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Solution Manual for Introductory Econometrics: A Modern Approach , 7th Edition

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    1 CHAPTER 1 The Nature of Econometrics and Economic Data

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    2 SOLUTIONS TO PROBLEMS 1.1 (i) Ideally, we could randomly assign students to classes of different sizes. That is, each student is assigned a different class size without regard to any student characteristics such as ability and family background. For reasons we will see in Chapter 2, we would like substantial variation in class sizes (subject, of course, to ethical considerations and resource constraints). (ii) A negative correlation means that a larger class size is associated with lower performance. We might find a negative correlation because a larger class size actually hurts performance. However, with observational data, there are other reasons we might find a negative relationship. For example, children from more affluent families might be more likely to attend schools with smaller class sizes, and affluent children generally might score better on standardized tests. Another possibility is that, within a school, a principal might assign the better students to smaller classes. Or, some parents might insist their children to be placed in smaller classes, and these same parents tend to be more involved in their children’s education. (iii) Given the potential for confounding factors – some of which are listed in (ii) – finding a negative correlation would not be strong evidence that smaller class sizes actually lead to better performance. Some way of controlling for the confounding factors is needed, and this is the subject of multiple regression analysis. 1.3 It does not make sense to pose the question in terms of causality. Economists would assume that students choose a mix of studying and working (and other activities, such as attending class, leisure, and sleeping) based on rational behavior, such as maximizing utility subject to the constraint that there are only 168 hours in a week. We can then use statistical methods to measure the association between studying and working, including regression analysis, which we cover starting in Chapter 2. But we would not be claiming that one variable “causes” the other. They are both choice variables of the student. SOLUTIONS TO COMPUTER EXERCISES C1.1 (i) The average of educ is about 12.6 years. There are two people reporting zero years of education and 19 people reporting 18 years of education. (ii) The average of wage in the sample is about $5.90, which seems low . (iii) Using Table B - 60 in the 2004 Economic Report of the President , the CPI was 56.9 in 1976 and 233 in 2013 . (iv) To convert 1976 dollars into 20 1 3 dollars, we use the ratio of the CPIs, which is 233 / 56.9 4.09 . Therefore, the average hourly wage in 2013 dollars is roughly 4.09 ($5.90) $ 24.13 , which is a reasonable figure.

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    3 (v) The sample contains 252 women (the number of observations with female = 1) and 274 men. C1.3 (i) The largest is 100, the smallest is 0. (ii) 289 out of 1,823, or about 15.85 percent of the sample. (iii) 17 (iv) The average of math4 is about 71.9 and the average of read4 is about 60.1. So, at least in 2001, the reading test was harder to pass. (v) The sample correlation between math4 and read4 is about .843, which is a very high degree of (linear) association. Not surprisingly, schools that have high pass rates on one test have a strong tendency to have high pass rates on the other test. (vi) The average of exppp is about $5,194.87. The standard deviation is $1,091.89, which shows rather wide variation in spending per pupil. [The minimum is $1,206.88 and the maximum is $11,957.64.] (vii) The percentage by which school A outspends school B is (vii) The percentage by which school A outspends school B is ( 6,000 − 5,500 ) 100 ∙ 5,500 ≈ 9.09 % .

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    4 When we use the approximation based on the difference in the natural logs we get a somewhat smaller number: 100 ∙ [log ( 6,000 ) − log ( 5,500 ) ] ≈ 8.71% . C1.5 (i) The smallest and largest values of children are 0 and 13, respectively. The average is about 2.27. (ii) Out of 4,358 women, only 611 have electricity in the home, or about 14.02 percent. (iii) The average of children for women without electricity is about 2.33, and for those with electricity it is about 1.90. So, on average, women with electricity have .43 fewer children than those who do not. (iv) We cannot infer causality here. There are many confounding factors that may be related to the number of children and the presence of electricity in the home; household income and level of education are two possibilities. For example, it could be that women with more education have fewer children and are more likely to have electricity in the home (the latter due to an income effect). C1.7 (i) The percentage of men in the sample report abusing alcohol is 9.9. The employment rate is 24.3. (ii) The employment rate of men who abuse alcohol is 22.6. (iii) The employment rate who do not abuse alcohol is 24.5 . (iv) The employ ment rate s of men who abuse alcohol and who do not are 22.6 and 24.5, respectively. The difference in these employ ment rates is very less, which means that alcoho l abuse does not cause unemploy ment.

