### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# 3. 250602220-Solutions-Chapter-3

### View Full Document

## About this Document

## 6

## 0

## Popular in

## Popular in Department

This 203 page Class Notes was uploaded by yeshuahanotzrivemelejhayehu Notetaker on Friday September 16, 2016. The Class Notes belongs to at 1 MDSS-SGSLM-Langley AFB Advanced Education in General Dentistry 12 Months taught by in Fall 2016. Since its upload, it has received 6 views.

## Reviews for 3. 250602220-Solutions-Chapter-3

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/16/16

CHAPTER 3 Exercise Solutions 31 Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 32 EXERCISE 3.1 (a) The required interval estimator is bt1 1 c( ). When b =83.416, tc = =(0.975,38).024 and se(b 1 = 43.410, we get the interval estimate: 83.416 ± 2.024 × 43.410 = (−4.46, 171.30) We estimate that β lies between −4.46 and 171.30. In repeated samples, 95% of similarly 1 constructed intervals would contain the true β1. (b) Ttest H :β 0= against H :β 0 ≠ we compute the t-value 0 1 1 1 b −β 83.416−0 t1= = 1 1 = 1.92 se(b1) 43.410 tiece t = 1.92 value does not exceed the 5% critical value tc = (0.975,38).024, we do not reject H 0. The data do not reject the zero-intercept hypothesis. (c) The p-value 0.0622 represents the sum of the areas under the t distribution to the left of t = −1.92 and to the right of t = 1.92. Since the t distribution is symmetric, each of the tail areas that make up the p-value are p/2 = 0.0622 2 = 0.0311. The level of significance, α, is given by the sum of the areas under the PDF for ||t|t,c so the area under the curve for tt> is α=/2 .025 and likewise for tt< − . Therefore not rejecting the null hypothesis c c because α< /2 p/2, or α< p, is the same as not rejecting the null hypothesis because −tc c< .From Figure xr3.1(a) we can see that having a p-value > 0.05 is equivalent to having −t<tt< . c c Figure xr3.1(a) Critical and observed t values for Exercise 3.1(c) Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 33 Exercise 3.1 (continued) (d) Testing H 0 10 = against H,1 10 > uses the same t-value as in part (b), t = 1.92. Because it is a one-tailed test, the critical valu e is chosen such that there is a probability of 0.05 in the right tail. That is, ct = (0.95,38).686 . Since t = 1.92 > tc= 1.69, H 0s rejected, the alternative is accepted, and we conclude that the intercept is positive. In this case p-value = P(t > 1.92) = 0.0311. We see from Figure xr3.1(b) that having the p-value < 0.05 is equivalent to having t > 1.69. 0.4 Rejection Region 0.3 PDF 0.2 0.1 1.92 0.0 -3 -2 -1 0 1 2 3 tc T Figure xr3.1(b) Rejection region and observed t value for Exercise 3.1(d) (e) The term "level of significance" is used to describe the probability of rejecting a true null hypothesis when carrying out a hypothesis test. The term "level of confidence" refers to the probability of an interval estimator yielding an interval that includes the true parameter. When carrying out a two-tailed test of the form H 0:β =k versus H 1:β,k≠ non-rejection of H implies c lies within the confidence interval, and vice versa, 0 providing the level of significance is equal to one minus the level of confidence. (f) False. The test in (d) uses the level of si gnificance 5%, which is the probability of a Type I error. That is, in repeated samples we have a 5% chance of rejecting the null hypothesis when it is true. The 5% significance is a probability statement about a procedure not a probability statement about β1 . It is careless and dangerous to equate 5% level of significance with 95% confidence, which relate s to interval estimation procedures, not hypothesis tests. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 34 EXERCISE 3.2 (a) The coefficient of EXPER indicates that, on average, a technical artist's quality rating goes up by 0.076 for every additional year of experience. 3.9 3.8 3.7 3.6 3.5 RATING 3.4 3.3 3.2 3.1 0 1 2 3 4 5 6 7 8 EXPER Figure xr3.2(a) Estimated regression function (b) Using the value tt = = 2.074 , the 95% confidence interval for β is given by c (0.975,22) 2 bt2 2 c±×( ) 0.076 2.074 0.044 ( 0.015, 0.167) We are 95% confident that the procedure we have used for constructing a confidence interval will yield an interval that includes the true parameter β . 2 (c) test H 0 20= against H,1 2 ≠ we use the test statistic t = b 2se(b 2 = 0.076/0.044 = 1.727. The t critical value for a two tail test with N − 2 = 22 degrees of freedom is 2.074. Since −2.074 < 1.727 < 2.074 we fail to reject the null hypothesis. (d) test H 0 20= against H 1 20 > we use the t-value from part (c), namely t =1.727 , but the right-tail critical value tc = (0.95,22)1.717 . Since 1.727 >1.717, we reject H and 0 conclude that β i2 positive. Experience has a positive effect on quality rating. