252y0211 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. 6,9~ Nx1. 2.2814 xP 20.383.0692.286914 zPzP 2026.2967.4993.83.0020.30 zPzP2. 97 xP 4962.067.2699697 zPzP3. 1313 xP 67.067.369136913 zPzP 7485.2486.4999.67.00067.3 zPzP4. 013 xP 50.167.36906913 zPzP 0667.4332.4999.050.1067.3 zPzP5. 17F (The cumulative probability up to 17) 17xP6917zP 33.1 zP 9082.4082.5.33.100 zPzP6. A symmetrical interval about the mean with 58% probability. We want two points 79.x and 21.x, so that 5800.2179. xxxP. From the diagram, if we replace x by z, 2900.021. zzP. The closest we can come is 2910.81.00 zP. So 81.021.z, and 86.49681.0921.zx, or 4.14 to 13.86. To check this note that 86.1314.4 xP6986.136914.4zP %585820.2910.281.00281.081.0 zPzP7. 095.x We want a point 095.x, so that 095.095.xxP. (This is the 90.5 percentile) From the diagram, if we replace x by z, 4050.0095. zzP. The closest we can come is 4049.31.10 zP. So 31.1095.z, and 86.79631.19095.zx, or 16.86 . To check this note that 86.16xP 6986.16zP 31.1 zP 4049.5.31.100 zPzP 095.0951. 252y0211 2/21/02 II. (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 2850 2 2380 3 2800 4 2860 5 2300 6 2650 7 2640a. 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 2850 8122500 2 2380 5664400 3 2800 7840000 4 2860 8179600 5 2300 5290000 6 2650 7022500 7 2640 6969600 18480 49088600,7n ,18480x490886002x2640718480nxx 6264074908860012222nxnxs 33.50233 or 13.224s.712.84713.224733.50233nssx Data divided by 1000 Row x x2 1 2.85 8.1225 2 2.38 5.6644 3 2.80 7.8400 4 2.86 8.1796 5 2.30 5.2900 6 2.65 7.0225 7 2.64 6.9696 18.48 49.0886,7n ,48.18x0886.492x640.2748.18nxx(thousands) 6640.270886.4912222nxnxs05023333.0 or 22413.0s.084712.0722413.0705023333.0nssxb. From the problem statement 10.. From Table 3 of the syllabus supplement, if the population variance is unknown xstx2 and .943.1605.12ttn So 6.1642640712.84943.12640 or 2475.4 to 2804.6.So 1646.0640.2084712.0943.1640.2 or 2.4754 to 2.8046 (thousands).2252y0211 2/21/02III. 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 2628. For your convenience the data are repeated below.Patient capacity 1 2850 2 2380 3 2800 4 2860 5 2300 6 2650 7 2640 Test to see if the mean breathing capacity is now above 2628 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)xstx21nD F0100:H:Hxsxt0xcvstx20a) You were asked to see if mean breathing capacity is now above 2628 (or 2.628 thousand). This gives us.2628 Since this does not contain an equality, it must be an alternative hypothesis, so we have2628:0Hand .2628:1Hb) Our facts, from the previous page are: ,7n ,2640x ,13.224s or better, 712.84xs (or 0.0847 thousand). In addition, 10.and .26280Since our alternate hypothesis is2628 and is one-sided, our one -sided critical value must be above 2628. Our value of t is .440.1610.1ttn 0.2750712.84440.126280xcvstx or 2.750 thousand . c) Make a diagram . Show an almost Normal curve with a center at 2628 and a 10% 'reject' zone above2750.0. Since 2640x is not in the 'reject' zone, do not reject 0H.3252y0211 2/21/02d) (i) The test ratio is .1417.0712.84262826400xsxt
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