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    5 otherwise on a password - protected website or school - approved learning management system for classroom use.

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    8 CHAPTER 2 The Simple Regression Model SOLUTIONS TO PROBLEMS 2.1 (i) Income, age, and family background (such as number of siblings) are just a few possibilities. It seems that each of these could be correlated with years of education. (Income and education are probably positively correlated; age and education may be negatively correlated because women in more recent cohorts have, on average, more education; and number of siblings and education are probably negativel y correlated.) (ii) Not if the factors we listed in part (i) are correlated with educ . Because we would like to hold these factors fixed, they are part of the error term. But if u is correlated with educ , then E( u|educ )  0, and so SLR.4 fails.

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    9 2.3 (i) Let y i = GPA i , x i = ACT i , and n = 8. Then x = 25.875, y = 3.2125, 1 n i   ( x i – x )( y i – y ) = 5.8125, and 1 n i   ( x i – x ) 2 = 56.875. From equation (2.19), we obtain the slope as 1 ˆ  = 5.8125/56.875  .1022, rounded to four places after the decimal. From (2.17), 0 ˆ  = y – 1 ˆ  x  3.2125 – (.1022)25.875  .5681. So we can write GPA = .5681 + .1022 ACT n = 8. The intercept does not have a useful interpretation because ACT is not close to zero for the population of interest. If ACT is 5 points higher, GPA increases by .1022(5) = .511. (ii) The fitted values and residuals — rounded to four decimal places — are given along with the observation number i and GPA in the following table: i GPA GPA ˆ u 1 2.8 2.7143 .0857 2 3.4 3.0209 .3791 3 3.0 3.2253 – .2253 4 3.5 3.3275 .1725 5 3.6 3.5319 .0681 6 3.0 3.1231 – .1231 7 2.7 3.1231 – .4231 8 3.7 3.6341 .0659 You can verify that the residuals, as reported in the table, sum to  .0002, which is pretty close to zero given the inherent rounding error. (iii) When ACT = 20, GPA = .5681 + .1022(20)  2.61. (iv) The sum of squared residuals, 2 1 ˆ n i i u   , is about .4347 (rounded to four decimal places), and the total sum of squares, 1 n i   ( y i – y ) 2 , is about 1.0288. So the R - squared from the regression is

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    10 R 2 = 1 – SSR/SST  1 – (.4347/1.0288)  .577. Therefore, about 57.7% of the variation in GPA is explained by ACT in this small sample of students. 2.5 (i) The intercept implies that when inc = 0, cons is predicted to be negative $124.84. This, of course, cannot be true, and reflects the fact that this consumption function might be a poor predictor of consumption at very low - income levels. On the other hand, on an annual basis, $124.84 is not so far from zero. (ii) Just plug 30,000 into the equation: cons = – 124.84 + .853(30,000) = 25,465.16 dollars. (iii) The MPC and the APC are shown in the following graph. Even though the intercept is negative, the smallest APC in the sample is positive. The graph starts at an annual income level of $1,000 (in 1970 dollars). 2.7 (i) When we condition on inc in computing an expectation, inc becomes a constant. So E( u | inc ) = E( inc  e | inc ) = inc  E( e | inc ) = inc  0 because E( e | inc ) = E( e ) = 0. inc 1000 10000 20000 30000 .7 .728 .853 APC MPC .9 APC MPC

