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WCU ECO 252 - ECO 252 Second Hour Exam

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Solution continues in252z99223/25/99 252y9922 ECO252 QBA2 Name SECOND HOUR EXAM Hour of Class Registered (Circle)March 23, 1999 MWF TR 10 12 12:30 2:00 night I. (14 points) Do all the following. Use diagrams! Show your work! 9,5~ Nx (Why do too many of you still believe that a probability can be negative?)1.  165 xP 22.109516955 zPzP(Diagrams!) 3888.2. P x0 3   22.056.0953950 zPzP   1252.0871.2123.022.0056.0  zPzP3. P x  2 0  56.078.0950952 zPzP   0700.2123.2823.056.0078.0  zPzP4.  3xP 22.0953 zPzP   5871.5000.0871.0022.0  zPzP5.  4F (The Cumulative probability) 4xP 954zP 1`1.0 zP   4562.0438.5000.011.00  zPzP6. A symmetrical interval about the mean with 64% probability.We want two points 82..18 and xx, so that 6400.18.82. xxxP. From the diagram, if we replace x by z,  3200.018. zzP. The closest we can come is  3212.92.00 zP. So 92.018.z(0.91 is acceptable and something in between even better), and  28.85992.0518.zx, or –3.28 to 13.28.check: 28.1328.3  xP 92.092.09528.139528.3 zPzP 6424.3212.2 7.055.xWe want a point 055.x, so that 055.055.xxP. From the diagram, if we replacex by z,  4450.0055. zzP. The closest we can come is  4452.60.10 zP . So 60.1055.z, and  4.145960.15055.zx, or 19.4.check: 4.19xP 60.1954.19 zPzP   4452.5000.60.100  zPzP 055.0548. 3/25/99 252y9922II. (6 points-2 point penalty for not trying part a.) Show your work! A new software package has been designed to help system analysts design information systems. We wish to see if there is a significant difference between the time required to develop a system using the new technology and using the current technology. Two independent samples are taken. The results are shown below. Assume that the data represent independent samples taken from populations with the normal distribution.x1x2Current Technology New Software300 286280 232344 310385 338372 200360 302288 I will be happy to nominate everyone who had trouble dealing with 21nn for the Bourbon prize for inflexibility. Samples only need to be the same size when data is paired and there is no reason to believe that this data is paired!a. Compute 2s, the standard deviation for time required to design a system using the new software. (3)Note that 812.42,714.33211 sx.b. Do a two-sided confidence interval for the difference between the means for the two technologies  01.. (2) Indicate the following: What assumptions did you make about the variances of the populations from which the samples were taken? Is there a significant difference between the means (You must tell why)? (1)Solution: 892.51800.26925000.27864771681000.278616682222222222222snxnxsnxx 4771681668Total9120430264000020051142443384961003103538242322817962861xItem222xb. Solution: Assume equal variances. From Table 3 of the Syllabus Supplement:Interval for Confidence IntervalHypotheses Test Ratio Critical ValueDifference between Two Means ( unknown, variances assumed equal) d t sd2s sn ndp 1 11 2DF n n  1 22HH :0:1     001 2tdsd0   111ˆ212222112nnsnsnspd t scvd 02 ,01.,714.54,800.2692,000.278,867.1832812.42,714.332212222211xxdsxsx106.3,11267211005.21 tnnDF.3/25/99 252y99222      746.222311000.13464202.1099711800.26925867.18326211212222112nnsnsnsp236.263022.6886171746.223111121221nnsnnssppd  5.817.54236.26106.3714.542dstd or –16.8 to 136.2. This interval includes 0. Since the interval includes zero, we can say that there is no significant difference between the means. Or if the null hypothesis is 0:0H or 0:210Hor 210:H, we cannot reject it.b. Alternate Solution: (Note - this method was not covered in Spring 2000 - If you have studied this material and want an extra credit question on it, please tell me in advance.) Assume unequal variances. From Table 3 of the Syllabus Supplement:Interval for Confidence IntervalHypotheses Test Ratio Critical ValueDifference between Two Means( unknown, variances assumed unequal) d t sd2ssnsnd 1212221122122222121222121nnnsnsDFnsnsHH :01:     001 2Same as HH :if 01:  1 21 200tdsd0d t scvd 026381.710800.4486800.26718381.2617867.1832222121222121nsnsnsns 656.266381.710222121nsnssd1122222121212222121nnsnnsnsnsDF    77.95800.44868381.2616381.710222, so use 9 degrees of freedom.250.39005.t, so   6.867.54656.26250.3714.542dstd or –31.9 to 141.4. Conclusion is the same as for a.33/25/99 252y9922III. 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 H0 and H1where applicable.1. A firm is currently contracting with firm 1 for delivery of raw materials. It will stay with firm 1 unless it can demonstrate that delivery times for firm 2 are faster. The statistics from the two samples are below. Assume that the parent populations are Normal. Firm 1 Firm 2days 3days 1450111sxn days 2days 5.1220222sxna. Test to


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