F F        

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    Since 0.225 is less than the critical F value of 3.23 for 2 degrees of freedom in the numerator and 40 degrees in the denominator (rounded using statistical tables), we cannot reject the null hypothesis of B 4 = B 5 = 0 at the 5% level. ( e ) How would you compute the output-labor and output capital elasticities for this model? Are they constant or variable? See answers to 2(b) and 2(c) above. Since the values of L and K are used in computing the elasticities, they are variable . 2.2. How would you compute the output-labor and output-capital elasticities for the linear production function given in Table 2.3? The Stata output for the linear production function given in Table 2.3 is: . reg output labor capital Source | SS df MS Number of obs = 51 -------------+------------------------------ F( 2, 48) = 1243.51 Model | 9.8732e+16 2 4.9366e+16 Prob > F = 0.0000 Residual | 1.9055e+15 48 3.9699e+13 R-squared = 0.9811 -------------+------------------------------ Adj R-squared = 0.9803 Total | 1.0064e+17 50 2.0127e+15 Root MSE = 6.3e+06 ------------------------------------------------------------------------------ output | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- labor | 47.98736 7.058245 6.80 0.000 33.7958 62.17891 capital | 9.951891 .9781165 10.17 0.000 7.985256 11.91853 _cons | 233621.6 1250364 0.19 0.853 -2280404 2747648 ------------------------------------------------------------------------------ The elasticity of output with respect to labor is: Q L B L L Q Q i i i i 2 / /    . It is often useful to compute this value at the mean. Therefore, evaluated at the mean values of labor and output, the output-labor elasticity is: .41535 0 07 + 4.32e 373914.5 47.98736 2   Q L B . Similarly, the elasticity of output with respect to capital is: Q K B K K Q Q i i i i 3 / /    . Evaluated at the mean, the output-capital elasticity is: .57965 0 07 + 4.32e 2516181 9.951891 3   Q K B . 2.3. For the food expenditure data given in Table 2.8, see if the following model fits the data well: SFDHO i = B 1 + B 2 Expend i + B 3 Expend i 2 and compare your results with those discussed in the text. The Stata output for this model gives the following: . reg sfdho expend expend2 Source | SS df MS Number of obs = 869 -------------+------------------------------ F( 2, 866) = 204.68 Model | 2.02638253 2 1.01319127 Prob > F = 0.0000

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    Residual | 4.28671335 866 .004950015 R-squared = 0.3210 -------------+------------------------------ Adj R-squared = 0.3194 Total | 6.31309589 868 .007273152 Root MSE = .07036 ------------------------------------------------------------------------------ sfdho | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- expend | -5.10e-06 3.36e-07 -15.19 0.000 -5.76e-06 -4.44e-06 expend2 | 3.23e-11 3.49e-12 9.25 0.000 2.54e-11 3.91e-11 _cons | .2563351 .0065842 38.93 0.000 .2434123 .2692579 ------------------------------------------------------------------------------ Similarly to the results in the text (shown in Tables 2.9 and 2.10), these results show a strong nonlinear relationship between share of food expenditure and total expenditure. Both total expenditure and its square are highly significant. The negative sign on the coefficient on “expend” combined with the positive sign on the coefficient on “expend2” implies that the share of food expenditure in total expenditure is decreasing at an increasing rate, which is precisely what the plotted data in Figure 2.3 show. The R 2 value of 0.3210 is only slightly lower than the R 2 values of 0.3509 and 0.3332 for the lin- log and reciprocal models, respectively. (As noted in the text, we are able to compare R 2 values across these models since the dependent variable is the same.) 2.4 Would it make sense to standardize variables in the log-linear Cobb-Douglas production function and estimate the regression using standardized variables? Why or why not? Show the necessary calculations. This would mean standardizing the natural logs of Y , K , and L . Since the coefficients in a log- linear (or double-log) production function already represent unit-free changes, this may not be necessary. Moreover, it is easier to interpret a coefficient in a log linear model as an elasticity. If we were to standardize, the coefficients would represent percentage changes in the standard deviation units. Standardizing would reveal, however, whether capital or labor contributes more to output. 2.5. Show that the coefficient of determination, R 2 , can also be obtained as the squared correlation between actual Y values and the Y values estimated from the regression model (= i Y  ), where Y is the dependent variable. Note that the coefficient of correlation between variables Y and X is defined as: 2 2 i i i i y x r x y     where ; i i i i y Y Y x X X     . Also note that the mean values of Y i and Y  are the same, namely, Y . The estimated Y values from the regression can be rewritten as: i i X B B Y 2 1 ˆ   . Taking deviations from the mean, we have: i i x B y 2 ˆ  . Therefore, the squared correlation between actual Y values and the Y values estimated from the regression model is represented by:

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    , ) ( ) ( ˆ ˆ 2 2 2 2 2 2 2 2 2 2 2 2                 i i i i i i i i i i i i i i i i x y x y x y B x y B x B y x B y y y y y r which is the coefficient of correlation. If this is squared, we obtain the coefficient of determination, or R 2 . 2.6. Table 2.18 gives cross-country data for 83 countries on per worker GDP and Corruption Index for 1998. ( a ) Plot the index of corruption against per worker GDP. 2 4 6 8 10 12 0 10000 20000 30000 40000 50000 gdp_cap index Fitted values ( b ) Based on this plot what might be an appropriate model relating corruption index to per worker GDP? A slightly nonlinear relationship may be appropriate, as it looks as though corruption may increase at a decreasing rate with increasing GDP per capita. ( c ) Present the results of your analysis. Results are as follows: . reg index gdp_cap gdp_cap2 Source | SS df MS Number of obs = 83 -------------+------------------------------ F( 2, 80) = 126.61 Model | 365.6695 2 182.83475 Prob > F = 0.0000 Residual | 115.528569 80 1.44410711 R-squared = 0.7599 -------------+------------------------------ Adj R-squared = 0.7539 Total | 481.198069 82 5.86826913 Root MSE = 1.2017 ------------------------------------------------------------------------------ index | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- gdp_cap | .0003182 .0000393 8.09 0.000 .0002399 .0003964 gdp_cap2 | -4.33e-09 1.15e-09 -3.76 0.000 -6.61e-09 -2.04e-09 _cons | 2.845553 .1983219 14.35 0.000 2.450879 3.240226 ------------------------------------------------------------------------------ ( d ) If you find a positive relationship between corruption and per capita GDP, how would you rationalize this outcome?

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    We find a perhaps unexpected positive relationship because of the way corruption is defined. As th e Transparency International website states, “Since 1995 Transparency International has published each year the CPI, ranking countries on a scale from 0 (perceived to be highly corrupt) to 10 (perceived to have low levels of corruption).” This means that higher values for the corruption index indicate less corruption. Therefore, countries with higher GDP per capita have lower levels of corruption. 2.7 Table 2.19 gives fertility and other related data for 64 countries. Develop suitable model(s) to expla in child mortality, considering the various function forms and the measures of goodness o f fit discussed in the chapter. The following is a linear model explaining child mortality as a function of the female literacy rate, per capita GNP, and the total fe rtility rate: . reg cm flr pgnp tfr Source | SS df MS Number of obs = 64 ------------- + ------------------------------ F( 3, 60) = 59.17 Model | 271802.616 3 90600.8721 Prob > F = 0.0000 Residual | 91875.3836 60 1531.25639 R - squared = 0.7474 ------------- + ------------------------------ Adj R - squared = 0.7347 Total | 363678 63 5772.66667 Root MSE = 39.131 ------------------------------------------------------------------------------ cm | Coef. Std. Err. t P>|t| [95% Conf. Interval] ------------- + ---------------------------------------------------------------- flr | - 1.768029 .2480169 - 7.13 0.000 - 2.264137 - 1.271921 pgnp | - .0055112 .0018782 - 2.93 0.005 - .0092682 - .0017542 tfr | 12.86864 4.190533 3.07 0.003 4.486323 21.25095 _cons | 168.3067 32.89166 5.12 0.000 102.5136 234.0998 ------------------------------------------------------------------------------ The results suggest that higher rates of female literacy and per capita GNP reduce child mortality, which one would expect. Moreover , as the fertility rate goes up, one might expect child mortality to go up, which we see. All results are statistically significant at the 1% level, and the value of r - squared is quite high at 0.7474. 2.8: Verify Equations (2.35), (2.36) and (2.37). Hint : Minimize: 2 2 2 ( ) i i u Y B X     ( ) i f i m f i R r R r u      (2.35) 2 i i i Y B X u   (2.36) 1 2 2 1 n i i i n i i X Y b X      (2.37)

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    var( b 2 ) = 2 2 1 n i i X    (2.38) 2 2 1 i e n      (2.39) We move from equation 2.35 to 2.36 by definition. (We have definied Y as R – r f and X as R m – r f .) There is no intercept in this model. Because of that, we can see that, in minimizing the sum of u i 2 with respect to B 2 and setting the equation equa l to zero , we obtain equation 2.37: (In this case, there is only one equation and one unknown.)                 2 2 2 2 2 2 2 2 2 0 0 ) ( X XY B X B XY X B XY X B Y X dB u d i i 2.9: Consider the following model without any regressors. 1 i i Y B u   How would you obtain an estimate of B 1 ? What i s the meaning of the estimated value? Does it make any sense? If you have a model without regressors, B 1 simply gives you the average value of Y. We can see this by using the data in Table 2.19 (from Exercise 2.7) and running a regression of with only a “dependent” variable, child mortality: . reg cm Source | SS df MS Number of obs = 64 ------------- + ------------------------------ F( 0, 63) = 0.00 Model | 0 0 . Prob > F = . Residual | 363678 63 5772.66667 R - squared = 0.0000 ------------- + ------------------------------ Adj R - squared = 0.0000 Total | 363678 63 5772.66667 Root MSE = 75.978 ------------------------------------------------------------------------------ cm | Coef. Std. Err. t P>|t| [95% Conf. Interval] ------------- + ---------------------------------------------------------------- _cons | 141.5 9.497258 14.90 0.000 122.5212 160.4788 ------------------------------------------------------------------------------ This is clearly not very useful and does not make much sense. B 1 , the intercept, gives you the mean val ue of child mortality. Summarizing this variable would give us the same value:

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