Midterm_W2015_solutions - Midterm Exam ECON 3210 O Use of Economic Data Laura Salisbury York University This exam contains 4 questions and 8

# Midterm_W2015_solutions - Midterm Exam ECON 3210 O Use of...

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Midterm ExamECON 3210 O: Use of Economic DataLaura SalisburyYork UniversityFebruary 12, 2015This exam contains 4 questions and 8 pages (including this one), for a total of 70 points.Read questions carefully, and answer all questions in the booklets provided.Formulas andtables are provided on the last three pages of the exam. You may use a calculator. You have90 minutes. Good luck!1
1. You are studying the relationship between time spent preparing for exams (TIME) andGPA. You collect data from four of your friends on average time spent studying for asingle exam (in hours) and GPA (on a scale from 0 to 4). You use this data to estimatethe following econometric model:GPA=β1+β2TIME+eHere is the data you collect:TIMEGPA103.921.752.993.5(a)(5 pts)Calculateb2, and interpret its meaning.I’ll useTas shorthand for TIME andGas shorthand for GPA. Start by calculatingT= (10+2+5+9)/4 = 6.5andG= (3.9+1.7+2.9+3.5)/4 = 3. Then, the easiestway to do this is to create the following table:TG(T-T)(G-G)(T-T)(G-G)(T-T)2103.93.50.93.1512.2521.7-4.5-1.35.8520.2552.9-1.5-0.10.152.2593.52.50.51.256.254i=110.441Now you have enough information to calculateb2:b2=4i=1(T-T)(G-G)4i=1(T-T)2=10.441= 0.2537In words, the meaning ofb2is the following:if you increase average time spentstudying by 1 hour, this is expected to increase your GPA by 0.2537 points.(b)(5 pts)Calculateb1and interpret its meaning.You calculatedb2,TandGin the previous part of the question.This gives youenough information to calculateb1:b1=G-b2T= 3-0.2537(6.5) = 1.3512The interpretation is this coefficient is:if you study for 0 hours, your GPA isexpected to be 1.3512.2
(c)(5 pts)Based on the estimates calculated in parts (a) and (b), it turns out that4i=1ˆe2i= 0.1220. Use this information and the data provided to calculate\var(b2).The formula for\var(b2)is:\var(b2) =ˆσ2Ni=1(Ti-T)2You have already computedNi=1(Ti-T)2= 41. To computeˆσ2, use the informationyou have just been given (4i=1ˆe2i= 0.1220) and the formula forˆσ2:ˆσ2=4i=1ˆe2iN-2=0.12202= 0.061So:\var(b2) =ˆσ2Ni=1(Ti-T)2=0.06141= 0.001488(d)(5 pts)Perform a two-tail test of significance ofβ2at the 5% level. What do youconclude?You are testingH0:β2= 0againstH1:β26= 0. Your test statistic is:t=b2-0se(b2).Because this is a two-sided test at the 5% level, you will rejectH0ift > t0.975,N-2ort <-t0.975,N-2. Recall thatN-2 = 2. Reading from the t-table provided, noticethatt0.975,2= 4.303. So, you will rejectH0ift >4.303ort <-4.303. To computeyour test statistic, you need to find the standard error ofb2, which is just equal toq\var(b2) =0.001488 = 0.0386. So, the value of your test statistic is:t=b2se(b2)=0.25370.0386= 6.5762This is greater than 4.303, so it is in the rejection region. Thus, you rejectH0. Youconclude thatβ2is significant at the 5% level.
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