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 35 Exercise 3.2 (continued) (e) The p-value of 0.0982 is given as the sum of the areas under the t-distribution to the left of −1.727 and to the right of 1.727. We do not reject H 0 because, for α= 0.05, p-value > 0.05. We can reject, or fail to reject, the null hypothesis just based on an inspection of the p-value. Having the p-value > α is equivalent to havingtt<=c 2.074. Figure xr3.2(b) P-value diagram Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 36 EXERCISE 3.3 (a) Hypotheses: H :β 0= against H :β 0 ≠ 0 2 1 2 Calculated t-value: t = 0.310 0.082 3.78 Critical t-value: ±t±c (0.995,22) 2.819 Decision: Reject H 0ecause t = t 3.78 >= c 2.819. (b) Hypotheses: H :0 2= against H :1 2 > Calculated t-value: t = 0.310 0.082 3.78 Critical t-value: tc = (0.99,22)2.508 Decision: Reject H 0ecause t = t 3.78 >= c 2.508. (c) Hypotheses: H :0 2= against H 1 2 < Calculated t-value: t = 0.310 0.082 3.78 Critical t-value: tt = = 1.717 c (0.05,22) Decision: Do not reject H because t = t 3.78 >−= 1.717. 0 c Figure xr3.3 One tail rejection region (d) Hypotheses: 5 H. 0 2= against5 H.1 2 ≠ Calculated t-value: t = (−0−10 =0.5) 0.082 2.32 Critical t-value: ±t±c (0.975,22) 2.074 Decision: Reject H because t=t−2<.32 2.074. 0 c (e) A 99% interval estimate of the slope is given by bt±b se( ) = 0.310 ± 2.819 × 0.082 = (0.079, 0.541) 2 2 c estieate β2to lie between 0.079 and 0.541 using a procedure that works 99% of the time in repeated samples. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 37 EXERCISE 3.4 (a) bt=×b se( ) = 1.257 × 2.174 = 2.733 1 1 24 20 16 12 MIM 8 4 0 0 10 20 30 40 50 60 70 80 90 100 PMHS Figure xr3.4(a) Estimated regression function (b) se(btb) = = 0.18=0 5.754 0.0313 2 2 (c) p-value = 2 × (−< P(t 1.257) ) = 2 × (1 − 0.8926) = 0.2147 Figure xr3.4(b) P-value diagram (d) The estimated slope b = 0.18 indicates that a 1% increase in males 18 and older, who are 2 high school graduates, increases average income of those males by $180. The positive sign is as expected; more education should lead to higher salaries. (e) Using tc = =(0.995,49)2.68 , a 99% confidence interval for the slope is given by bt±b se( ) = 0.180 ± 2.68 × 0.0313 = (0.096, 0.264) 2 2 c Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 38 Exercise 3.4 (continued) (f) tosrting 2 H.0 2= aga,ns2 H.1 2 ≠ we calculate 0.180−0.2 − t = = 0.639 0.0313 The critical values for a two-tailed test with a 5% significance level and 49 degrees of freedom are ±t =±c2.01. Since t = −0.634 lies in the interval ( −2.01, 2.01), we do not reject H .0The null hypothesis suggests that a 1% increase in males 18 or older, who are high school graduates, leads to an increase in average income for those males of $200. Non-rejection of H me0ns that this claim is compatible with the sample of data. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 39 EXERCISE 3.5 (a) The linear relationship between life insurance and income is estimated as INSURANCE = 6 +.8550 3.8802INCOME (se) ( ) ( ).3835 0.1121 Figure xr3.5 Fitted regression line and mean (b) The relationship in part (a) indicates that, as income increases, the amount of life insurance increases, as is expected. If taken literally, the value of b1 = 6.8550 implies that if a family has no income, then they would purchase $6855 worth of insurance. However, given the lack of data in the region whereINCOME = 0, this value is not reliable. (i) If income increases by $1000, then an estimate of the resulting change in the amount of life insurance is $3880.20. (ii) The standard error of 2 is 0.1121. To test a hypothesis about2β the test statistic is b2 2 se b ~ ( )2 ()2 An interval estimator for β 2s ⎣b22 2 cc( ), se( ) ⎦ , where tcis the critical value for t with)2 − degrees of freedom at the α level of significance. (c) To test the claim, the relevant hypotheses are 0 : 2 = 5 versus H1: 2 ≠ 5. The alternative β ≠ 5 has been chosen because, before we sample, we have no reason to suspect β > 5 or 2 2 β2 < 5. The test statistic is that given in part (b) (ii) β2set equal to 5. The rejection region (18 degrees of freedom) is | t | > 2.101. The value of the test statistic is b2− 5 3.8802 − t = = − = 9.99 se()2 0.1121 As t =−−<.99 2.101, we reject the null hypothesis and conclude that the estimated relationship does not support the claim. