252y0221 3/29/02 ECO252 QBA2 Name KEY SECOND HOUR EXAM Hour of Class Registered (Circle)(Open this document in 'Page Layout') March 19, 2002 MWF 10 11 TR 12:30 2:00 Hour of Class Attended (If Different) __I. (14 points) Do all the following. Diagrams will help! 9,5~ Nx Probabilities still can't be negative!1. 163 xP 22.189.09516953 zPzP 7021.3888.3133.22.10089.0 zPzP2. 50 xP 056.0955950 zPzP2123.3. 1425 xP 00.133.395149525 zPzP 8409.3413.4996.00.10033.3 zPzP4. 42 xP 11.078.0954952 zPzP 2385.0438.2823.011.0078.0 zPzP5. 2F (The Cumulative probability up to 2) . 2xP 952zP 33.0 zP 3707.1293.5.033.00 zPzP6. A symmetrical interval about the mean with 37% probability. 6. We want two points 315.698. and xx, so that 3700.315685. xxxP. Make a diagram Showing 5 in the middle at the center of a 37% region split into two areas with probabilities of 18.5%. From the diagram, if we replace x by z, 1850.0315. zzP. The closest we can come is 1844.48.00 zP or 1879.49.00 zP. Since neither of these is much closer than the other, use 485.0315.z, and 365.459485.05315.zx, or 0.635to 9.365. To check this note that 365.9635.0 xP95365.995635.0zP %373723.21879.1844.2485.002485.0485.0 zPzP. Of course 48.0315.z or 49.0315.z is perfectly acceptable.7. 02.x - We want a point 02.x, so that 02.02.xxP. Make a diagram of z showing zero in the middle, .4800 between 0 and 02.z and .02 above 02.z. From the diagram, if we replace x by z, .4800.002. zzP The Normal table says ,4798.05.20 zPwhich is the closest we cancome to.4800. So 05.202.z, and .45.23905.2502.zx To check this note 9545.2345.23 zPxP 05.2 zP .02.0202.4798.5.05.200 zPzP2252y0221 3/29/02II. (6 points-2 point penalty for not trying part a.) Show your work! Do not answer part b with a yes orno unless you have stated your hypotheses!a. According to your text, a study was made to compare ages of purchasers of Crest with nonpurchasesrs, yielding the following results. These are two independent samples taken from an approximately normal population. Row crest nocrest 1x 2x 1 34 28 2 35 22 3 23 44 4 44 33 5 52 55 6 46 63 7 28 45 8 48 31The Minitab 'describe' function gave the following results for the 'nocrest' column.Variable N Mean Median StDevnocrest 8 40.12 38.50 14.11a. Compute the standard deviation, 1s, for the 'crest' column. Show your work! (3)b. Compute a 90% confidence interval for the difference betweeen the two population means 1 and2 on the assumption that these are independent samples taken from approximately normal populationswith similar variances. According to your confidence interval, is there a significant difference between the population means? Why? (3) Solution: a) .81n Row 1x 21x 1 34 1156 2 35 1225 3 23 529 4 44 1936 5 52 2704 6 46 2116 7 28 784 8 48 2304 310 1275475.388310111nxx 92857.10575.741775.388127541212112121nxnxs29161.101sb) 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 2tdsd0 211ˆ212222112nnsnsnspd t scvd 02 3252y0221 3/29/0237.112.4075.3829161.10,92857.105,75.38211211xxdssx 1428820921.199,11.14,12.40212222nnDFssx ,10. 211ˆ212222112nnsnsnsp= 510335.15220921.19992857.105140921.199792857.1057 761.11405.ts sn ndp 1 11 2 17475.612758375.3825.510335.1528181510335.152 Confidence Interval: dstd2 87.1037.117475.6761.137.1 or -12.24 to9.50. The interval includes 0, so there is no significant difference between the means. Formally, our hypotheses are 21100:H0:H or 211210:H:H or 0:H0:H211210 We do not reject H0.4252y0221 3/29/02III. 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) . You must do problem 2a! Show your work! State H0 and H1where applicable. Do not answer a question 'yes' or 'no' without citing a statistical test. Use a 95% confidence level unless another level is specified.1. For your convenience, data is repeated from the previous page. Use a 90% confidence level in this problem. Row crest nocrest 1x 2x 1 34 28 2 35 22 3 23 44 4 44 33 5 52 55 6 46 63 7 28 45 8 48 31Variable N Mean Median StDevnocrest 8 40.12 38.50 14.11a. Test the hypothesis that the mean age of Crest buyers is lower* than the mean for those who did not buyCrest. Assume that these are independent samples taken from approximately normal populations with similar variances and .(i) State your null and alternate hypotheses. (2)(ii) Find a critical value for the difference between the sample means and use it to test your hypothesis. (2)(iii)
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