Midterm 1 - Solutions

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1 Ec Aalyi of Ecoomic Data Uiverity of Califoria - Davi Jauary 28, 2010 Itructor: Joh Parma Midterm 1 - Solutio You have util 10:20am to complete thi exam. Pleae remember to put your ame, ectio ad ID umber o both your catro heet ad the exam. Fill i tet form A o the catro heet. Awer all multiple choice quetio o your catro heet. Chooe the igle bet awer for each multiple choice quetio. Awer the log awer quetio directly o the exam. Keep your awer complete but cocie. For the log awer quetio, you mut how your work where appropriate for full credit. Name: ID Number: Sectio: (POTENTIALLY) USEFUL FORMULAS x = 1 i=1 x i 2 = 1 1 i=1 (x i x) 2 CV = x kew = kurt = µ = E(X) z = x µ σ t = x µ ( 1)( 2) i=1 ( x i x ) 3 (+1) ( 1)( 2)( 3) i=1 ( x i x ) 4 3( 1)2 ( 2)( 3) P r[t 1 > t α, 1 ] = α P r[ T 1 > t α 2, 1 ] = α i=1 a = a i=1 (ax i) = a i=1 x i i=1 (x i + y i ) = i=1 x i + i=1 y i 2 = x(1 x) for proportio data t α, 1 = T INV (2α, 1) P r( T 1 t ) = T DIST ( t, 1, 2) P r(t 1 t ) = T DIST (t, 1, 1) (POTENTIALLY) USEFUL EXCEL OUPUT TINV(.005,999)=2.81 TINV(.01,999)=2.58 TINV(.02,999)=2.33 TINV(.025,999)=2.24 TINV(.05,999)=1.96 TINV(.10,999)=1.65 TINV(.20,999)=1.28 TINV(.005,99)=2.87 TINV(.01,99)=2.63 TINV(.02,99)=2.36 TINV(.025,99)=2.28 TINV(.05,99)=1.98 TINV(.10,99)=1.66 TINV(.20,99)=1.29

2 2 Midterm 1 - Solutio SECTION I: MULTIPLE CHOICE (60 poit) 1. Suppoe we take a ample of 400 people ad meaure their height. If we meaure height i iche, the coefficiet of variatio will be: (a) Larger tha if we meaure height i meter. (b) Smaller tha if we meaure height i meter. (c) The ame a whe we meaure height i meter. (d) (a), (b) or (c) could be true depedig o the value of the ample mea. (c) Chagig the uit of meauremet will recale the ample mea ad the ample tadard deviatio by the ame factor. Whe dividig the ew ample mea by the ew tadard deviatio, thi calig factor will cacel out ad we will be left with the origial coefficiet of variatio. 2. Which of the followig i ot a radom variable? (a) The ample mea. (b) The ample variace. (c) The populatio variace. (d) All of the above are radom variable. (c) The populatio variace i a cotat parameter. The ample mea ad ample variace ca take o differet value for differet ample. 3. Suppoe we ca reject the ull hypothei that µ 50 at the 5% igificace level. Which of the followig tatemet i defiitely true? (a) We ca reject the ull hypothei that µ = 50 at the 5% igificace level. (b) We ca reject the ull hypothei that µ 50 at the 10% igificace level. (c) The value we obtaied for t wa poitive. (d) The value we obtaied for t wa greater tha the critical value for a upper oe-tailed hypothei tet at the 5% igificace level. (b) If we rejected the ull hypothei that µ 50 at the 5 igificace level, our value for t mut have bee le tha t.05, 1. The magitude of t.1, 1 i maller tha the magitude of t.05, 1 o it will alo be the cae that t < t.1, The ample variace of 100 obervatio of mothly uemploymet rate will be the ample variace of 100 obervatio of the uemploymet rate coverig the ame time period for which each moth uemploymet rate i a average of that moth rate ad the previou two moth rate. (a) Le tha or equal to. (b) Le tha. (c) Equal to. (d) Greater tha or equal to. (d) I geeral, uig a movig average will ted to reduce the amout of variace i a data erie. There are ome pecial cae where it will leave the variace uchaged (for example, thik about a ample with zero variace).