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    11 (ii) Again, when we condition on inc in computing a variance, inc becomes a constant. So Var( u | inc ) = Var( inc  e | inc ) = ( inc ) 2 Var( e | inc ) = 2 e  inc because Var( e | inc ) = 2 e  . (iii) Families with low incomes do not have much discretion about spending; typically, a low - income family must spend on food, clothing, housing, and other necessities. Higher - income people have more discretion, and some might choose more consumption while others more saving. This discretion suggests wider variability in saving among higher income families. 2.9 (i) We follow the hint, noting that 1 c y = 1 c y (the sample average of 1 i c y is c 1 times the sample average of y i ) and 2 c x = 2 c x . When we regress c 1 y i on c 2 x i (including an intercept) , we use equation (2.19) to obtain the slope: From (2.17), we obtain the intercept as 0  = (c 1 y ) – 1  ( c 2 x ) = ( c 1 y ) – [( c 1 / c 2 ) 1 ˆ  ]( c 2 x ) = c 1 ( y – 1 ˆ  x ) = c 1 0 ˆ  ) because the intercept from regressing y i on x i is ( y – 1 ˆ  x ). (ii) We use the same approach from part (i) along with the fact that 1 ( ) c y  = c 1 + y and 2 ( ) c x  = c 2 + x . Therefore, 1 1 ( ) ( ) i c y c y    = ( c 1 + y i ) – ( c 1 + y ) = y i – y and ( c 2 + x i ) – 2 ( ) c x  = x i – x . So c 1 and c 2 entirely drop out of the slope formula for the regression of ( c 1 + y i ) on ( c 2 + x i ), and 1  = 1 ˆ  . The intercept is 0  = 1 ( ) c y  – 1  2 ( ) c x  = ( c 1 + y ) – 1 ˆ  ( c 2 + x ) = ( 1 ˆ y x   ) + c 1 – c 2 1 ˆ  = 0 ˆ  + c 1 – c 2 1 ˆ  , which is what we wanted to show. (iii) We can simply apply part (ii) because 1 1 log( ) log( ) log( ) i i c y c y   . In other words, replace c 1 with log( c 1 ), replace y i with log( y i ), and set c 2 = 0. (iv) Again, we can apply part (ii) with c 1 = 0 and replacing c 2 with log( c 2 ) and x i with log( x i ). If 0 1 ˆ ˆ and   are the original intercept and slope, then 1 1 ˆ    and 0 0 2 1 ˆ ˆ log( ) c      .