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 40 Exercise 3.5 (continued) (d) To test the hypothesis that the slope of the relationship is one, we proceed as we did in part (c), using 1 instead of 5. Thus, our hypotheses are H 0: β2= 1 versus H :1β 2 1. The rejection region is | t | > 2.101. The value of the test statistic is t = =8802−1 25.7 0.1121 Since tt 25.7 c 2.101, we reject the null hypothesis. We conclude that the amount of life insurance does not increase at the same rate as income increases. (e) Life insurance companies are interested in household characteristics that influence the amount of life insurance cover that is purchased by different households. One likely important determinant of life insurance cover is household income. To see if income is important, and to quantify its effect on insurance, we set up the model INSURANCE =β +β INCOME+ e i i i 1 2 where INSURANCE is life insurance cover by the i-th household, INCOME is i household income, β and β are unknown parameters that describe the relationship, and e 1 2 i is a random uncorrelated error that is assumed to have zero mean and constant variance σ . To estimate our hypothesized relationship, we take a random sample of 20 households, collect observations on INSURANCE and INCOME and apply the least-squares estimation procedure. The estimated equation, with standard errors in parentheses, is INSURANCE = 6 +.8550 3.8802INCOME ()e ( 7.3()5 0.11)1 The point estimate for the response of life-insurance coverage to an income increase of $1000 (the slope) is $3880 and a 95% interval estimate for this quantity is ($3645, $4116). This interval is a relatively narrow one, suggesting we have reliable information about the response. The intercept estimate is not significantly different from zero, but this fact by itself is not a matter for concern; as mentione d in part (b), we do not give this value a direct economic interpretation. The estimated equation could be used to assess likely requests for life insurance and what changes may occur as a result of income changes. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 41 EXERCISE 3.6 (a) A 95% interval estimator for β is b ± t × se(b ). Using our sample of data the 2 2 (0.975,14) 2 corresponding interval estimate is −0.3857 ± 2.145 × 0.03601 = (−0.4629, −0.3085) If we used the interval estimator in repeated samples, then 95% of interval estimates like the above one would contain β . Thu2, β is lik2ly to lie in the range given by the above interval. (b) We set up the hypotheses H : β =00 v2rsus H : β < 0.1The2alternative β < 0 is chos2n because we would expect the unit costs of production to decline as cumulative production increases if there is learning. The test statistic, given H 0s true, is b2 t t ~ (14) se(b2) The rejection region is t < −1.761. The value of the test statistic is −0.3857 − t = 0.03601 10.71 Since t = −10.71 < −1.761, we reject H and conclude that learning does exist. We 0 conclude in this way because −10.71 is an unlikely value to have come from the t distribution which is valid when there is no learning. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 42 EXERCISE 3.7 (a) We set up the hypotheses H 0β 1j= versus H 1β1 j . The economic relevance of this test is to test whether the return on the firm’s stock is risky relative to the market portfolio. Each beta measures the volatility of the stock relative to the market portfolio and volatility is often used to measure risk. A beta value of one indicates that the stock’s volatility is the same as that of the market portfolio. The test statistic given0H is true, is b − 1 t=t j ~ ()8 se() j The rejection region ist <− 1.980 and t >1.980 , where t =1.980 . (0.975,118) The results for each company are given in the following table: Stock rult-value Decision 0.9593−1 Disney − t= = 0.287 Since −1.98<< t 1.98, fail to reject H0 0.1420 0.9830 −1 GE − t= = 0.162 Since −1.98<< t 1.98, fail to reject H0 0.1047 1.0744 −1 GM t= = 0.478 Since −1.98<< t 1.98, fail to reject H0 0.1558 1.2683−1 IBM t= = 1.726 Since −1.98<< t 1.98, fail to reject H0 0.1554 Microsoft t= =.4299 −1 2.284 Since t >1.98 , reject H 0.1882 0 0.4030 −1 Mobil-Exxon − t= = 7.256 Since t < −1.98, reject H 0 0.08228 For Disney, GE, GM and IBM, we failed to reject the null hypothesis, indicating that the sample data are consistent with the conjecture that the Disney, GE, GM and IBM stocks have the same volatility as the market po rtfolio. For Microsoft and Mobil-Exxon, we rejected the null hypothesis, and concluded that these two stocks do not have the same volatility as the market portfolio. