Midterm1PracticeShortAnswer：1个简短的回答中的实践(可编辑)

Midterm 1 I Practice -ShortAnswer: 1个简短的回答中的实 践（可编辑）（文档可以直接使用，也可根据实际需要修改使用 ，可编辑推荐下载）Practice Questi ons for Exam 11. The height of male stude nts at your college/u ni versity is no rmally distributed with a mean of 70 in ches and a sta ndard deviati on of 3.5 in ches. If you had a list of teleph one n umbers for male stude nts for the purpose of con duct ing a survey, what would be the probability of ran domly call ing one of these stude nts whose height is: taller than 60?(b) between 53 and 65?(c) shorter tha n 57, the mean height of female stude nts?(d) shorter than 50?(e) taller than Shaquille ONeal, the cen ter of the Bost on Celtics, who is 71 tall?Compare this to the probability of a woma n being preg nant for 10 mon ths (300 days), where days of pregnancy is normally distributed with a mean of 266 days and a standard deviation of 16 days.An swer:(a) Pr(Z 0.5714) = 0.2839;(b) Pr( Z 2) = 0.9545 or approximately 0.95;(c) Pr(Z 2.99 =(d) Pr(Z 4.2857) = 0.000009 (the text does not show values above 2.99 standard deviations, Pr(Z 0.0014) and Pr(Z 2.1250) = 0.0168.2) Adult males are taller, on average, than adult females. Visiting two recent American Youth Soccer Organi zati on (AYSO) un der 12 year old (U12) soccer matches on a Saturday, you do not observe an obvious differe nee in the height of boys and girls of that age. You suggest to your little sister that she collect data on height and gen der of childre n in 4 th to 6th grade as part of her scie nee project. The accompa nying table shows her findin gs.Height of You ng Boys and Girls, Grades 4-6, in in chesBoysGirls召够!i| SGids57.83.95558.44.257(a) Let your n ull hypothesis be that there is no differe nee in the height of females and males at this age level. Specify the alter native hypothesis.(b) Find the differe nee in height and the sta ndard error of the differe nee.(c) Gen erate a 95% con fide nee in terval for the differe nee in height.(d) Calculate thet-statistic for comparing the two means. Is the differenee statistically significant at the 1%level? Which critical value did you use? Why would this n umber be smaller if you had assumed a one sided alter native hypothesis? What is the in tuiti on beh ind this?An swer:(a) H0 :也门护-叱袱=0 vs. H 1 :戸羯戸-叱漁 工0I 2 I 77 T! 3.94.2(b) 、晒-Y晟=-0.6, SE( 碍忻-、G泳)= .77.(c) -0.6 1.96 X0.77 = (-2.11, 0.91).(d) t = -0.78, so t 40.(a) Given your knowledge about the population, find the probability that in a random sample of 40, Bush would receive a share of 40% or less.(b) How would this situation change with a random sample of 100?(c) Give n your an swers in (a) and (b), would you be comfortable to predict what the voti ng inten ti ons forthe entire population are if you did not knowp but had polled 10,000 individuals at random andcalculated ? ? Expla in.(d) This result seems to hold whether you poll 10,000 people at random in the Netherlands or the UnitedStates, where the former has a population of less than 20 million people, while the United States is 15times as populous. Why does the population size not come into play?An swer:0.40 0.50 Pr( p 0.40) = Pr(Z =) = Pr(Z -1.26) 0.104. In roughly every 10 th sample of this size,摩 40Bush would receive a vote of less tha n 40%, although in truth, his share is 50%.c0.40 0.50O.25 I 100(b) Pr( ? 0.40) = Pr(Z ) = Pr(Z 45$488.87$328.64507Age 45$412.20$276.631237Test whether or not the differe nee in average earnings is statistically sig nifica nt. Give n your kno wledge of age -earning profiles, does this result make sense?An swer:294.67(a) The con fide nee in terval for mea n weekly earni ngs is 434.492.58 x = 434.49 18.20心744=(416.29, 452.69). Based on the sample at hand, the best guess for the population mea n is $434.49.However, because of ran dom sampli ng error, this guess is likely to be wrong. In stead, the in terval estimate for the average earnings lies between $416.29 and $452.69. Committing to such an interval repeatedly implies that the result ing stateme nt is in correct 1 out of 100 times. For a 90% con fide nce in terval, the only cha nge in the calculati on of the con fide nce in terval is to replace 2.58 by 1.64. Hence the con fide nce in terval is smaller. A smaller in terval implies, give n