252y0121 3/26/01 ECO252 QBA2 Name SECOND HOUR EXAM Hour of Class Registered (Circle) March 20, 2001 MWF 10 11 TR 12:302:00 I. (14 points) Do all the following. (Make diagrams!!!) 9,6~ Nx1. 163 xP 11.100.19616963 zPzP 7078.3665.3413.11.10000.1 zPzP2. 50 xP 11.067.0965960 zPzP 2048.0438.2486.011.0067.0 zPzP3. 1424 xP 89.033.396149624 zPzP 8129.3133.4996.89.00033.3 zPzP4. 42 xP 22.089.0964962 zPzP 2262.0871.3133.022.0089.0 zPzP5. 2F (The Cumulative probability up to 2) . 2xP 962zP 44.0 zP 3300.1700.5.044.00 zPzP6. A symmetrical interval about the mean with 79% probability. We want two points 105.895. and xx, so that 7900.105895. xxxP. Make a diagram showing 6 in the middle at the center of a 79% region split into two areas with probabilities of .3950. From the diagram, if we replace x by z, 3950.0105. zzP. The closest we can come is 3944.25.10 zP or 3962.26.10 zP. Sinceneither of these is much closer than the other, use 255.1105.z, and 295.1169255.16105.zx, or -5.295 to 17.295. To check this note that 295.17295.5 xP96295.1796295.5zP 7900.3950.2255.1022558.1295.1 zPzP7. 008.x We want a point 008.x, so that 008.008.xxP. Make a diagram of zshowing zero in the middle, .4920 between 0 and 008.z and .008 above 008.z. From the diagram, if we replace x by z, .4920.0008. zzP The Normal table says 4920.41.20 zP. So 41.2008.z, and .69.27941.26008.zx To check this note 9669.2769.27 zPxP 41.2 zP 008.4920.5.41.200 zPzP 252y0121 3/19/01II. (6 points-2 point penalty for not trying part a.) Show your work! A test shopper goes to Miller's Supermarkets 8 times and to Albert's Supermarkets 8 times. The amount that was spent using a standardized shopping list is shown below. You can regard these data as two independent random samples from populations with a normal distribution. For Miller's, the sample mean is 115.25 and the sample standard deviation is 1.75255. mx axObs Miller's Albert's 1 112 119 2 115 121 3 115 122 4 117 120 5 117 122 6 117 124 7 115 122 8 114 122a. Compute as, the standard deviation for Albert's. (3)b. Compute a 99% confidence interval for the difference between the population means 1 and 2 assuming that the variances are equal for the two parent populations. According to your confidence interval, is there a significant difference between the population means (You must tell why!)? (3) c. (Extra Credit) (i) Redo the confidence interval on the assumption that the variances are not equal. (6) (ii) Use an F-test to tell whether you should have assumed that the variances were or were not equal. (2) Solution: a) Use 2x for Albert's and 1x for Miller's.25.1152x and 75255.12s, so07143.322s. Line 1x 21x 1 119 14161 2 121 14641 3 122 14884 4 120 14400 5 122 14884 6 124 15376 7 122 14884 8 122 14884 972 11811450.1218972111nxx 2857.2716750.12181181141212112121nxnxs51186.11sb) From Table 3 of the Syllabus Supplement: Interval for Confidence IntervalHypotheses Test Ratio Critical Value2Difference between Two Means ( unknown, variances assumed equal) d t sd2s sn ndp 1 11 2DF n n 1 22HH :0:1 001 2tdsd0 211ˆ212222112nnsnsnspd t scvd 023252y0121 3/19/0125.625.11550.12151186.1,2857.2,50.121211211xxdssx 14288275255.1,0714.3,25.115212222nnDFssx,01. 211ˆ212222112nnsnsnsp= 977.267955.220714.32857.2140714.372857.2714005.t s sn ndp 1 11 2 818467.06698875.025.67955.2818167955.2 Confidence Interval: dstd2 4366.225.681847.0977.225.6 or 3.81 to 8.69. The interval does not include 0, so there is a significant difference between the means.Formally, our hypotheses are 21100:H0:H or 211210:H:H or 0:H0:H211210 We reject H0.Solution: b) From the formula table;Interval for Confidence IntervalHypotheses Test Ratio Critical ValueDifference between Two Means( unknown, variances assumed unequal)dstd2222121nsnssd1122122222121222121nnnsnsDFnsnsHH :01: 001 2Same as HH :if 01: 1 21 200dsdt0dcvstd2025.625.11550.12151186.1,2857.2,50.121211211xxdssx 01.75255.1,0714.3,25.1152222ssx4669643.0383929.080714.3285714.082857.2222121222121nsnsnsns 8183172.0669643.0222121nsnssd 1122222121212222121nnsnnsnsnsDF 7052.137383929.428571.2669643.222, so use 13 degrees of freedom. 5252y0121 3/19/01 012.313005.tConfidence Interval: The 2-sided interval is dstd2 669643.0012.325.6 25.62.01 or 4.24 to 8.26c) F test. 211210::HH. To test this at the 1% significance level, test 7442.00714.32857.22221ss and343.17442.012122ss against 7,7005.F.
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