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    12 2.11 (i) We would want to randomly assign the number of hours in the preparation course so that hours is independent of other factors that affect performance on the SAT. Then, we would collect information on SAT score for each student in the experiment, yielding a data set {( , ) : 1,..., } i i sat hours i n  , where n is the number of students we can afford to have in the study. From equation (2.7), we should try to get as much variation in i hours as is feasible. (ii) Here are three factors: innate ability, family income, and general health on the day of the exam. If we think students with higher native intelligence think they do not need to prepare for the SAT, then ability and hours will be negatively correlated. Family income would probably be positively correlated with hours , because higher income families can more easily afford preparation courses. Ruling out chronic health problems, health on the day of the exam should be roughly uncorrelated with hours s pent in a preparation course. (iii) If preparation courses are effective, 1  should be positive ; other factors equal, an increase in hours should increase sat . (iv) The intercept, 0  , has a useful interpretation in this example: because E( u ) = 0, 0  is the average SAT score for students in the population with hours = 0. 2.13 (i) Since 𝑥 𝑖 is a binary variable, it is equal to either 0 or 1. Thus, the number of observations with 𝑥 𝑖 = 0 will be 𝑛 0 = ∑ ( 1 − 𝑥 𝑖 ) since the value in the summation is equal to 1 whenever 𝑥 𝑖 = 0 and equal to 0 whenever 𝑥 𝑖 = 1 . Similarly, 𝑛 1 = ∑ 𝑥 𝑖 will give us the number of observations with 𝑥 𝑖 = 1 since we are only going to be counting instances in which 𝑥 𝑖 is not equal to 0. We know that 𝑥 ̅ = 1 𝑛 ∑ 𝑥 𝑖 . We also have shown that ∑ 𝑥 𝑖 = 𝑛 1 . Thus, 𝑥 ̅ = 𝑛 1 𝑛 . We can write 1 − 𝑥 ̅ = 1 − 𝑛 1 𝑛 = 𝑛 − 𝑛 1 𝑛 = 𝑛 0 𝑛 since 𝑛 0 + 𝑛 1 = 𝑛 . 𝑥 ̅ = 𝑛 1 𝑛 tells us the fraction of observations for which 𝑥 𝑖 = 1 . (ii) We know that 𝑦 ത 0 = 𝑛 0 − 1 ∑ 𝑦 𝑖 𝑛 0 𝑖 = 1 , where we are just looking at values of 𝑦 𝑖 for which 𝑥 𝑖 = 0 . One way we can formalize this is by recognizing that ( 1 − 𝑥 𝑖 ) 𝑦 𝑖 = 𝑦 𝑖 if 𝑥 𝑖 = 0 and 0 otherwise. Thus, 𝑦 ത 0 = 𝑛 0 − 1 ∑ 𝑦 𝑖 𝑛 0 𝑖 = 1 = 𝑛 0 − 1 ∑ ( 1 − 𝑥 𝑖 ) 𝑦 𝑖 𝑛 𝑖 = 1 . Similarly, we can show that 𝑦 ത 1 = 𝑛 1 − 1 ∑ 𝑦 𝑖 𝑛 1 𝑖 = 1 = 𝑛 1 − 1 ∑ 𝑥 𝑖 𝑦 𝑖 𝑛 𝑖 = 1 , since 𝑥 𝑖 𝑦 𝑖 = 𝑦 𝑖 if 𝑥 𝑖 = 1 and 0 otherwise. (iii) We can express 𝑦 ത = 1 𝑛 ∑ 𝑦 𝑖 𝑛 𝑖 = 1 = 1 𝑛 ൫ ∑ 𝑦 0 , 𝑖 𝑛 0 𝑖 = 1 + ∑ 𝑦 1 , 𝑖 𝑛 1 𝑖 = 1 ൯ = 𝑛 0 𝑛 𝑦 ത 0 + 𝑛 1 𝑛 𝑦 ത 1 We have shown that 1 − 𝑥 ̅ = 𝑛 0 𝑛 and 𝑥 ̅ = 𝑛 1 𝑛 . Thus, 𝑦 ത = ( 1 − 𝑥 ̅ ) 𝑦 ത 0 + 𝑥 ̅ 𝑦 ത 1

