252x0521 4/01/05 (Page layout view!) ECO252 QBA2 Name SECOND HOUR EXAM Hour of Class Registered April 4, 2005 Circle 9am 10am Show your work! Make Diagrams! Exam is normed on 50 points. Answers without reasons are not usually acceptable. I. (8 points) Do all the following. Make diagrams! x ~ N 21, 6 - If you are not using the supplement table, make sure that I know it. 1. P0 x 21 .00 2. Px 10 .22 3. P7.00 x 30 .4 4. x.035 252x0521 4/01/05 (Page layout view!) II. (24+ points) Do all the following? (2points each unless noted otherwise). 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. 1. Computer Problem(Bassett et. al.) 400 children were divided into two groups. The first and larger group were taught Mathematics by traditional methods. The second group was taught by an experimental method. Test scores were recorded and are available. The computer analysis of the data is shown in the three tests below. The First Test was done using method 2. The researcher was reprimanded by her supervisor for assuming that the population variances were equal, so she ran the Second Test without assuming equal variances. Because she was very annoyed at her supervisor she ran the third test for equal variances. The output is below. Assume a significance level of 1%. Do not do any unnecessary computations. #First Test:: MTB > TwoSample c1 c2; SUBC> Pooled; SUBC> Alternative -1. Two-Sample T-Test and CI: x1, x2 Two-sample T for x1 vs x2 N Mean StDev SE Mean x1 250 68.45 7.96 0.50 x2 150 70.62 7.06 0.58 Difference = mu (x1) - mu (x2) Estimate for difference: -2.17521 95% upper bound for difference: -0.87525 T-Test of difference = 0 (vs <): T-Value = -2.76 P-Value = 0.003 DF = 398 P-Value = 0.002 DF = 343 #Second Test: MTB > TwoSample c1 c2; SUBC> Alternative -1. Two-Sample T-Test and CI: x1, x2 Two-sample T for x1 vs x2 x1 x2 N 250 150 Mean 68.45 70.62 StDev 7.96 7.06 SE Mean 0.50 0.58 Difference = mu (x1) - mu (x2) Estimate for difference: -2.17521 95% upper bound for difference: -0.91322 T-Test of difference = 0 (vs <): T-Value = -2.84 #Third Test: MTB > VarTest c1 c2; SUBC> Unstacked. Test for Equal Variances: x1, x2 F-Test (normal distribution) Test statistic = 1.27, p-value = 0.107 a) Turn in your first computer assignment (2) b) Look only at the first test. (i) What are the null and alternative hypotheses? (1) (ii) Can we conclude that the experimental method is better? What are the numbers in the output that bring you to this conclusion? (2) c) Make a drawing of an (almost) Normal curve. Label the center of the curve with a zero and show the area under the curve that is the p-value. (1) 2 252x0521 4/01/05 (Page layout view!) d) Look at the results of the second test. Do they look different to you from the results of the first test? Why? (2) e) Look at the results of the third test? What do you think were the null and alternative hypotheses? Was her supervisor right that she should not have assumed equal variances? Why(2) [8] Questions 2-5 refer to Exhibit 1. Exhibit 1: (Schiffler, Adams) A new feed is supposedly superior to what you have used in the past to feed your pigs. You divide your pigs into two troops of 60 pigs each. After one month the results are as follows. You want to decide if the new feed is actually better. Assume that the sample data comes from two Normal populations with equal population variance. Old Feed x1 175 .9 , s1 12 , n1 60 New Feed x 2 180 .2 , s 2 19 , n 2 60 2. What is the alternative hypothesis? a) 1 2 b) c) 1 2 1 2 1 2 d) e) None of the above. 3. What is s d ? a) b) c) d) e) 0.5 0.7 1.5 2.9 8.4 4. If we do not reject the Null hypothesis, do we decide that there is a reason to switch to the new feed? (1) 5. Change your assumptions to assume that both samples have a population standard deviation of 15. Find a 93% two-sided confidence interval for the difference between the means. (3) Is there a significant difference between the means? Why? (1) [17] 3 252x0521 4/01/05 (Page layout view!) Questions 6 and 7 refer to Exhibit 2. Exhibit 2: (Lees) The net income figures for seven regions in which Smelly-Welly Dirt Devourer is sold are given before and after a reorganization. Region Before After Difference Reorganization Reorganization 1 40 62 -22 2 35 49 -14 3 42 39 3 4 30 28 2 5 55 55 0 6 63 66 -3 7 36 40 -4 The researcher decides that the Wilcoxon Signed Rank test for paired samples is appropriate. The region with a tie is dropped from consideration leaving 6 pairs for this test only. The test is one sided. Minitab gives the following sample statistics for the data for use in Problem 7. (Use .05 ) Description Variable Before x1 n 7 Mean 43.00 After 7 48.43 5.15 13.62 7 -5.43 3.49 9.24 Difference x2 d SE Mean StDev 4.46 11.80 6. In the Wilcoxon Signed Rank Test, the number that you compare to the values in the Wilcoxon Signed Rank Test Table is (3) a. 3 b. 3.5 c. 17 d. 17.5 e. 18 f. 18.5 g. None of the above – write in the correct number and show your work. [20] 7. If we change our assumptions to state that the underlying distribution is Normal, we should not be using the Wilcoxon Signed Rank Test. If we use a test based on the mean we have all of the following: a. Confidence Interval: D d t 2 s d . b. Test Ratio: t d D0 sd c. Critical Value: d cv D0 t 2 s d . On the basis of the information above, find d , s d and the numbers of degrees of freedom. If you do any calculations make sure that I know