252y0212 2/27/02 ECO252 QBA2 Name ___key_____________ FIRST HOUR EXAM Hour of class registered _____ February 19, 2002 Class attended if different ____Show your work! Make Diagrams! How many of you looked at "Things You Should Never Do" before this exam?I. (14 points) Do all the following. 6,8~ Nx1. 2.2815 xP 37.317.1682.286815 zPzP 1206.3790.4996.17.1037.30 zPzP2. 89 xP 4977.083.2688689 zPzP3. 1414 xP 00.167.368146814 zPzP 8412.3413.4999.00.10067.3 zPzP4. 014 xP 33.167.36806814 zPzP 0917.4082.4999.033.1067.3 zPzP5. 18F (The cumulative probability up to 18) 18xP6818zP 67.1 zP 9525.4525.5.67.100 zPzP6. A symmetrical interval about the mean with 68% probability. We want two points 84.x and 16.x, so that 6800.1684. xxxP. From the diagram, if we replace x by z, 3400.016. zzP. The closest we can come is 3389.99.00 zP or 3413.00.10 zP. So 995.016.z, and 97.586995.0816.zx, or 2.03 to 13.97. To check this note that 97.1303.2 xP6897.136803.2zP %585820.29102995.002995.0995.0 zPzP7. 085.x We want a point 085.x, so that 085.085.xxP. (This is the 91.5 percentile) From the diagram, if we replace x by z, 4150.0085. zzP. The closest we can come is 4147.37.10 zP. So 37.1085.z, and 22.88637.18085.zx, or 16.22 .To check this note that 22.16xP 6822.16zP 37.1 zP 4147.5.37.100 zPzP 085.0853. 252y0212 2/20/02II. (6 points-2 point penalty for not trying part a.) A new product is tried on seven patients. Their breathing capacity after using the product is shown below (Note: You may want to move the decimal point to the left and work in thousands.).Patient capacity 1 2840 2 2380 3 2800 4 2860 5 2300 6 2850 7 2650 a. Compute the sample standard deviation, s, of the breathing capacity. Show your work! (3)b. Compute a 90% confidence interval for the mean breathing capacity, .(3)Solution: a) Original data Row x x2 1 2840 8065600 2 2380 5664400 3 2800 7840000 4 2860 8179600 5 2300 5290000 6 2850 8122500 7 2650 7022500 18680 50184600,7n ,18680x5012846002x6.2668718680nxx 66.266875018460012222nxnxs 71.55769 or 16.236s.2586.89716.236771.55769nssx Data divided by 1000 Row x x2 1 2.84 8.0656 2 2.38 5.6644 3 2.80 7.8400 4 2.86 8.1796 5 2.30 5.2900 6 2.85 8.1225 7 2.65 7.0225 18.68 50.1846,7n ,68.18x1846.502x6686.2768.18nxx(thousands) 66686.271846.5012222nxnxs0557697.0 or 2362.0s.08927.072362.070557697.0nssxb. From the problem statement 10.. From Table 3 of the syllabus supplement, if the population variance is unknown xstx2 and .943.1605.12ttn So 4.1736.26682586.89943.16.2668 or 2495.2 to 2842.0.So 1735.06686.208927.0943.16686.2 or 2.495 to 2.842 (thousands).2252y0212 2/20/02III. Do at least 3 of the following 4 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. You have not done a hypothesis test unless you have stated your hypotheses, run the numbers and stated your conclusion. Use a 95% confidence level unless another level is specified.1. The population mean for similar patients to those mentioned on the previous page who had not used thenew medicine was 2630. For your convenience the data are repeated below.Patient capacity 1 2840 2 2380 3 2800 4 2860 5 2300 6 2850 7 2650 Test to see if the mean breathing capacity is now above 2630 using the sample mean and standard deviation you found in part II. a. State the null and alternate hypothesis (2)b. Find a critical value appropriate for this problem, using a confidence level of 90%.(3)c. Use your critical value to test the hypothesis. State clearly whether you reject the null hypothesis. (2)d. Repeat the test using (i) a test ratio (2) and (ii) a confidence interval. (2). Each time state clearly whether you reject the null hypothesis and why.e. Do a 90% two - sided confidence interval for the variance. (3)f. (Extra credit) Assume that the data does not come from a normal distribution. (i) State a confidence interval for the median using the highest and lowest values and give the confidence level.(4) (ii) Do the same using the second highest and second lowest values and give the confidence level. (3)Solution: From Table 3 of the Syllabus Supplement:Interval for Confidence IntervalHypotheses Test Ratio Critical ValueMean ( Unknown)xstx21nDF0100:H:Hxsxt0xcvstx20a) You were asked to see if mean breathing capacity is now above 2630 (or 2.63 thousand). This gives us.2630 Since this does not contain an equality, it must be an alternative hypothesis, so we have2630:0Hand .2630:1Hb) Our facts, from the previous page are: ,7n ,6.2668x ,16.236s or better,2586.89xs (or 0.0847 thousand). In addition, 10.and .26300Since our alternate hypothesis is 2630 and is one-sided, our one -sided critical value must be above 2630. Our value oft is .440.1610.1ttn 6.27582586.89440.126300xcvstx or 2.759 thousand . c) Make a diagram . Show an almost Normal curve with a center at 2630 and a 10% 'reject' zone above2758.6. Since 6.2668x is not in the 'reject' zone, do not reject 0H.3252y0212
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