the same average ear nings and the sta ndard deviati on, that the stateme nt will be false more ofte n. The larger the con fide nce in terval, the more likely it is to contain the population value.488.87 412.20(b) Assum ing un equal populati on varia nces,t = 2= 4.62, which is statistically1328.64276.6350712.7sig nifica nt at conven ti on al levels whether you use a two-sided or one -sided alter native. Hence the n ullhypothesis of equal average earnings in the two groups is rejected. Age-ear ning profiles typically take onan in verted U -shape. Maximum earnings occur in the 40s, depe nding on some other factors such as years of educati on, which are not con sidered here. Hence it is not clear if the alter native hypothesis should be one -sided or two -sided. In such a situati on, it is best to assume a two -sided alter native hypothesis.5) Sir Fra ncis Galt on, a cous in of James Darwin, exam ined the relati on ship betwee n the height of childre n and their parents towards the end of the 19 th century. It is from this study that the name regression originated. You decide to update his findings by collecting data from 110 college students, and estimate the followi ng relati on ship:;也=19.6 + 0.73 XMidparh , R2 = 0.45, SER = 2.0where Studenth is the height of students in inches, and Midparh is the average of the parental heights. (Followi ng Galt ons methodology, both variables were adjusted so that the average female height was equal to the average male height.)(a) In terpret the estimated coefficie nts.(b) What is the meaning of the regressionR2?(c) What is the predicti on for the height of a child whose pare nts have an average height of 70.06 in ches?(d) What is the interpretation of the SER here?(e) Given the positive intercept and the fact that the slope lies between zero and one, what can you say about the height of students who have quite tall parents? Those who have quite short parents?An swer:(a) For every one inch in crease in the average height of their pare nts, the stude nts height in creases by 0.73 of an in ch. There is no reas on able in terpretati on for the in tercept.(b) The model expla ins 45 perce nt of the variati on in the height of stude nts.(c) 19.6 + 0.73 X70.06 = 70.74.(d) The SER is a measure of the spread of the observations around the regression line. The magnitude of the typical deviation from the regression line or the typical regression error here is two inches.(e) Tall parents will have, on average, tall students, but they will not be as tall as their parents. Short pare nts will have short stude nts, although on average, they will be somewhat taller tha n their pare nts.6) The baseball team n earest to your home tow n is, once aga in, not doing well. Give n that your knowledge of what it takes to win in baseball is vastly superior to that of management, you want to find out what it takes to win in Major League Baseball (MLB). You therefore collect the winning percentage of all 30 baseball teams in MLB for 1999 and regress the winning perce ntage on what you con sider the primary determ inant for wins, which is quality pitch ing (team ear ned run average). You find the follow ing in formati on on team performa nee:Summary of the Distributi on of Winning Perce ntage andTeam Earned Run Average for MLB in 1999AverageStan dard deviationPerce ntile10%25%40%50% (media n)60%75%90%TeamERA4.710.533.844.354.724.784.915.065.25Winning Perce ntage0.500.080.400.430.460.480.490.590.60(a) What is your expected sig n for the regressi on slope? Will it make sense to in terpret the in tercept? Ifnot, should you omit it from your regression and force the regression line through the origin?(b) OLS estimati on of the relati on ship betwee n the wi nning perce ntage and the team ERA yield the followi ng:隠:品=0.9 -0.10 Xteamera , R2=0.49, SER = 0.06,where winpct is measured as wins divided by games played, so for example a team thatwon half of itsgames would have Winpet = 0.50. In terpret your regressi on results.(c) It is typically sufficient to win 90 games to be in the playoffs and/or to win a division. Winning over100 games a seas on is excepti on al: the Atla nta Braves had the most wins in 1999 with 103. Teams play atotal of 162 games a year. Given this information, do you consider the slope coefficient to be large or small?