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    13 (iv) If 𝑥 𝑖 is binary, then 𝑥 𝑖 2 = 𝑥 𝑖 . Thus, 𝑛 − 1 ∑ 𝑥 𝑖 2 − 𝑥 ̅ 2 = 𝑛 − 1 ∑ 𝑥 𝑖 − 𝑥 ̅ 2 = 𝑥 ̅ − 𝑥 ̅ 2 = 𝑥 ̅ ( 1 − 𝑥 ̅ ) (v) First note that 𝑛 − 1 ∑ 𝑥 𝑖 𝑦 𝑖 𝑛 𝑖 = 1 = 𝑛 − 1 ∑ 𝑦 𝑖 𝑛 1 𝑖 = 1 = 𝑛 1 𝑛 𝑦 ത 1 since we will only be counting instances in which 𝑥 𝑖 = 1 . Using the result from part i, we show that 𝑛 − 1 ∑ 𝑥 𝑖 𝑦 𝑖 𝑛 𝑖 = 1 = 𝑥 ̅ 𝑦 ത 1 Therefore 𝑛 − 1 ∑ 𝑥 𝑖 𝑦 𝑖 𝑛 𝑖 = 1 − 𝑥 ̅ 𝑦 ത = 𝑥 ̅ 𝑦 ത 1 − 𝑥 ̅ 𝑦 ത = 𝑥 ̅ 𝑦 ത 1 − 𝑥 ̅ [ ( 1 − 𝑥 ̅ ) 𝑦 ത 0 + 𝑥 ̅ 𝑦 ത 1 ] , using the result from part iii. With some manipulation, we can thus show that 𝑛 − 1 ∑ 𝑥 𝑖 𝑦 𝑖 𝑛 𝑖 = 1 − 𝑥 ̅ 𝑦 ത = 𝑥 ̅ ( 1 − 𝑥 ̅ ) ( 𝑦 ത 1 − 𝑦 ത 0 ) (vi) Equation 2.74 states that 𝛽 መ 1 = 𝑦 ത 1 − 𝑦 ത 0 . This is derived in the same way as the usual OLS estimator: minimizing the sum of the squared residuals to get 𝛽 መ 1 = ∑ ( 𝑥 𝑖 − 𝑥 ̅ ) ( 𝑦 𝑖 − 𝑦 ത ) ∑ ( 𝑥 𝑖 − 𝑥 ̅ ) 2 Expanding the numerator and denominator: 𝛽 መ 1 = ∑ 𝑥 𝑖 𝑦 𝑖 − 𝑛 𝑥 ̅ 𝑦 ത ∑ 𝑥 𝑖 2 − 𝑛 𝑥 ̅ 2 Using the results from parts v and vi: 𝛽 መ 1 = 𝑥 ̅ ( 1 − 𝑥 ̅ ) ( 𝑦 ത 1 − 𝑦 ത 0 ) 𝑥 ̅ ( 1 − 𝑥 ̅ ) = 𝑦 ത 1 − 𝑦 ത 0 (vii) We know that 𝛽 መ 0 = 𝑦 ത − 𝛽 መ 1 𝑥 ̅ . Using the results from parts iii and vi, we have: 𝛽 መ 0 = ( 1 − 𝑥 ̅ ) 𝑦 ത 0 + 𝑥 ̅ 𝑦 ത 1 − ( 𝑦 ത 1 − 𝑦 ത 0 ) 𝑥 ̅ = 𝑦 ത 0 − 𝑥 ̅ 𝑦 ത 0 + 𝑥 ̅ 𝑦 ത 1 − 𝑥 ̅ 𝑦 ത 1 + 𝑥 ̅ 𝑦 ത 0 = 𝑦 ത 0 2.15 (i) The population average treatment effect 𝜏 𝑎௧𝑒 = 𝐸 [ 𝑦 ( 1 ) − 𝑦 ( 0 ) ] . If we are able to observe 𝑦 𝑖 ( 0 ) and 𝑦 𝑖 ( 1 ) for each observation 𝑖 (e.g. we can observe the same observation in both states of the world), then 𝐸 ቂ 1 𝑛 ∑ 𝑦 𝑖 ( 1 ) ቃ = 1 𝑛 ∑ 𝐸 [ 𝑦 𝑖 ( 1 ) ] = 1 𝑛 ∗ 𝑛𝐸 [ 𝑦 ( 1 ) ] = 𝐸 [ 𝑦 ( 1 ) ] . Similarly, 1 𝑛 ∑ 𝑦 𝑖 ( 0 ) is an unbiased estimator for 𝐸 [ 𝑦 ( 0 ) ] . Therefore 𝐸 ቄ 1 𝑛 ∑ [ 𝑦 𝑖 ( 1 ) − 𝑦 𝑖 ( 0 ) ] ቅ = 𝐸 [ 𝑦 ( 1 ) − 𝑦 ( 0 ) ] (ii) We can write 𝑦 𝑖 = ( 1 − 𝑥 𝑖 ) 𝑦 𝑖 ( 0 ) + 𝑥 𝑖 𝑦 𝑖 ( 1 ) . Then, 𝑦 ത 0 = 1 𝑛 ∑ [ ( 1 − 𝑥 𝑖 ) 𝑦 𝑖 ( 0 ) + 𝑥 𝑖 𝑦 𝑖 ( 1 ) ] 𝑛 𝑖 = 1 = 1 𝑛 0 ∑ 𝑦 𝑖 ( 0 ) 𝑛 0 𝑖 = 1 since 𝑥 𝑖 = 0 for all observations used to calculate 𝑦 ത 0 . By a similar logic, 𝑦 ത 1 = 1 𝑛 1 ∑ 𝑦 𝑖 ( 1 ) 𝑛 1 𝑖 = 1 since 𝑥 1 = 1 for all observations used to calculate 𝑦 ത 1 . Thus, both 𝑦 ത 0 and 𝑦 ത 1 are computed using a subset of the sample. By contrast 𝑦 ത ( 0 ) and 𝑦 ത ( 1 ) would be computed using the entire sample, using the outcomes of each observation in both states of the world. 2.17 (i) 𝑉𝑎𝑟 ( 𝑢 𝑖 | 𝑥 𝑖 ) = 𝑉𝑎𝑟 [ ( 1 − 𝑥 𝑖 ) 𝑢 𝑖 ( 0 ) + 𝑥 𝑖 𝑢 𝑖 ( 1 ) ] = ( 1 − 𝑥 𝑖 ) 2 𝜎 0 2 + 𝑥 𝑖 2 𝜎 1 2 . Recall that we are conditioning on 𝑥 𝑖 , so we can treat it as a deterministic variable. Finally, note that since 𝑥 𝑖 is binary, we can write 𝑉𝑎𝑟 ( 𝑢 𝑖 | 𝑥 𝑖 ) = ( 1 − 𝑥 𝑖 ) 𝜎 0 2 + 𝑥 𝑖 𝜎 1 2 . The conditional variance is a weighted average of the variances from the two different states of the world. (ii) There are three ways in which 𝑉𝑎𝑟 ( 𝑢 𝑖 | 𝑥 𝑖 ) could be constant: 1. 𝑥 𝑖 = 0 for all 𝑖 2. 𝑥 𝑖 = 1 for all 𝑖 3. 𝜎 0 2 = 𝜎 1 2