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 43 Exercise 3.7 (continued) (b) We set up the hypotheses H :β1 ≥ versus H :β 1 < . The relevant test statistic, given 0 j 1 j H 0 is true, is bj− 1 t = ~ ()18 se() j The rejection region is t < −1.658 where ct = (0.05,118)1.658 . The value of the test statistic is 0.4030−1 − t = = 7.256 0.08228 Since t = −7.256 < t c −1.658, we reject H and0conclude that Mobil-Exxon’s beta is less than 1. A beta equal to 1 suggests a stock's variation is the same as the market variation. A beta less than 1 implies the stock is less volatile than the market; it is a defensive stock. (c) We set up the hypotheses H 0β1 j versus H 1β1 j . The relevant test statistic, given H 0 is true, is t t bj−1 ~ se b ()18 () j rejhriionn t > 1.658 where tc = (0.95,118)658 . The value of the test statistic is 1.4299 −1 t = = 2.284 0.1882 Since t = 2.284 > t c = 1.658, we reject H an0 conclude that Microsoft’s beta is greater than 1. A beta equal to 1 suggests a stock's variation is the same as the market variation. A beta greater than 1 implies the stock is more volatile than the market; it is an aggressive stock. (d) A 95% interval estimator for Microsoft’s beta is bt± b×se( ) . Using our sample j j(0.975,118) of data the corresponding interval estimate is 1.4299 ± 1.980 × 0.1882 = (1.057, 1.803) Thus we estimate, with 95% confidence, that Mi crosoft’s beta falls in the interval 1.057 to 1.803. It is possible that Microsoft’s beta falls outside this interval, but we would be surprised if it did, because the procedure we used to create the interval works 95% of the time. The problem with the interval estimate is that it is wide. We feel sure that Microsoft is more volatile than the market, but how much more is not known precisely. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 44 Exercise 3.7 (continued) (e) The two hypotheses are H :0 α j= 0 versus H 1 α j≠ 0. The test statistic, given0H is true, is a tt= j ~ se() j ()8 The rejection region is t <− 1.980 and t >1.980 , where t(0.975,118)980 . The results for each company are given in the following table: Stock rult-value Decision −0.0010 Disney − t = = 0.152 Since −1.98<< t 1.98, fail to rejectH 0 0.0067 t = =0059 1.199 Since −1.98<< t 1.98 , fail to reject H GE 0.0049 0 −0.0023 GM − t = = 0.317 Since −1.98<< t 1.98, fail to reject H0 0.0073 0.0068 IBM t = = 0.940 Since −1.98<< t 1.98, fail to reject H0 0.0073 Microsoft t = =0102 1.156 Since −1.98<< t 1.98, fail to reject H 0.0088 0 0.0073 Mobil-Exxon t = = 1.904 Since −1.98<< t 1.98, fail to reject H0 0.0039 We do not reject the null hypothesis for any of the stocks. This indicates that the sample data is consistent with the conjecture from ec onomic theory that the intercept term equals 0. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 45 EXERCISE 3.8 (a) We set up the hypotheses H : β = 0 versus H : β < 0. The alternative β < 0 is chosen 0 2 1 2 2 because an inverse relationship is one wher e the dependent variable increases as the independent variable decreases, and visa versa. Thus, a negative β2suggests an inverse relationship between variables. The test statistic, given H0is true, is t t b2 ~ se(b ) (182) 2 The rejection region is−tt< (0.05,182) 1.653. The value of the test statistic is −194.233 − t = = 19.031 10.2061 Since t = −19.03 < −1.653, we reject the null hypothesis that β2= 0 and accept the alternative that β< 0. We conclude that there is a statistically significant inverse 2 relationship between the number of house starts and the 30-year fixed interest rate. (b) We set up the hypotheses H 0 2=− 150 versus H1 2 ≠− 150 . The test statistic, given H 0 is true, is t t b2 2 ~ se(b ) (182) 2 The rejection region is t <− 1.973 and t >1.973 , with t(0.975,182)973 . The value of the test statistic is −194.233 150 − t = = 4.334 10.2061 Since t = −4.334 < −1.973, we reject the null hypothesis β2=− 150 and accept the alternative that β2− 150 . The data indicate that, if the 30-year fixed interest rate increases by 1%, house starts will not fall by 150,000. (c) A 95% interval estimate of the slope from the regression estimated in part (a) is: −1±×4.=33 1.973 10.2061 ( 214.4, 174.1) This interval estimate suggests that, with 95 % confidence, an increase in the 30-year fixed interest rate by 1% will result in a drop in house starts of between 174,100 to 214,400 houses. We would be surprised if the true value of β d2d not lie in this interval. In part (b) we tested, at a 5% level of significance, whether β2=−150, and we came to the conclusion that β≠2− 150 . This conclusion is consiste nt with our interval estimate because at a 95% level of confidence, − 150 lies outside the interval. Remember the relationship between confidence intervals and hypothesis testing: At a ()α level of confidence and an α level of significance, we will no t reject a null hypothesis for a hypothesized value if it falls inside the confidence interval. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 46 EXERCISE 3.9 (a) We set up the hypotheses H : β 0 0 v2rsus H : β > 0.1The2alternative β> 0 is chosen2 because we assume that growth, if it does infl uence the vote, will do so in a positive way. The test statistic, given H 0s true, is b t t 2 ~ se(b2) (29) The rejection region is tt =.699 (0.95,29)The value of the test statistic is 0.6599 t = = 4.0460 0.1631 Since t = 4.0460 > 1.699, we reject the null hypothesis that β 2 0 and accept the alternative that β> 2 . We conclude that economic growth has a positive effect on the percentage vote. (b) A 95% interval estimate for β 2 from the regression in part (a) is: b2± = (0.975,29))2 0.6599 ± 2.045 × 0.1631 = (0.3264, 0.9934) This interval estimate suggests that, with 95% confidence, the true value of β is between 2 0.3264 and 0.9934. Since β2represents the change in percentage vote due to economic growth, we expect that a 1% increase in the gr owth rate will increase the percentage vote by an amount between 0.3264 to 0.9934 percent. (c) We set up the hypotheses H 0: β 2 0 versus H :1 β2< 0 . The alternative β<20 is chosen because we assume that inflation, if it does influence the vote, will do so in a negative way. The test statistic, given H 0is true, is b t t 2 ~ (29) se(b2) The rejection region is t t− 1.699= . The value of the test statistic is (0.05,29) −0.4450 − t = = 0.856 0.5197 Since −−0>.856 2.045 , we do not reject the null hypothesis. There is not enough evidence to suggest inflation has a negative effect on the vote. (d) A 95% interval estimate for β from the regression in part (c) is: 2 b2± = (0.975,29))2 −0.4450± 2.045 × 0.5197 = ( −1.508, 0.618) This interval estimate suggests that, with 95% confidence, the true value of β is between 2 −1.508 and 0.618. It suggests that a 1% increase in the inflation rate could increase or decrease or have no effect on the percentage vote. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 47 EXERCISE 3.10 (a) Tchoeefficient β2represents the increase in price from an extra square foot of living area. We can refer to it as the marginal price per square foot. (i) A 95% interval estimate of β 2or all houses is: b2± ×(0.975,1078)( )2 92.747 ± 1.962 × 2.4105 = (88.02, 97.48) We estimate, with 95% confidence, that the marginal price per square foot for all houses lies between $88.02 and $97.48. (ii) A 95% interval estimate of β 2or town houses is: bt± ×= b se( ) 55.585 ± 1.995 × 7.0999 = (41.42, 69.75) 2 (0.975,68) 2 We estimate, with 95% confidence, that the marginal price per square foot for town houses lies between $41.42 and $69.75. (iii)A 95% interval estimate of β for French style houses is: 2 b2± ×(0.975,95)e( 2 184.167 ± 1.985 × 10.1626 = (163.99, 204.34) We estimate, with 95% confidence, that the marginal price per square foot for French style houses lies between $163.99 and $204.34. These confidence interval estimates tell us th at town houses have a lower marginal price per square foot compared to the average, and also that French style houses have a much higher marginal price per square foot than all houses. Furthermore, we see that the narrowest confidence interval is that for all houses, reflecting the fact that the larger sample size provides more information, leading to a smaller standard error and more precise estimation. (b) The results for testing the hypotheses H 0: β2=80 versus H : β 1 802are given in the following table. In each case the test statistic is tb= ( −80 b) se( ) which has a t 2 2 (2 − distribution if H0 is true. The rejection region is tt< − cand tt> cwhere tc N (0.975−2). Sample t-value N − 2 tc Decision rule All t = =2.7473−80 5.29 1078 1.962 t > 1.962, reject H houses 2.4105 0 Town 55.5853−80 − t= = 3.44 68 1.995 t < −1.995 , rejectH 0 houses 7.0999 French 184.1667−80 Style t = = 10.25 95 1.985 t > 1.985, reject H 0 10.1626 All cases lead to the rejection of the null hypothesis. We conclude that an additional square foot does not add $80 to the average sale price of all houses, the sale price of town houses, nor the sale price of French style houses. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 48 EXERCISE 3.11 (a) For all houses in sample: Hypotheses: H00 2= against H0:1 28 ≠ Calculated t-value: t = 81.38=90 80) 1.9185 0.724 Critical t-value: ±±c (0.975,878)1.963 Decision: Do not reject H 0ecause −1.963< 0 <.724 1.963. We conclude that the data is consistent with the conjecture that an additional square foot of living space is associated with an increase in the sale price of the house by $80. (b) For houses that are vacant at time of sale: Hypotheses: H0:β8= against H0:β 8 ≠ 0 2 1 2 Calculated t-value: t =(69.90=80 80) 2.2675 4.45 Critical t-value: ±±c (0.975,463)1.965 Decision: Reject H 0ecause −4.45<− 1.965 We conclude that, for houses that are vacant at time of sale, an additional square foot of living space is not associated with an increase in the sale price of the house by $80. (c) For houses that are occupied at time of sale: Hypotheses: H00 2= against H01 28 ≠ Calculated t-value: t = 89.25=88 80) 3.0394 3.05 Critical t-value: ±±t ± 1.966 c (0.975,413) Decision: Reject H because 3.05 >1.966 . 