DIAGRAM2/17/99 252y9911 ECO252 QBA2 Name FIRST HOUR EXAM Hour of Class Registered (Circle) February 23, 1999 MWF TR 10 12 12:30 2:00Show your work! Make Diagrams!I. (14 points) Do all the following. 6,7~ Nx1. 31 xP 67.033.1673671 zPzP 1596.2486.4082.067.0033.1 zPzP2. 1111 xP 67.000.367116711 zPzP 74725.2486.49865.67.00000.3 zPzP3. 914 xP 33.050.36796714 zPzP 6291.1293.4998.33.00050.3 zPzPFor 3.5 see SS Table 18.4. 0xP 17.1670 zPzP 8790.5000.3790.0017.1 zPzP5. 5.129 xP 92.033.0675.12679 zPzP 1919.1293.3212.33.0092.00 zPzP6. A symmetrical interval about the mean with 94% probability.We want two points 03.97. and xx, so that 9400.03.97. xxxP. From the diagram, if we replace x by z, 4700.003. zzP. The closest we can come is 4699.88.10 zP. So 88.103.z, and 28.117688.1703.zx, or –4.28 to 18.28.7. 11.xWe want a point 11.x, so that 11.11.xxP. From the diagram, if we replace x by z, 3900.011. zzP. The closest we can come is 3907.23.10 zP. So 23.111.z, and 38.77623.1711.zx or 14.38 .2/17/99 252y9911II. (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 Days 1 48 2 63 3 88 4 101 5 31 a. Compute the sample variance, s, of the number of days it takes to sell a house. Show your work.(3)b. Compute a 90% confidence interval for the mean time, , that it takes to sell a house.(3)Solution: a. Home x (Days) x2 1 48 2304 2 63 3969 3 88 7744 4 101 10201 5 31 961 Total 331 25179. b. From the problem statement 10.. From Table 3 of the syllabus supplement, if the population variance is unknown x t sx2 and 132.2405.12ttn.7805.1255780.2857.816nssx. So 25.272.667805.12132.22.66 or 38.95 to 93.45. 2.665331nxx 42.6652517912222nxnxs 7.816 or 5780.28s.22/17/99 252y9911III. 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 0H and 1H where appropriate.1. The broker claims that the average time to sell a house is at most 50 days. a. State the Hypotheses that you are testing. (2)b. Using the data from the previous problem (page 2.), test the hypotheses (95% 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 90% confidence interval for the standard deviation.(3)Solution: From Table 3 of the Syllabus Supplement:Interval for Confidence IntervalHypotheses Test Ratio Critical ValueMean ( known)xzx20100:H:Hxxz0xcvzx20Mean ( unknown)xstx21nDF0100:H:Hxsxt0xcvstx20a.50:H50:H10 132.2,05.,41,50405.10ttnDFnFrom the previous page: 2.66x and 7805.1255780.2857.816nssx.b. (i) Test Ratio: 2676.17805.12502.660xsxt. This is in the ‘accept’ regionbelow 2.132, so do not reject H0. (ii) Critical Value: Since this is a one-sided test, xcvstx0 24.27507805.12132.250 or 77.24. This means that we reject H0if the sample mean is above 77.24. since 2.66xis belowthis critical value, do not reject H0. (iii) Confidence Interval: Since this is a one-sided test, xstx 24.272.667805.121323.22.66 or 95.38. This does not contradict50:H0, because any mean in the range 38.95 to 50 satisfies both statements, so do not reject H0.c. From the t table t= 1.2676 is smaller than 533.1410.tbut is larger than 190.1415.t, so that, forthis 1-sided test, we can say 10.15. pvalue. 2676.1 tPpvalued. From page 1 of the Syllabus Supplement: n s n s 1 122222122 .41 nDF and10.. Since 05.2 and 95.21 , look up 4677.94205. and 7107.04295.. 3So 295.22205.211snsn or 7107.07.81644877.97.81642 or 59.459632.3442. Finally, taking square roots, 80.67555.18 .2/17/99 252y99112. 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 5 716 297 538 549 34Assuming that the distribution is not normal, a. Find a confidence level for the following interval for the median: 11329 . (3)b. Test the hypotheses that the median is at most 50 at the 10% level. (5)c. If we have a sample of 300 numbers in order and take the 15th 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 300 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 2nd number from the end. If 9nand 2k, we find from the binomial table (with 5.p), 1211 kxP 96094.01953.21121 xP. 81 xPxPb. From the outline:Hypotheses about a median Hypotheses about a proportionIf p is the proportion above0If p is the proportion
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