3 Midterm 1 - Solutio 3 5. Which of the followig i ot a meaure of cetral tedecy? (a) The ample rage. (b) The ample media. (c) The ample midrage. (d) The ample mea. (a) The ample rage i a meaure of diperio. 6. Suppoe we have data o the iteret rate, i t. Which of the followig could we ue to get the percet chage i the iteret rate from year t to year t + 1 (expreed a a decimal)? (a) i t+1 i t. (b) l(i t+1 i t ). (c) i t+1 i t. (d) l(i t+1 ) l(i t ). (d) Percet chage i a variable ca be approximated by a differece log. 7. Icreaig the ample ize ued for a two-tailed hypothei tet will ted to: (a) Decreae the probability of a Type I error. (b) Decreae the probability of a Type II error. (c) Icreae the probability of a Type I error. (d) Icreae the probability of a Type II error. (b) The probability of a Type I error i equal to the value of the igificace level α. Icreaig the ample ize will chage the critical value that correpod to the igificace level of α ad the rage of ample mea for which we would get a Type I error but the probability of a Type I error will till be α. The probability of a Type II error ca be reduced by icreaig the ample ize which will decreae the variace of the ditributio of the ample mea. 8. Suppoe that we have data o coi flip. The variable X i equal to oe if the coi flip i head ad zero if the coi flip i tail. I a ample of 100 coi flip, the ample mea of X tur out to be.6. The value of the kewe for the ample will be: (a) Poitive. (b) Negative. (c) Zero. (d) Not eough iformatio. (b) The ig of the kewe will deped o the ig of (x i x) 3. Note that for 60 of our obervatio, the value of (x i x) 3 i (1.6) 3, or.064. For the other 40 obervatio, the value of (x i x) 3 i (0.6) 3, or So (x i x) 3 i equal to which i Which of the followig tatemet i ot true? (a) Two radom variable ca have the ame mea but differet media. (b) Two radom variable ca have the ame mea but differet variace. (c) The mea of the um of two radom variable i alway equal to the um of their mea.

4 4 Midterm 1 - Solutio (d) The variace of the um of two radom variable i alway equal to the um of their variace. (d) The variace of the um of two radom variable will be equal to the um of their variace plu a additioal term depedet o the covariace of the two variable. 10. If the ditributio of X i ymmetric ad reache it highet poit at the exact middle of the ditributio: (a) The mode of X will be equal to the media of X. (b) The mea of X will be equal to the media of X. (c) The media of X will be at the exact middle of the ditributio. (d) All of the above are true. (d) If the ditributio i ymmetric, the mea ad media will both be at the exact middle of the ditributio. If the highet poit of the ditributio i at the exact middle, the the mode i at the middle of the ditributio. 11. Which of the followig would icreae the probability of a Type I error? (a) Decreaig the ample ize. (b) Icreaig the ample ize. (c) Decreaig the igificace level. (d) Icreaig the igificace level. (d) The probability of a Type I error i equal to the igificace level. Icreaig the igificace level will icreae the probability of Type I error. 12. Suppoe that the 95% cofidece iterval for the populatio mea hour of tudyig per week i (6.5, 6.9). Which of the followig tatemet i true? (a) You would reject the ull hypothei that µ 6.5 at the 10% igificace level. (b) You would reject the ull hypothei that µ = 7.1 at the 5% igificace level. (c) You would reject the ull hypothei that µ 6.9 at the 5% igificace level. (d) All of the above are true. (b) Note that 7.1 fall outide of the 95% cofidece iterval. That tell u that we would reject the ull hypothei that µ = 7.1 at a 5% igificace level. 13. Suppoe you have a dataet of the uemployemet rate for 100 differet citie for the moth of December, with oe obervatio per city. Thee are: (a) Cro-ectioal data. (b) Pael data. (c) Time-erie data. (d) Both (a) ad (c). (a) Thee are cro-ectioal data. We are obervig a cro-ectio of citie at a igle poit i time.