0 We conclude that, for houses that are occupied at time of sale, an additional square foot of living space is not associated with an increase in the sale price of the house by $80. (d) For houses that are occupied at time of sale: Hypotheses: H00 2≤ against H01 28 > Calculated t-value: t = 89.25=88 80) 3.0394 3.05 Critical t-value: ct = (0.95,413).649 Decision: Reject H b0cause 3.05 >1.649 We conclude that, for houses that are occupied at time of sale, an additional square foot of living space increases the sale price of the house by more than $80. (e) For houses that are vacant at time of sale: Hypotheses: H00 2≥ against H01 28 < Calculated t-value: t =(69.90=80 80) 2.2675 4.45 Critical t-value: −ct = (0.05,463)1.648 Decision: Reject H b0cause −4.45<− 1.648 We conclude that, for houses that are vacant at time of sale, an additional square foot of living space increases the sale price by less than $80. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 49 Exercise 3.11 (continued) (f) (i) A 95% interval estimate for β 2 from the full sample is given by bt2 ×=(0.975,878)e( )2 81.389 ± 1.963 × 1.9185 = (77.62, 85.15) (ii) A 95% interval estimate for β 2or houses vacant at the time of sale is given by bt2 ×=(0.975,463)e( )2 69.908 ± 1.965 × 2.2675 = (65.45, 74.36) (iii) A 95% interval estimate for β 2or houses occupied at the time of sale is given by bt2 ×=(0.975,413)e( )2 89.259 ± 1.966 × 3.039 = (83.28, 95.23) Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 50 EXERCISE 3.12 (a) Estimateeduation: n WAGE =8 +.6658 0.0824EXPER (se)( ) ( )3787 0.0173 (t) ())2.88 4.77 The estimated equation tells us that with every year of experience the associated increase in hourly wage is $0.0824. Furthermore, it tells us that the average wage for those without experience is $8.6658. The relatively large t -values suggest that the least squares estimates are statistically significant at a 5% level of significance. 70 60 50 40 WAGE 20 10 0 0 10 20 30 40 50 60 EXPER Figure xr3.12(a) Fitted regression line and observations (b) Hypotheses: H :β 0= against H :β 0 > 0 2 1 2 The test statistic, given H0is true, is b t t 2 ~ (998) se(b2) Calculated t-value: t = (=.0824) 0.0173 4.769 Critical t-value: tt = = 1.646 c (0.95,998) Decision: Reject H because 4.769 >1.646 0 We conclude that the slope of the relationship, β , is statistically significant. There is a 2 positive relationship between the hourly wage and a worker’s experience. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 51 Exercise 3.12 (continued) (c) (i) For females, the estimated equation is: WAGE =8 +.4747 0.0209EXPER (se)( ) ( )4797 0.0218 ( ) (().67 0.958 With every extra year of experience the associated increase in average hourly wage for females is $0.0209. This estimate is not si gnificantly different from zero, however. The average wage for females without experience is $8.4747. 50 40 30 WAGE 20 10 0 0 5 10 15 20 25 30 35 40 45 EXPER Figure xr3.12(b) Fitted regression line and observations for females (c) (ii) For males, the estimated equation is: WAGE =8 +.8200 0.1448EXPER (se)( ) ( )5549 0.0254 ( ) (().89 5.698 With every extra year of experience, the associated increase in average hourly wage for males is $0.1448. The average wage for males without experience is $8.8200. 70 60 50 40 WAGE 20 10 0 0 10 20 30 40 50 EXPER Figure xr3.12(c) Fitted regression line and observations for males Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 52 Exercise 3.12(c) (continued) (c) (iii) For blacks, the estimated equation is: WAGE = 6 +.0054 0.1197EXPER () ( 0.997) 0.)461 () ( 6.022 ) (2.594 ) With every extra year of experience, the associated increase in average hourly wage for blacks is $0.1197. The average wage for blacks without experience is $6.0054. 28 24 20 16 12 WAGE 8 4 0 0 10 20 30 40 50 EXPER Figure xr3.12(d) Fitted regression line and observations for blacks (c) (iv) For white males, the estimated equation is: WAGE = 9 +.0315 0.1451EXPER (s( )( ).5808 0.0266 ( ) (()).55 5.452 With every extra year of experience the associated increase in average hourly wage for white males is $0.1451. The average wage for white males without experience is $9.0315. 70 60 50 40 WAGE 20 10 0 0 10 20 30 40 50 60 EXPER Figure xr3.12(e) Fitted regression line and observations for white males Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 53 Exercise 3.12(c) (continued) (c) Comparing the estimated wage equations for the four categories, we find that experience counts the most, or leads to the largest increase in wages, for white males. The effect is only slightly less for males in general. It is less for blacks and very small for females. For those with no experience the wage ranking is white males, males, females, blacks. (d) Residu pallots 50 40 30 20 RESID 10 0 -10 0 10 20 30 40 50 60 EXPER Figure xr3.12(f) Plotted residuals for full sample regression 40 30 20 RES10 0 -10 0 4 8 12 16 20 24 28 32 36 40 44 48 EXPER Figure xr3.12(g) Plotted residuals for