what they are. (4) [24] 4 252x0521 4/01/05 (Page layout view!) 8. A marble machine is recalibrated, and the owner is afraid that it is producing marbles that are too small. The standard size is 12mm. The following results pop out after 105 diameters are fed to the computer.. One-Sample T: x1 Test of mu = 12 vs < 12 Variable x1 N 105 Mean 12.0150 StDev 0.0498 95% Upper SE Mean Bound T P 0.0049 12.0231 3.09 0.999 Make a diagram showing p-value. Suppose that you were doing a 2-sided test with the same numbers, what would the p-value be.? [26] 9. (Lee) The number of people calling in sick during a certain week is below. M 42 T 33 W 35 R 25 F 45 The null hypothesis is that people are equally likely to call in sick on each day. This is a chi-square test of (1) a. Homogeneity b. Independence c. Uniformity d. Normal distribution e. Poisson Distribution 10. (Lee) The number of people calling in sick during a certain week is below. M 42 T 33 W 35 R 25 F 45 The null hypothesis is that people are equally likely to call in sick on each day. What is the chi-square value? (3) a. 11.9023 b. 6.889 c. 3.296 d. 4.198 e. 7.895 11. (Lee) The number of people calling in sick during a certain week is below. M 42 T 33 W 35 R 25 F 45 The null hypothesis is that people are equally likely to call in sick on each day. Do we reject the null hypothesis at a 5% significance level? No answer will be accepted without a reason. 12. (Lee) The number of people calling in sick during a certain week is below. M 42 T 33 W 35 R 25 F 45 The null hypothesis is that people are equally likely to call in sick on each day. Can we do this by another method than Chi Squared? Do it! (4) [36] 5 252x0521 4/01/05 (Page layout view!) (Mostly Blank page) Location - Normal distribution. Compare means. Location - Distribution not Normal. Compare medians. Paired Samples Method D4 Independent Samples Methods D1- D3 Method D5b Method D5a Proportions Method D6 Variability - Normal distribution. Compare variances. Method D7 6 252x0521 4/01/05 (Page layout view!) ECO252 QBA2 SECOND EXAM April 4, 2005 TAKE HOME SECTION Name: _________________________ Student Number: _________________________ III. Neatness Counts! Show your work! Always state your hypotheses and conclusions clearly. (19+ points) 1) An industrial plant is trying to figure out whether gas or electric fuel is cheaper per delivered quadrillion btus. Random samples of 11 electricity-using plants and 16 gas using plants are taken. The results appear below. The columns labeled x1 and x2 are the data and the rx1 and rx2 columns are the ranks of the numbers within their own column, which you may find useful. Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 x1 45.14 10.11 29.38 19.65 16.25 29.46 8.13 45.63 24.49 12.71 37.04 x2 9.55 38.76 16.65 19.00 17.00 29.01 12.34 11.18 12.15 14.40 8.00 16.19 33.46 18.37 9.86 rx1 10 2 7 5 4 8 1 11 6 3 9 rx2 2 15 9 12 10 13 6 4 5 7 1 8 14 11 3 Minitab gives the following information, which also may help you. Descriptive Statistics: x1, x2 Variable x1 x2 N 11 15 N* 0 0 Mean 25.27 17.73 SE Mean 4.02 2.35 StDev 13.33 9.11 Minimum 8.13 8.00 Q1 12.71 11.18 Median 24.49 16.19 Q3 37.04 19.00 Maximum 45.63 38.7 Before you start, personalize the data as follows: If the second to last number of your student number is 0-4 add it to the second to last digit of the numbers in x1. If the second to last number of your student number is 5-9, divide it by 100 and subtract it from x1. ( If the number is 2, the first two numbers become 45.14 + .20 = 45.34 and 10.21.) (If the number is 6, the first two numbers become 45.14 - .06 = 45.08 and 10.05.) Use .10 a) Compute a (mean and) standard deviation for the electric plants, show your work. Excessive rounding will be penalized throughout this exam. (1) b) Test to see if x1 is Normally distributed. (3) c) Test to see if the standard deviations of the two samples are equal. (1) d) Test to see if the means of the two samples differ significantly on the assumption that your answers to b) and c) showed equal variances and Normal distributions. Use a test ratio, critical value or a confidence interval (4) or all three (6). Your answers to all three should be almost identical.[9] e) Assume that the tests showed unequal variances but Normal distributions, repeat the test (4 extra credit) f) Assume that the tests showed that the distributions were not Normal, repeat the test. (4) [13] 7 252x0521 4/01/05 (Page layout view!) 2) A national survey categorized 600 responses about federal government regulation according to the income of the respondees. Too little Just enough Too much regulation regulation regulation Low income 125 48 27 Medium income 103 58 39 High income 72 69 59 Personalize the data as follows: Subtract the second digit of your student number from the upper lefthand number and add it to the lower right-hand number. Do the following .05 : a) Is there a relation between incomes and views of government regulation? (4) b) Use a method for comparing two proportions to compare the proportion of low income people and medium income people who feel that there there is too little regulation. (3) c) Use the the Marascuilo procedure to compare the proportions of the three income groups that say there is just enough regulation. (Note that you are now dividing responses into ‘just enough and ‘not just enough’ and that this cuts down your degrees of freedom) (4) 8