5 Midterm 1 - Solutio Suppoe that rather tha the ample mea, we ue aother tatitic that we ll call X to etimate the populatio mea. If the ditributio of X i ymmetric, what mut be true for X to be a ubiaed etimator for the populatio mea? (a) The ditributio of X mut get arrower a the ample ize icreae. (b) The variace of X hould ot deped o the ample ize. (c) The ditributio of X mut be cetered at the populatio mea. (d) The value of X hould approach the populatio mea a the ample ize goe to ifiity. (c) To be a ubiaed etimator of the populatio mea, the expected value of X ha to be equal to the populatio mea. Ue the figure above to awer quetio 15 through 17. The graph o the left how the ditributio of radom variable X ad the graph o the right how the ditributio of radom variable Y. 15. The media of X i ad the media of Y i. (a) Poitive, poitive. (b) Poitive, egative. (c) Negative, egative. (d) Not eough iformatio. (a) The media i the value at which the area uder to curve to the left of the value i 50%. From the graph, we ca tell that thi i a poitive value i both cae. 16. Which of the followig tatemet i true? (a) kewe x > kewe y. (b) kewe y > kewe x. (c) kewe x < 0 ad kewe y < 0. (d) kewe x = 0. (b) The ditributio of x i left kewed ad would have a egative value for kewe. The ditributio of y i right kewed ad will have a poitive value for the kewe. So kewe y > kewe x.

6 6 Midterm 1 - Solutio 17. The ditributio of the ample mea of Y will be: (a) Right kewed. (b) Left kewed. (c) Symmetric. (d) Cetered at 0. (c) The ample mea will be ditributed ormally ad cetered at the mea of Y (which i ot ecearily equal to zero). 18. Which of the followig would arrow the cofidece iterval for the populatio mea? (a) Decreaig the ample ize. (b) Decreaig the igificace level. (c) Both (a) ad (b) would arrow the cofidece iterval. (d) Neither (a) or (b) would arrow the cofidece iterval. (d) Decreaig the ample ize or decreaig the ize of α will both icreae the width of the cofidece iterval for the populatio mea. 19. The ditributio of the ample mea of X (uig a ample ize of 10) ha a: (a) Larger variace tha the ditributio of X. (b) Smaller variace tha the ditriubutio of X. (c) Variace equal to the variace of X. (d) Noe of the above. (b) The variace of the ditributio of the ample mea i σ2 the variace of the ditributio of x (σ 2 ). which i maller tha 20. The probability of a Type I error whe doig a two-tailed hypothei tet at a 5% igificace level i: (a) Equal to the probability of a Type I error whe doig a oe-tailed hypothei tet at a 2.5% igificace level. (b) Equal to the probability of a Type I error whe doig a oe-tailed hypothei tet at a 5% igificace level. (c) Equal to the probability of a Type I error whe doig a oe-tailed hypothei tet at a 10% igificace level. (d) Equal to the probability of a Type I error whe doig a oe-tailed hypothei tet at a 20% igificace level. (b) The probability of a Type I error i equal to the value of α whether we are doig a oe-tailed or two-tailed tet.

7 Midterm 1 - Solutio 7 SECTION II: SHORT ANSWER (40 poit) 1. (14 poit) A urvey i give to 1000 college tudet akig them how may year they thik it will take them to graduate from college. Repoe raged from three year to eve year. The ditributio of tudet by their repoe i give i the table below: Year to graduate Number of tudet (a) Calculate a 90% cofidece iterval for the proportio of tudet i the populatio who believe they will graduate i exactly four year. Firt we eed to kow the proportio of tudet i the ample that believe they will graduate i exactly four year: x = x = =.4 Now we ca ue our hortcut for proportio data to get the ample tadard deviatio: 2 = x(1 x) 2 =.4(1.4) =.24 =.4899 Now we have everythig we eed to calculate a cofidece iterval: x ± t α 2, 1.4 ± t.10 2, The value of t.05,999 i give by TINV(.10,999) which i So our cofidece iterval i: ± ±.0256 (.3744,.4256) (b) Suppoe a reearcher wat to ue thee data to create a 95% cofidece iterval for the average umber of year it take tudet to graduate from college. At what value would thi cofidece iterval be cetered?

