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2/22/99 252y9912 ECO252 QBA2 FIRST HOUR EXAM February 23, 1999 Name Hour of Class Registered (Circle) MWF TR 10 12 12:30 2:00 Show your work! Make Diagrams! I. (14 points) Do all the following. x ~ N 6,7 36 1 6 z 1. P1 x 3 P P 1.00 z 0.43 7 7 P1.00 z 0 P0.43 z 0 .3413 .1664 .1749 11 6 11 6 z 2. P11 x 11 P P 2.43 z 0.71 7 7 P2.43 z 0 P0 z 0.71 .4925 .2611 .7536 96 16 6 z 3. P16 x 9 P P 3.14 z 0.43 7 7 P3.14 z 0 P0 z 0.43 .4992 .1664 .6656 For 3.14 see SS Table 18. 06 4. Px 0 P z Pz 0.86 P 0.86 z 0 Pz 0 7 .3051 .5 .8051 12 .5 6 96 z 5. P9 x 12.5 P P0.43 z 0.93 7 7 P0 z 0.93 P0 z 0.43 .3238 .1664 .1574 6. A symmetrical interval about the mean with 96% probability. We want two points x .98 and x.02 , so that Px.98 x x .02 .9600 . From the diagram, if we replace x by z, P0 z z.02 .4800 . The closest we can come is P0 z 2.05 .4798 or P0 z 2.06 .4803 . So z.02 2.054 , though 2.05 or 2.06 are fine, and x z.02 6 2.054 7 6 14.38 , or –8.38 to 20.38. x.13 We want a point x .13 , so that Px x .13 .13 . From the diagram, if we replace x by z, P0 z z.13 .3700 . The closest we can come is P0 z 1.13 .3708 . So z .13 1.13 , and x z.11 6 1.137 6 7.91 , or 13.91 . 7. 2/22/99 252y9912 II. (6 points-2 point penalty for not trying part a.) A sample of the number of days it took a broker to sell houses appears below. Assume that we were sampling from a normal distribution. Home 1 2 3 4 5 Days 49 69 88 99 33 a. Compute the sample variance, s , of the number of days it takes to sell a house. Show your work.(3) b. Compute an 80% confidence interval for the mean time, , that it takes to sell a house.(3) Solution: a. Home x (Days) x2 x 338 1 49 2401 x 67 .6 n 5 2 69 4761 2 2 3 88 7744 2 x nx 25796 567 .6 4 99 9801 s2 n 1 4 5 33 1089 736 .8 or s 27.1441. Total 338 25796 . b. From the problem statement .20 . From Table 3 of the syllabus supplement, if the x t sx and t n21 t .410 1.533 . population variance is unknown sx s n 2 736 .8 27 .1441 12 .1392 . So 67.6 1.533 12.1372 67.6 18.61 or 48.99 to 86.21. 5 5 2 2/22/99 252y9912 III. Do at least 3 of the following 5 Problems (at least 10 each) (or do sections adding to at least 30 points Anything extra you do helps, and grades wrap around) . Show your work! State H 0 and H 1 where appropriate. 1. The broker claims that the average time to sell a house is at most 48 days. a. State the Hypotheses that you are testing. (2) b. Using the data from the previous problem (page 2.), test the hypotheses (99% confidence level) using: (i) A test ratio (2) (ii) Critical values (2) (iii) A confidence interval (2) c. Find an approximate p-value for the null hypothesis. (1) d. Now find a 99% confidence interval for the standard deviation.(3) Solution: From Table 3 of the Syllabus Supplement: Interval for Confidence Hypotheses Test Ratio Critical Value Interval xcv 0 z 2 x Mean ( x z 2 x x 0 H0 : 0 z Known) H : 1 x t 2 s x Mean ( Unknown) DF n 1 H 0 : 48 a. H 1 : 48 x 0 H0 : 0 t H1 : 0 x 0 sx xcv 0 t 2 s x 4 0 48, DF n 1 4, .01, tn1 t .01 3.747 From the previous page: x 67.6 and s x s 736 .8 27 .1441 12 .1392 . 5 5 n x 0 67 .6 48 1.6146 . This is in the ‘accept’ region b. (i) Test Ratio: t sx 12 .1392 below 3.747, so do not reject H 0 . (ii) Critical Value: Since this is a one-sided test, xcv 0 t sx 48 3.747 12.1392 50 45.49 or 95.49. This means that we reject H 0 if the sample mean is above 95.49. since x 67.6 is below this critical value, do not reject H 0 . (iii) Confidence Interval: Since this is a one-sided test, x t s x 67.6 3.747 12.1392 67.6 45.49 or 22.11. This does not contradict H 0 : 48 , because any mean in the range 22.11 to 48 satisfies both statements, so do not reject H 0 . 4 4 c. From the t table t = 1.6146 is smaller than t .05 2.132 but is larger than t .10 1.533 , so that, for this 1- sided test, we can say .10 pvalue .05 . d. From page 1 of the Syllabus Supplement: n 1 s 2 Since .005 and 1 .995 , look up 2 2 n 1s 2 .2005 2 n 1s 2 .2995 or 2 2 2 4 .005 2 2 1 . DF n 1 4 and .01 . 