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    14 The first two scenarios imply that we have no observations in either the treatment or control groups and thus the experiment is not of any use. The last case suggests that outcomes for non - treated state have exactly the same variability as outcomes for the treated state. While this is certainly a possibility, it would imply that the only effect of the treatment is to change the mean outcome, not any other aspects of its distribution. SOLUTIONS TO COMPUTER EXERCISES C2.1 (i) The average prate is about 87.36 , and the average mrate is about .732. (ii) The estimated equation is prate = 83.08 + 5.86 mrate n = 1,534, R 2 = .075. (iii) The intercept implies that, even if mrate = 0, the predicted participation rate is 83.08 percent. The coefficient on mrate implies that a one - dollar increase in the match rate – a fairly large increase – is estimated to increase prate by 5.86 percentage points. This assumes, of course, that this change prate is possible (if, say, prate is already at 98, this interpretation makes no sense). (iv) If we plug mrate = 3.5 into the equation , we get ˆ prate = 83.08 + 5.86(3.5) = 103.59. This is impossible, as we can have at most a 100 percent participation rate. This illustrates that, especially when dependent variables are bounded, a simple regression model can give strange predictions for extreme values o f the independent variable. (In the sample of 1,534 firms, only 34 have mrate  3.5.) (v) mrate explains about 7.5% of the variation in prate . This is not much and suggests that many other factors influence 401(k) plan participation rates. C2.3 (i) The estimated equation is sleep = 3,586.4 – .151 totwrk n = 706, R 2 = .103. The intercept implies that the estimated amount of sleep per week for someone who does not work is 3,586.4 minutes, or about 59.77 hours. This comes to about 8.5 hours per night.