8 8 Midterm 1 - Solutio The cofidece iterval will be cetered at the ample mea of the umber of year to graduate: x = 1 xi x = 1 ( ) 1000 x = 4.8 So the cofidece iterval will be cetered at 4.8 year. (c) Why might the reearcher reach a icorrect cocluio? Fully explai your awer icludig whether the reearcher would be likely to overetimate or uderetimate the average umber of year it take tudet to graduate. The variable we are obervig i ot the variable the reearcher i tryig to make iferece about. We oberve the umber of year tudet thik it will take them to graduate, ot the umber of year it will actually take them to graduate. Studet mot likely do t accout the poibility of bad evet occurrig that would delay graduatio (failig clae, family emergecie, etc.). If thi i the cae, tudet expected year to graduate will ted to be lower tha the actual year it take to graduate. The ample mea from the data above would therefore provide a uderetimate of the populatio mea of iteret to the reearcher.

9 Midterm 1 - Solutio 9 2. (14 poit) The true populatio mea of a radom variable X i 50. Suppoe that you draw a ample of 100 obervatio of X that ha a ample tadard deviatio of 10 ad you ue thi to tet the followig et of hypothee uig a 5% igificace level: H o : µ = 55 H a : µ 55 (a) Write dow a formula for the tet tatitic you would ue for the tet. The oly variable i your expreio hould be the ample mea, x (you hould plug i the appropriate umerical value for ay other variable or cotat). The formula for the tet tatitic i: Pluggig i our value give u: t = x µ o t = x t = x 55 (b) For what rage of value of x would you ed up committig a Type II error? We will commit a Type II error if we fail to reject the ull hypothei eve though the ull hypothei i fale. So we eed to figure out the rage of value for x for which we would ot reject the ull hypothei. To do thi, we firt eed to figure out the critical value we would ue. For a two-ided tet with a 5% igificace level, the critical value would be give by: c upper = t α 2, 1 = t.025,99 c lower = c upper = t.025,99 The value for t.025,99 i give by TINV(.05,99) ad i So we will fail to reject the ull hypothei if t i betwee ad Thi combied with our equatio for t above allow u to figure out the rage of x for which we will commit a Type II error: c lower < t < c upper 1.98 < x 55 < < x < So we will make a Type II error wheever the ample mea i betwee ad

10 10 Midterm 1 - Solutio 3. (12 poit) For each ceario below, write dow the ull ad alterative hypothee you would ue, the formula for the tet tatitic you would ue, the critical value you would ue, ad the bai o which you would reject the ull hypothei. For the tet tatitic ad the critical value, iclude pecific umber wheever poible. (a) You have a ample of 100 book. You wat to tet whether or ot the average umber of page i a book i greater tha 200 page uig a 5% igificace level. If you chooe to ue a upper oe-tailed tet: H o : µ 200 H a : µ > 200 t = x µ o = x c = t α, 1 = t.05,99 = T INV (.1, 99) = 1.66 reject the ull if t > c If you chooe to ue a lower oe-tailed tet: H o : µ 200 H a : µ < 200 t = x µ o = x c = t α, 1 = t.05,99 = T INV (.1, 99) = 1.66 reject the ull if t < c (b) You take the temperature of 100 people to tet whether the average body temperature i equal to 98.6 degree. You wat to ue a 5% igificace level. H o : µ = 98.6 H a : µ 98.6 t = x µ o = x c = t α 2, 1 = t.025,99 = T INV (.05, 99) = 1.98 reject the ull if t > c (c) You have a ample of 100 people wage. You wat to ue a lower oe-tailed tet whether the average wage i below $20 a hour uig a 10% igificace level. H o : µ 20 H a : µ < 20

11 Midterm 1 - Solutio 11 t = x µ o = x c = t α, 1 = t.10,99 = T INV (.2, 99) = 1.29 reject the ull if t < c

Confidence Intervals. Our Goal in Inference. Confidence Intervals (CI) Inference. Confidence Intervals (CI) x $p s

Confidence Intervals. Our Goal in Inference. Confidence Intervals (CI) Inference. Confidence Intervals (CI) x $p s Cofidece Iterval Iferece We are i the fourth ad fial part of the coure - tatitical iferece, where we draw cocluio about the populatio baed o the data obtaied from a ample choe from it. Chapter 7 1 Our