(d) What would be the effect on the slope, the in tercept, and the regressi onR2 if you measured Win pct inperce ntage poin ts, i.e., as (Win s/Games) X100?(e) Are you impressed with the size of the regressi onR2? Give n that there is 51% of un expla ined variati onin the winning perce ntage, what might some of these factors be?An swer:(a) You expect a n egative relati on ship, since a higher team ERA implies a lower quality of the in put. No team comes close to a zero team ERA, and therefore it does not make sense to in terpret the in tercept. Forc ing the regressi on through the orig in is a false implicati on from this in sight. I nstead the in tercept fixes the level of the regressi on.(b) For every one point in crease in Team ERA, the wi nning perce ntage decreases by 10 perce ntage poi nts, or 0.10. Roughly half of the variati on in winning perce ntage is expla ined by the quality of team pitch ing.(c) The coefficie nt is large, since in creas ing the winning perce ntage by 0.10 is the equivale nt of winning 16 more games per year. Since it is typically sufficient to win 56 percent of the games to qualify for the playoffs, this difference of 0.10 in winning percentage turns can easily turn a loosing team into a winning team.(d) Clearly the regressi onR2 will not be affected by a cha nge in scale, since a descriptive measure of thequality of the regression would depend on whim otherwise. The slope of the regression will compensatein such a way that the interpretation of the result is unaffected, i.e., it will become 10 in the above example. The in tercept will also cha nge to reflect the fact that if X were 0, the n the depe ndent variable would now be measured in percentage, i.e., it will become 94.0 in the above example.(e) It is impressive that a sin gle variable can expla in roughly half of the variati on in wi nning perce ntage. An swers to the sec ond questi on will vary by stude nt, but will typically in clude the quality of hitt ing, field ing, and man ageme nt. Salaries could be in cluded, but should be reflected in the in puts.7) In 2001, the Arizo na Diamo ndbacks defeated the New York Ya nkees in the Baseball World Series in 7 games. Some players, such as Bautista and Finley for the Diam on dbacks, had a substa ntially higher batt ing average dur ing the World Series tha n duri ng the regular seas on. Others, such as Brosius and Jeter for the Yan kees, did substa ntially poorer. You set out to in vestigate whether or not the regular seas on batting average is a good indicator for the World Series batting average. The results for 11 players who had the most at bats for the two teams are:* “匚a * 二= -.347 + 2.290 AZSeasavg , R2=0.11, SER = 0.145,AZWirt? = 0.134 + 0.136 NYSeasavg , R2=0.001, SER = 0.092,where Wsavg and Seasavg in dicate the batt ing average dur ing the World Series and the regular seas on respectively.(a) Focus ing on the coefficie nts first, what is your in terpretati on?(b) What can you say about the expla natory power of your equati on? What do you con clude from this?An swer:(a) The two regressi ons are quite differe nt. For the Diam on dbacks, players who had a 10 point higher batt ing average dur ing the regular seas on had roughly a 23 point higher batt ing average duri ng the World Series. Hence top performers did relatively better. The opposite holds for the Yankees.(b) Both regressi ons have little expla natory power as see n from the regressi onR2. Hence performa nceduri ng the seas on is a poor forecast of World Series performa nce.8) You have obta ined a sample of 14,925 in dividuals from the Curre nt Populati on Survey (CPS) and are in terested in the relati on ship betwee n average hourly earnings and years of educati on. The regressi on yields the follow ing result:ahfe=-4.58 + 1.71 Xeduc , R2 = 0.182, SER = 9.30where ahe and educ are measured in dollars and years respectively.a.Interpret the coefficients and the regressionR2.b. Is the effect of educati on on earnings large?c. Why should educati on matter in the determ in ati on of earnin gs? Do the results suggest that there is a guara ntee for average hourly ear nings to rise for every one as they receive an additi onal year of