2 4 14 .8603 and .2995 0.2070 . So 4736 .8 2 4736 .8 14 .8603 n 1 s 2 0.2070 or 198 .33 2 14237 .68 . Finally, taking square roots, 14.083 119.32 . 3 2/22/99 252y9912 2. A new broker opens in town and a sample of nine houses gives the following days to sale: Home Days 1 62 2 28 3 114 4 113 5 71 6 29 7 53 8 54 9 34 Assuming that the distribution is not normal, a. Find a confidence level for the following interval for the median: 34 71 . (3) b. Test the hypotheses that the median is at most 33 at the 10% level. (5) c. If we have a sample of 350 numbers in order and take the 15 th from each end, what would the confidence level be? (3) d. (Extra credit) What numbers would we use if we wanted a 99% confidence interval for the median and had a sample of 350 numbers? (3) Solution: a. If we put the numbers in order, we get 28 29 34 53 54 62 71 113 114. Thus we want the 3rd number from the end. If n 9 and k 3 , we find from the binomial table (with p .5 ), 1 1 2Px k 1 1 2Px 2 1 2.08984 .82032 . Px 2 Px 7 b. From the outline: Hypotheses about a median Hypotheses about a proportion If p is the proportion above 0 If p is the proportion below 0 H 0 : 0 H 1 : 0 H 0 : p .5 H 1 : p .5 H 0 : p .5 H1 : p .5 H : 33 So, our hypotheses are 0 and If p is the proportion above 33, our hypotheses become H 1 : 33 H 0 : p .5 . There are 7 numbers above 33, and according to the binomial table H 1 : p .5 pvalue Px 7 1 Px 6 1 .91016 .08984 . Reject H 0 . c. From the outline: 2 P x k 1 1 2 n 350 and k 15 , 1 1 2Px k 1 d. np k .5 .5n . If P z npq .5 n P z k 1 1 2 14 .5 .5350 161 .5 1 2 P z 1 2 Px 14 1 2 P z 1 2Pz 25 .42 1 9.354 .5 350 . n 1 z . 2 n From the outline: k . z z.005 2.576 , so 2 2 350 1 2.576 350 k 126 .4 . So use the 126th number from each end. 2 4 2/22/99 252y9912 3. A consumer service desk claims that the average waiting time is less than 11 minutes. A sample of 300 customers gives us the following: x 10.25, s 6.01. Assume that the parent distribution is Normal and that our confidence level is 95%. a. State the Hypotheses to be tested. (1) b. Find a critical value or values for x and test the hypotheses. (3) c. Graph a power function for the test showing most of your calculations. (8) d. Find a p-value for the Null Hypothesis if the value of x is 12. (2) Solution: H 0 : 11 299 a. 0 11, DF n 1 299 , .05, tn 1 t .05 z .05 1.645 H 1 : 11 6.012 0.34699 xcv 0 z sx 11 1.645 (0.34699 ) 10.4292 . 300 n 300 Since 10.25 is below the critical value, reject H 0 . c. Since we do not reject H 0 when x is above 11.4292, the probability of a type II error is b. s x s 6.01 Px 10.4292 1 , when 1 is any value below 11. Since the critical value is about 10.4, the values I will use are 11, 10.7, 10.4292, 10.1 and 9.8. Operating Characteristic Power 1 .05 Px 10.4292 11 .95 10 .4292 10 .7 Px 10.4292 10.7 P z Pz 0.78 .2823 + .5 = .7823 0.34699 .22 Px 10.4292 10.4292 .50 .50 10 .4292 10 .1 Px 10.4292 10.1 P z Pz 0.95 .5 - .3289 = .1711 0.34699 .83 11 .3085 10 .60 Px 10.4292 9.8 P z Pz 2.01 .5 - .4778=.0222 0.35276 .98 DIAGRAM 12 11 d. Since this is a left-sided test and 0 11 , Px 12 P z 0.34699 Pz 2.88 .5 .4980 .9980 . 5 2/22/99 252y9912 4. A Senatorial candidate believes that she must get at least 78% of the vote in Philadelphia and Duquesne County to win the election. A sample survey of 2000 voters indicates that 1499 planned to vote for her. Using a Hypothesis-testing procedure, assess her chances of winning. a. State the Hypotheses to be tested. (1) b. Do a test of this hypothesis at the 99% level. (3) c. Do the same test at the 87% level. (2) d. Do a 2-sided 98% confidence interval for the proportion that will vote for her. (2) e. Do a 2 sided 96% confidence interval for the proportion that will vote for her. (2) f. (Extra Credit) If the true proportion that will vote for her is 76%, what is the power of your test? (3) Solution: From Table 3: Interval for Confidence Hypotheses Test Ratio Critical Value Interval Proportion p p z 2 s p p p0 p cv p 0 z 2 p H 0 : p p0 z p H 1 : p p0 pq p0 q 0 sp p n n q 1 p H : p .78 x 1499 .7495 , so that a. 0 Note that p 0 .78, n 2000 , p n 2000 H : p . 