252y0762 10/30/07 ECO252 QBA2 Name KEY SECOND EXAM Class________________________Nov 1-5 2007 Student Number_______________Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not usually acceptable. 9,12~ Nx - If you are not using the supplement table, make sure that I know it.1. 2121 xP 9122191221zP 00.167.3 zP 00.10067.3 zPzP8412.3413.4999. For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the area between -3.67 and 1.00. Because this is on one side of zero we must add the area between -3.67 and zero to the area between zero and 1.00. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 12. Indicate the mean by a vertical line! Shade the area between -21 and 21. This area is on both sides of the mean (12) so we add to get our answer. 2. 14xP91214zP 22.0 zP 4129.0871.5.22.000 zPzPFor z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the area above 0.22. Because this is on one side of zero we must subtract the area between zero and 0.22 from the area above zero. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 12. Indicate the mean by a vertical line! Shade the above 14. This area is on one side of the mean (12) so we subtract to get our answer. 3. 05.100 xP91205.109120zP 22.033.1 zP 022.0033.1 zPzP3211.0871.4082. For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the area between -1.33 and -0.22. Because this is on one side of zero we must subtract the area between -0.22 and zero from the larger area between -0.22 and zero. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 12. Indicate the mean by a vertical line! Shade the area between zero and 10.05. This area is on one side of the mean (12) so we subtract to get our answer. 4. 055.x (Do not try to use the t table to get this.) For z make a diagram. Draw a Normal curve witha mean at 0. 055.z is the value of z with 5.5% of the distribution above it. Since 100 – 5.5 = 94.5, it is also the 94.5th percentile. Since 50% of the standardized Normal distribution is below zero, your diagram should show that the probability between 055.z and zero is 94.5% - 50% = 44.5% or .4450.0055. zzP The closest we can come to this is .4452.60.10 zP (1.59 is alsoacceptable here.) So .60.1055.z To get from 055.z to 055.x, use the formula zx , which is the opposite of xz . 40.26960.112 x. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 12. Show that 50% of the distribution is below the mean (12). If 5.5% of the distribution is above 055.x, it must be above the mean and have 44.5% of the distribution between it and the mean.252y0762 10/30/07Check: 40.26xP91240.26zP 055.0548.4452.5.60.10060.1 zPzPzP252y0762 10/30/07II. (5+ points) Do all the following. Look them over first – There is a section III in the in-class exam and the computer problem is at the end. Show your work where appropriate. There is a penalty for not doing Problem 1a.Note the following:1. This test is normed on 50 points, but there are more points possible including the take-home. You are unlikely to finish the exam and might want to skip some questions.2. A table identifying methods for comparing 2 samples is at the end of the exam.3. If you answer ‘None of the above’ in any question, you should provide an alternative answer and explain why. You may receive credit for this even if you are wrong.4. Use a 5% significance level unless the question says otherwise.5. Read problems carefully. A problem that looks like a problem on another exam may be quite different. 6. Make sure that you state your null and alternative hypothesis, that I know what method you are using and what the conclusion is when you do a statistical test.1. (Groebner) We wish to compare the amount of time men and women spend in the supermarket. The two columns below, 1x and 2x represent two independent samples with 7 shoppers in each sample. You mayassume that the parent distributions are Normal. 21xxd Men Women DifferenceRow 1x 2x d 1 32 33 -1 2 42 33 9 3 22 26 -4 4 28 41 -13 5 32 33 -1 6 36 48 -12 7 25 44 -19Minitab computes the following.Variable N Mean SE Mean StDev Minimum Q1 Median Q3 Maximumx1 7 31.00 2.55 6.76 22.00 25.00 32.00 36.00 42.00x2 7 36.86 2.91 7.69 26.00 33.00 33.00 44.00 48.00d 7 -5.86 -19.00 -13.00 -4.00 -1.00 9.00a. Compute the sample variance for the d column – Show your work! (2)b. Is there a significant difference between the variances for men and women? State your hypotheses and your conclusion clearly! (2)c. Test to see there is a difference between the average amount of time men and women shop. (3)d. Using the sample means and standard deviations computed above and changing each sample size from 7 to 100, find an 89% 2-sided confidence interval for the difference between the amount of time men and women shop. Does it indicate a significant difference between men’s and women’s times? Why? (3) [10]Solution: a. Compute the sample variance for the d column – Show your work! (2)Row d 2d 1 -2 4 2 8 64 3 -5 25 4 -14 196 5 -2 4 6 -13 169 7 -20 400 -48 862So 7n, 48d, 8622dFirst86.586.3600.3185714.674821xxndd252y0762 10/30/07The formula for the sample variance is 1222nxnxs. For the difference this becomes1222ndndsd 685714.577732 614286.240773 80952.88685714.53212ndd. 4239.980952.88 ds.b. Is there a significant difference between the variances for men
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