female regression 50 40 30 20 RESID 0 -10 -20 0 10 20 30 40 50 60 EXPER Figure xr3.12(h) Plotted residuals for male regression Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 54 Exercise 3.12(d) (continued) (d) 20 15 10 5 RESID 0 -5 -10 0 10 20 30 40 50 60 EXPER Figure xr3.12(i) Plotted residuals for black regression 50 40 30 20 RESID0 0 -10 -20 0 10 20 30 40 50 60 EXPER Figure 3.12(j) Plotted residuals for white male regression The main observation that can be made from all the residual plots is that the pattern of positive residuals is quite different from the pattern of negative residuals. There are very few negative residuals with an absolute magnitude larger than 10, whereas the positive residuals are often larger than 10, with a few very large ones, and one over 40. These characteristics suggest a distribution of the e rrors that is not normally distributed, but skewed to the right. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 55 EXERCISE 3.13 (a) Estimateeduation: n WAGE =8 +.5837 0.0842EXPER (se) 0.1738 0.0078 ( ) ( ) ( ) 49.40 10.76 (()) With every extra year of experience the associated increase in hourly wage is $0.0842. The average wage for those without experience is $8.5837. The relatively large t-values imply the least squares estimates are statistically significant at a 5% level of significance. 80 70 60 50 40 WAGE 30 20 10 0 0 10 20 30 40 50 EXPER Figure xr3.13(a) Fitted regression line and observations using all data (b) Hypotheses: H 0 20= against H 1 20 > . The test statistic, given0H is true, is b t t 2 ~ (4731) se(b2 Calculated t-value: t = =.0842) 0.0078 10.76 Critical t-value: tt = = 1.645 c (0.95,4731) Decision: Reject H because t t 10.76 >= 1.645 0 c We conclude that the slope of the relationship, β2, is statistically significant. There is a positive relationship between the hourly wage and a worker’s experience. Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 56 Exercise 3.13 (continued) (c) (i) For females, the estimated equation is: WAGE =8 +.0375 0.0501EXPER (se( )( ).2285 0.0103 ( ) (()).18 4.856 With every extra year of experience the associated increase in average hourly wage for females is $0.0501. The average wage for females without experience is $8.0375. 80 70 60 50 40 WA30 20 10 0 0 10 20 30 40 50 EXPER Figure 3.13(b) Fitted regression line and observations for females (c) (ii) For males, the estimated equation is: WAGE = 9 +.1170 0.1153EXPER (se( ) ( )2510 0.0113 (t) 36.32 10.216 ( ()) With every extra year of experience the associated increase in average hourly wage for males is $0.1153. The average wage for males without experience is $9.1170. 80 70 60 50 40 W30E 20 10 0 0 10 20 30 40 50 EXPER Figure xr3.13(c) Fitted regression line and observations for males Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 57 Exercise 3.13(c) (continued) (c) (iii) For blacks, the estimated equation is: WAGE = 7 +.3825 0.0667EXPER (se( ) ( )5002 0.0233 ( ) (()).76 2.860 With every extra unit of experience the associated increase in average hourly wage for blacks is $0.0667. The average wage for blacks without experience is $7.3825. 40 30 20 WAGE 10 0 0 10 20 30 40 50 EXPER Figure xr3.13(d) Fitted regression line and observations for blacks (c) (iv) For white males, the estimated equation is: n WAGE = 9 +.2606 0.1164EXPER (se( )( ).2644 0.0118 ( ) (()).02 9.847 With every extra year of experience the associated increase in average hourly wage for white males is $0.1164. The average wage for white males without experience is $9.2606. 80 70 60 50 40 WAGE 30 20 10 0 0 10 20 30 40 50 60 EXPER Figure xr3.13(e) Fitted regression line and observations for white males Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 58 Exercise 3.13(c) (continued) (c) Comparing the estimated wage equations fo r the four categories, we find that experience counts the most, or leads to the largest increase in wages, for white males. The effect is only slightly less for males in general. For bl acks experience is worth slightly more than half of what it is for white males. For females experience is worth slightly less than half of what it is for white males. For those with no experience the wage ranking is white males, males, females, blacks. (d) Residu pallots 70 60 50 40 30 20 RESID 10 0 -10 -20 0 10 20 30 40 50 60 EXPER Figure xr3.13(f) Plotted residuals for full sample regression 70 60 50 40 30 RESID 20 10 0 -10 0 10 20 30 40 50 60 EXPER Figure xr3.13(g) Plotted residuals for female regression Chapter 3, Exercise Solutions, Principles of Econometrics, 3e 59 Exercise 3.13(d) (continued) (d) 70 60 50 40 30 RESID 10 0 -10 -20 0 10 20 30 40 50 60 EXPER Figure xr3.13(h) Plotted residuals for male regression 40 30 20 RES10 0 -10 0 10 20 30 40 50 60 EXPER Figure xr3.13(i) Plotted residuals for black regression 70 60 50 