educati on? Do you thi nk that the relati on ship betwee n educati on and average hourly ear nings is lin ear?d. The average years of educati on in this sample is 13.5 years. What is mean of average hourly earnings in the sample?e. In terpret the measure SER. What is its un it of measureme nt.An swer:a. A pers on with one more year of educati on in creases her ear nings by $1.71. There is no meaning attached to the in tercept, it just determ ines the height of the regressi on. The model expla ins 5 perce nt of the variation in average hourly earnings.b. The differe nee betwee n a high school graduate and a college graduate is four years of educati on. Hence a college graduate will ear n almost $7 more per hour, on average ($6.84 to be precise). If you assume that there are 2,000 work ing hours per year, the n the average salary differe nee would be close to $14,000 (actually $13,680). Depe nding on how much you have spe nt for an additi onal year of educati on and how much in come you have forgone, this does n ot seem particularly large.c. In gen eral, you would expect to find a positive relati on ship betwee n years of educati on and average hourly earnings. Education is considered investment in human capital. If this were not the case, then itwould be a puzzle as to why there are stude nts in the econo metrics course surely they are not there tojust find themselves (which would be quite expensive in most cases). However, if you consider educati on as an in vestme nt and you wan ted to see a return on it, the n the relati on ship will most likely n ot be lin ear. For example, a con sta nt perce nt return would imply an exp onen tial relati on ship whereby the additi onal year of educati on would bring a larger in crease in average hourly earnings at higher levels of educati on. The results do not suggest that there is a guara ntee for earnings to rise for every one as they become more educated since the regressi onR2 does not equal 1. In stead the result holds on average.d. Si nee= Y - X ? Y =+ 角 X . Substituti ng the estimates for the slope and the in tercept the nresults in a mean of average hourly ear nings of roughly $18.50.e. The typical prediction error is $9.30. Since the measure is related to the deviation of the actual and fitted values, the unit of measurement must be the same as that of the dependent variable, which is in dollars here.9) You have obtained measurements of height in inches of 29 female and 81 male students (Studenth ) atyour uni versity. A regressi on of the height on a con sta nt and a binary variable (BFemme), which takes avalue of one for females and is zero otherwise, yields the following result:71.0 -4.84 XBFemme , R2 = 0.40, SER = 2.0 (0.3) (0.57)(a) What is the interpretation of the intercept? What is the interpretation of the slope? How tall are females, on average?(b) Test the hypothesis that females, on average, are shorter tha n males, at the 1% level.(c) Is it likely that the error term is homoskedastic here?An swer:(a) The in tercept gives you the average height of males, which is 71 in ches in this sample. The slope tells you by how much shorter females are, on average (almost 5 in ches). The average height of females is therefore approximately 66 in ches.(b) The t-statistic for the difference in means is -8.49. For a one -sided test, the critical value is-2.33. Hencethe differe nce is statistically sig nifica nt.(c) It is safer to assume that the varia nces for males and females are differe nt. In the un derly ingsample thesta ndard deviati on for females was smaller.10) You have collected 14,925 observations from the Current Population Survey. There are 6,285 females in the sample, and 8,640 males. The females report a mean of average hourly earnings of $16.50 with a standard deviation of $9.06. The males have an average of $20.09 and a standard deviation of $10.85. The overall mean average hourly ear nings is $18.58.a. Using the t-statistic for testing differences between two means (section 3.4 of your textbook), decide whether or not there is sufficient evidence to reject the null hypothesis that females and males have iden tical average hourly ear nin gs.b. You decide to run two regressions: first, you simply regress average hourly earnings on