Scheffé Confidence IntervalTukey Confidence Intervali. A Single Confidence Intervalii. Scheffé Confidence Intervaliii. Bonferroni Confidence Interval – not worth the effort.iv. Tukey Confidence Interval252y0631 11/28/06 (Page layout view!) ECO252 QBA2 Name KEY THIRD HOUR EXAM Hour of Class Registered (Circle) Nov 30 and Dec 1, 2006 MWF 1 MWF 2 TR 12:30 TR 2I. (8 points) Do all the following (2points each unless noted otherwise). Make Diagrams! Show your work! 3.9,15~ Nx 1. 20xP 54.03.91520 zPzP 54.000 zPzP2946.2054.5. For z make a diagram. Draw a Normal curve with a mean at 0. Indicate the mean by a vertical line! Shade the area above 0.54. Because this is on one side of zero we must subtract the area between 0 and 0.54 from the area above zero. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 15. Indicate the mean by a vertical line! Shade the area above 20. This area is totally above the mean so we must subtract the area between the mean and 20 from the area %50above the mean. This is how we usually find a p-value when the distribution of the test ratio is Normal. 2. 140 xP 11.061.13.915143.9150 zPzP 011.0061.1 zPzP4025.0438.4463. 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 -1.61 and -0.11. Because this is on one side of zero we must subtract the area between -0.11 and zero from the area between -1.61 and zero. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 15. Indicate the mean by a vertical line! Shade the area between zero and 14. This area is totally below the mean so we must subtract the area between 14and the mean (15) from the area between zero and the mean. 3. 1616 xP 11.033.33.915163.91516 zPzP 11.00033.3 zPzP5434.0438.4996. 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 -3.33 and 0.11. Because this is on both sides of zero we must add together the areabetween -3.33 and zero and the area between zero and 0.11. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 15. Indicate the mean by a vertical line! Shade the area between -16 and 16. This area includes the mean (15), and areas to either side of it so we add together these two areas.4. 42.x(Find 42.z first) Solution: (Do not try to use the t table to get this.) For z make a diagram. Draw a Normal curve with a mean at 0. 42.z is the value of z with 42% of the distribution above it. Since 100 – 42 = 58, it is also the 58th percentile. Since 50% of the standardized Normal distribution is below zero, your diagram should show that the probability between 42.z and zero is 58% - 50% = 8% or .0800.042. zzP The closest we can come to this is .0783.20.00 zP (0.21 is also acceptable here.) So .20.042.z To get from 42.z to 42.x, use the formula zx , whichis the opposite of xz . 86.163.920.015 x. If you wish, make a completely separate diagram for x. Draw a Normal curve with a mean at 20. Show that 50% of the distribution is below the mean (7). If 42% of the distribution is above 42.x, it must be above the mean and have 8% of the 1252y0631 11/28/06 (Page layout view!)distribution between it and the mean. Note that there will be some questions on the Final where you will need odd values of z like 125.z.Check: 86.16xP 20.03.91586.16 zPzP 20.000 zPzP0793.5. %424207. 2252y0631 11/28/06 (Page layout view!)II. (22+ points) Do all the following (2points each unless noted otherwise). Do not answer question ‘yes’ or ‘no’ without giving reasons. Show your work in questions that are not multiple choice. Look them over first. The exam is normed on 50 points.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. 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.3. Use a 5% significance level unless the question says otherwise.4. Read problems carefully. A problem that looks like a problem on another exam may be quite different.5. Use statistical tests! Just because two means or two proportions look different to you does not mean that they are significantly different unless you prove that the probability of getting the observed difference if the null hypothesis is true is very small.1. Turn in your computer problems 2 and 3 marked to show the following: (5 points, 2 point penalty for not doing.)a) In computer problem 2 – what is tested and what are the results?b) In computer problem 3 – what coefficients are significant? What is your evidence?c) In the last graph in computer problem 3, where is the regression line? [5]2. (Abronovic) The distance that a baseball travels after being hit is a function of the velocity (in mph) of the pitched ball. A ball is pitched to a batter with a 35 inch, 32 oz bat that is swung at 70mphfrom the waist and at an angle of 35%. The experiment is repeated 9 times. A partial Minitab printoutappears below. Use 01.throughout this problem.DIST = ……… + ……… VELOCPredictor Coef StDev t-ratio pConstant 31.311 0.999 …………… …………VELOC 0.74667 0.01529 …………… …………s = 1.185 R-sq = ……… Rsq(adj) = ………Analysis of VarianceSOURCE DF SS MS F pRegression 1 3345.1 3345.1 ……… …………Error 7 9.8 1.4Total 8 3354.9 a) The fastest pitchers can throw at about 100 mph. How far will such a pitch be hit? (2)b) What is the value of R-squared? (2) c) Fill in the F space in the ANOVA and explain specifically what is tested and what are the conclusions. (3)d) Is the constant (31.311) significant? Why? (2) [14]Solution: a) The
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