40 30 RESID 10 0 -10 -20 0 10 20 30 40 50 60 EXPER Figure xr3.13(j) Plotted residuals for white male regression In all residual plots the pattern of positive re siduals is quite different from the pattern of negative residuals. There are very few negativ e residuals with an absolute magnitude larger than 10, whereas the positive residual s are often larger than 10, with a few very large ones, and one over 40. These characteristics suggest a distribution of the errors that is not normally distributed, but skewed to the right. CHAPTER 4 Exercise Solutions 60 Chapter 4, Exercise Solutions, Principles of Econometrics,61e EXERCISE 4.1 2 2 ∑ ei 182.85 (a) R =−1 =2 1 = 0.71051 ∑ ()i− 631.63 2 (b) colculate R2 we need ∑ ()i− y , 2 2 2 2 ∑ ∑()i−i y−y N= −× =930.94 20 16.035 788.5155 Therefore, 2 SSR 666.72 R = = = 0.8455 SST 788.5155 (c) From e2 (NK − σ2 R =− ∑ =− SST SST hawee, 2 SS−T(1− ×2) 552.36 (1 0.7911) σ= = = 6.4104 NK− − (20 2) Chapter 4, Exercise Solutions, Principles of Econometrics, 3e 62 EXERCISE 4.2 ˆ ∗ (a) y = 5.83 8.69 x where x =∗ x (1.23) (1.17) 10 y = +0.583 0.0869 x y (b) where y = (0.123) (0.0117) 10 y = +0.583 0.869 x ∗ ∗y x (c) where y x and = (0.123) (0.117) 10 10 2 vTlues R remain the same in all cases. Chapter 4, Exercise Solutions, Principles of Econometrics, 3e EXERCISE 4.3 (a) y0 1 02 =+×= 115 6 ) 1 5 ( ⎛ ⎞ 1 1()x− 2 ⎛ ⎞ − 2 (b) var( f =σ ˆ2⎜ ⎟ + 0 2= 5.3333 ⎜ ⎟ = 14.9332 ⎝ ⎠ Nxx ∑ ( i − 0 51 ⎝ ⎠ se( f ) = =14.9332 3.864 (c) Using se( f )from part (b) antt = = 3.182, c (0.975,3) −yt±=f=se( ) 6 3.182 3.864 ( 6.295,18.295) 0 c (d) Using se( f )om part (b) and tt = = 5.841, c (0.995,3) −y0=fcse( ) 6 5.841 3.864 ( 16.570,28.570) (e) Using xx== 0 1, the prediction iy0 =111 2 , and 2 2 ) 1 1 n( 2 ⎛ ⎞ 1 1 ()0− ⎛ ⎞ − var( f ) =σ ⎜ ⎟ Nxx ( )− 2= 50 513+1 ⎜ ⎟ = 6.340 ⎝ ⎠ ∑ i ⎝ ⎠ se( f ) = =6.340 2.530 yt0±fcse( ) 2 3.182 2.530 ( 6.050,10.050) Width in part (c)−=−8.295 ( )6.295 24.59 Width in part (e)−=−0.050 ( )6.050 16.1 The width in part (e) is smaller than the width in part (c), as expected. Predictions are more precise when made for x values close to the mean. Chapter 4, Exercise Solutions, Principles of Econometrics, 3e 64 EXERCISE 4.4 (a) Whe entimating E,y)0 we are estimating the average value of y for all observational units with an x-value of x0. When predicting y , 0e are predicting the value of y for one observational unit with an x-value of x . The first exercise does not involve the random 0 error e0; the second does. (b) ( ) E((b102b +1xβ = 201 02 var(bx+ = + xvar(+ ) x 2 bar( ) 2 cov( , ) 102 1 0 2 0 12 2 2 22 2 = + −σ ∑ xi x σ σx0x 0 2 Nxx ( ) (x)x ( ) xx 2 ∑ ∑ ∑i i i 2 2 2 σ+−(∑) ()i xN σ−2(20 0 = + 2 2 Nx∑ ∑) i− i (xx ⎛ ⎞ 2 ⎛ 12 1xx 2 ( xx 2 − ⎜ ⎟=σ ⎜ ⎟+ 0 0 2 =σ2 + ⎝ ⎠ ⎝ ⎠x xNx∑ ∑ i − −i ( ) (c) It is not appropriate to say thaE yy0 0 because y0 is a random variable. y yex=βx+β ≠ β +β + = [ 0 10 2 10 02 0 [ ] We need to include y 0n the expectation so that β =+ β −ˆ0 0 + =0− =)E0 1 021 02 x0 x(()E.e0 ) Chapter 4, Exercise Solutions, Principles of Econometrics, 3e 65 EXERCISE 4.5 (a) If we multiply the x values in the simple linear regression model y =β+β +xe by 10, 1 2 the new model becomes ⎛⎞ e=β +1 ⎜⎟2 () 10+ ⎝⎠0 =β +1 2 *x+ * ∗ wher2 2β =β 10 a xnx = × 10 The estimated equation becomes b y=+ ⎛⎞ x× () 10 1 ⎝⎠10 Thus, β and b do not change and β and b becomes 10 times smaller than their 1 1 2 2 original values. Since e does not change, the variance of the error term var(e) =σ 2 is unaffected. (b) Multiplying all the y values by 10 in the simple linear regression model y =β 1 2x +e gives the new model e y×1x0 ( ())×10 10 ( 10 ) 1 2 or ∗ ∗ * * y =β 1 2x +e where yy=× 10, ββ × * *β=×10, =× 10, ∗ 10 1 1 2 2 The estimated equation becomes ∗ y =× y =10bx()1 210 ( 10 ) Tbots, β1and β ar2 affected. They are 10 times larger than their original values. Similarly, b1and b 2re 10 times larger than their original values. The variance of the new error term is ∗ 2 var(e ) ==ar() 1e0 100 var( ) 100 Thus, the variance of the error term is 100 times larger than its original value. Chapter 4, Exercise Solutions,Principles of Econometrics, 3e 66 EXERCISE 4.6 (a) The least squares estimator for1β is1 2 − . Thus, y = 1 2 , and hence (,x ) lies on the fitted line. ˆ (b) Consider the fitted linei ib1 2 . Averaging over N, we obtain y = ∑ yi= 1 1 ( )x =N bb x +b( ) =+bb =+ ∑ i NN ∑ ∑ 1 Ni i 1 2 1 2 N 1 2 From part (a), we also havey = b1 2 . Thus, y = y . Chapter 4, Exercise Solutions,Principles of Econometrics, 3e 67 EXERCISE 4.7 (a) y = bx 002 (b) Using the solution from Exercise 2.4 part (f) SSE = = e ˆ +(2.0659 + 2.1319 1.1978 0

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "I bought an awesome study guide, which helped me get an A in my Math 34B class this quarter!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.