78 1 p .78 .22 .009263 2000 b. (i) Test Ratio: z , or s p p p0 p .7495 .2505 .009689 . q 1 p . 0 0 2000 .7495 .78 3.2927 . Since .01 and this is a one-sided test, use .009263 z .01 2.326 . Since –3.2927 is less than z .01 , reject H 0 . Or pvalue Pz 3.2927 .5 .4995 .0005 , so reject H 0 . Or: (ii) Critical Value: pcv p0 z p .78 2.326.009263 .7585 . Since p .7495 is below this value, reject H 0 . Or: (iii) Confidence Interval: p p z s p .7495 2.326.009689 .7720 . Since p .7720 contradicts H 0 : p .78 , reject H 0 . c. If .13 , we know that z .13 1.13 from the first page of this exam. (i) Test Ratio: z 3.2927 . Since –3.2927 is less than z .13 , reject H 0 . Or pvalue .0005 , so reject H 0 . Or: (ii) Critical Value: pcv p0 z p .78 1.13.009263 .7695 . Since p .7495 is below this value, reject H 0 . Or: (iii) Confidence Interval: p p z s p .7495 1.13.009689 .7604 . Since p .7604 contradicts H 0 : p .78 , reject H 0 . d. Since .02 and z z.01 2.326 , p p z s p .7495 2.326.009689 .7495 .025 , or .727 2 2 to .772. e. Since .04 , z z.02 and z.02 2.054 from the first page of this exam, p p z s p 2 2 .7495 2.054 .009689 .7495 .0199 or .730 to .769. 6 2/22/99 252y9912 f. If we use the critical value from b., we do not reject H 0 if p is above .7585. So the power is .7585 .76 1 Pz 0.16 1 .5 .0636 1 .5636 .4364 . 1 P p .7585 p .76 1 P z .009263 Actually p .76 .24 .009550 2000 , but it does not affect the power. . 7 2/22/99 252y9912 5. a. b. c. d. e. f. g. Solution: A Student wishes to estimate the mean amount that members of city governments earn. The student guesses that the standard deviation is about $1000. What sample size is required to keep the error in estimating the mean at $100 if a confidence level of 98% is used. (3) The student also wants to estimate the proportion of cities that have private refuse collectors. What is the required sample size if no estimate of the proportion is available before the study, the confidence level is to be 98% and the permitted error is .05. (3) A finance expert wants to test whether new legislation has reduced the volatility of stock prices. He assumes that the appropriate parent distribution is Normal and that the variance of stock prices before the legislation was 9.0, and on the basis of 23 daily measurements after the legislation finds that s 2 5.3 . State the Hypotheses to be tested. (1) Test the Hypotheses in c. ( .01 ) (3) What if the sample variance in c. were taken from a sample of 85 days? (3) If we claim that a population has a Poisson distribution with a mean of 12 and an actual value is found to be 5, do a 2-sided 95% hypothesis test on the claim. (3) If in f. the mean was 120 and the number found was 50, repeat the test. (3) a. Since .02 , z z.01 2.326 . From the outline n 2 z 2 2 e2 2.326 2 1000 2 100 2 541 .02 . So use n 542 . b. Since the significance level is the same, z z.01 2.326 . We do not know p , so we use p .5 . 2 From the outline n pqz 2 e2 .5.52.326 2 .05 2 541 .02 . So use n 542 . H 0 : 2 9.0 c. H 1 : 2 9.0 d. 02 9.0 , s 2 5.3 , n 23 and .01 . Since DF n 1 22 and the test is one-sided, use .29922 9.5425 . From the outline, 2 n 1s 2 02 22 5.3 12.956 . Since this is above 9.5425, do not 9.0 reject H 0 . e. 02 9.0 , s 2 5.3 , n 85 and .01 . DF n 1 84 and the test is one-sided. From the outline, 2 n 1s 2 02 84 5.3 49 .467 . Since this sample is too large for the 9.0 2 Table, we must use z 2 2 2DF 1 249 .467 284 1 9.9465 12 .8228 2.88 . We compare this to z .01 2.326 . Since it is below –2.326, reject H 0 . f. Since this is a two-sided test, with .05 , use the Poisson (12.0) table to compute a p-value. pvalue 2Px 5 2.02034 .04068 . Since pvalue , reject H 0 . g. This is still a Poisson problem, but because of the large mean and the lack of a suitable Poisson table, use the Normal approximation to the Poisson distribution, i.e. x ~ N m, m , where m is the mean. The easiest way to do this is z z.025 xm 50 120 6.39 . Since this is a two-sided test, z should be between m 120 1.960 . It is not, so reject H 0 . 8