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    15 (ii) If someone works two more hours per week , then  totwrk = 120 (because totwrk is measured in minutes), and so sleep  = – .151(120) = – 18.12 minutes. This is only a few minutes a night. If someone were to work one more hour on each of five working days, sleep  = – .151(300) = – 45.3 minutes, or about five minutes a night. C2.5 (i) The constant elasticity model is a log - log model: log( rd ) = 0  + 1  log( sales ) + u , where 1  is the elasticity of rd with respect to sales . (ii) The estimated equation is log( ) rd = – 4.105 + 1.076 log( sales ) n = 32, R 2 = .910. The estimated elasticity of rd with respect to sales is 1.076, which is just above one. A one percent increase in sales is estimated to increase rd by about 1.08%. C2.7 (i) The average gift is about 7.44 Dutch guilders. Out of 4,268 respondents, 2,561 did not give a gift, or about 60 percent. (ii) The average mailings per year is about 2.05. The minimum value is .25 (which presumably means that someone has been on the mailing list for at least four years) , and the maximum value is 3.5. (iii) The estimated equation is 2 2.01 2.65 4,268, .0138. gift mailsyear n R     (iv) The slope coefficient from part (iii) means that each mailing per year is associated with – perhaps even “causes” – an estimated 2.65 additional guilders, on average. Therefore, if each mailing costs one guilder, the expected profit from each mailing is estimated to be 1.65 guilders. This is only the average, however. Some mailings generate no contributions, or a contribution less than the mailing cost; other mailings generated much more than the mailing cost. (v) Because the smallest mailsyear in the sample is .25, the smallest predicted value of gifts is 2.01 + 2.65(.25)  2.67. Even if we look at the overall population, where some people have

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    16 received no mailings, the smallest predicted value is about two. So, with this estimated equation, we never predict zero charitable gifts. C2.9 (i) In 1996, 1,051 counties had zero murders. Out of 2,197 counties, 31 counties had at least one execution and the largest number of executions is 3. (ii) The estimated equation is 𝑚𝑢𝑟𝑑𝑒𝑟𝑠 = 5 . 46 + 58 . 56 𝑒𝑥𝑒𝑐𝑠 𝑛 = 2197 , 𝑅 2 = 0 . 0439 . (iii) The slope coefficient on execs implies that if the number of executions increases by one, the estimated number of murders increases largely by about 59. No, the estimated equation does not suggest a deterrent effect of capital punishment. (iv) The smallest number of murders can be predicted by the equation is 5.46 , that is about 5 murders. The residual for a county with zero executions and zero murders is - 5.46. (v) This simple linear regression equation predicts that if the number of executions increases by one, the estimated number of murders increases largely by about 59, which means capital punishment does not have a deterrent effect on murders — capital punishment is not discouraging people from doing murders. The sign and magnitude of the estimate +58.56 makes us suspect that the error term u and the independent variable execs are correlated. Therefore, the regression model is not well suited for prediction. C2.11 (i) There are 141 students in the sample. The average college GPA is a 3.057, while the maximum GPA is a 4.0 (1 student with this GPA). (ii) 56 students (39.7% of the sample) owned their own PC. (iii) 𝑐𝑜𝑙 𝐺 ෠ 𝑃𝐴 = 2 . 989 ( 0 . 040 ) + 0 . 170 ( 0 . 063 ) ∗ 𝑃𝐶 . We estimates that students who do not own a PC have an average college GPA of 2.989. Those who own a PC have college GPA’s that are on average 0.17 points higher than those who do not. Both coefficient estimates are statistically significant at the 1% le vel. In terms of magnitude, a 0.17 point increase in GPA represents about a 5.7% increase from the mean college GPA for those without a computer. (iv) The 𝑅 2 = 0 . 05 . We are only explaining about 5% of the variation in college GPAs with computer ownership. Thus, there are a lot of other factors that influence college GPA.

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    17 (v) If we truly believed that PC ownership was randomly assigned, then we could infer a causal effect on college GPA. However, it is highly unlikely that we have random assignment in this experiment. There are a wide range of factors that could both influe nce college GPA and PC ownership that we are overlooking. For example, consider parent’s income. Students from wealthier families are more likely to own PCs. These same students are also less likely to have to work during college, freeing up more time for their studies and thus earning higher GPAs. As such, we cannot disentangle the effects of PC ownership from parental income. There are any number of other omitted variables that could cause us to violate the random assignment assumption.

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