STAT 1220 Common Final Exam SPRING 2016 May 5, 2016 PLEASE PRINT THE FOLLOWING INFORMATION: Name: Instructor: Student ID #: Section/Time: THIS EXAM HAS TWO PARTS. PART I. Part I consists of 30 multiple choice questions. Each correct answer is scored 2 points; each incorrect (or blank) answer is scored 0, so there is no penalty for guessing. You may do calculations on the test paper, but your answers must be marked on the OPSCAN sheet with a soft lead pencil (HB or No. 2 lead). Any question with more than one choice marked will be counted as incorrect. If more than one choice seems correct, choose the one that is most complete or most accurate. Make sure that your name and ID number are written and correctly bubbled on the OPSCAN sheet. PART II. Part II consists of 3 free response questions, with values as indicated. You must show all work in the space provided or elsewhere on the exam paper in a place that you clearly indicate. Work on loose sheets will not be graded. FOR DEPARTMENT USE ONLY: Part II. Question 1 2 Score Part I Part II TOTAL 3 Cumulative Binomial Probability P (X x) p n 5 x 0 1 2 3 4 0.05 0.7738 0.9774 0.9988 1.0000 1.0000 0.10 0.5905 0.9185 0.9914 0.9995 1.0000 0.25 0.2373 0.6328 0.8965 0.9844 0.9990 0.3 3 0.1317 0.4609 0.7901 0.9547 0.9959 0.50 0.0313 0.1875 0.5000 0.8125 0.9688 0.6 6 0.0041 0.0453 0.2099 0.5391 0.8683 0.75 0.0010 0.0156 0.1035 0.3672 0.7627 0.90 0.0000 0.0005 0.0086 0.0815 0.4095 0.95 0.0000 0.0000 0.0012 0.0226 0.2262 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 0.3585 0.7358 0.9245 0.9841 0.9974 0.9997 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.1216 0.3917 0.6769 0.8670 0.9568 0.9887 0.9976 0.9996 0.9999 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.0032 0.0243 0.0913 0.2252 0.4148 0.6172 0.7858 0.8982 0.9591 0.9861 0.9961 0.9991 0.9998 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 0.0003 0.0033 0.0176 0.0604 0.1515 0.2972 0.4793 0.6615 0.8095 0.9081 0.9624 0.9870 0.9963 0.9991 0.9998 1.0000 1.0000 1.0000 1.0000 1.0000 0.0000 0.0000 0.0002 0.0013 0.0059 0.0207 0.0577 0.1316 0.2517 0.4119 0.5881 0.7483 0.8684 0.9423 0.9793 0.9941 0.9987 0.9998 1.0000 1.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0009 0.0037 0.0130 0.0376 0.0919 0.1905 0.3385 0.5207 0.7028 0.8485 0.9396 0.9824 0.9967 0.9997 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0002 0.0009 0.0039 0.0139 0.0409 0.1018 0.2142 0.3828 0.5852 0.7748 0.9087 0.9757 0.9968 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0004 0.0024 0.0113 0.0432 0.1330 0.3231 0.6083 0.8784 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0003 0.0026 0.0159 0.0755 0.2642 0.6415 STAT 1220 Page 1 SPRING 2016 Part I Problems 1-5 pertain to the data set of 90 measurements represented by the following stem and leaf diagram, in which the stems in the tens place and leaves are the digits in the ones place. P are the digits P For these data x = 6, 600 and x2 = 500, 940. 9 8 7 6 5 4 0 0 0 0 0 6 0 0 0 1 2 6 1 1 0 1 3 7 2 1 1 2 5 8 3 1 1 2 5 9 4 2 1 3 5 9 4 2 1 3 6 5 3 2 4 6 6 4 2 4 7 7 4 2 6 8 7 5 3 6 8 5 3 6 5 3 7 6 4 7 6 4 8 7 5 9 8 5 8 6 9 7 9 7 7 8 8 8 9 9 1. The sample mean of this data set is about (a) 75.4 (b) 76.8 (c) 73.3 (d) 70.5 (e) 66.0 2. The sample standard deviation of this data set is about (a) 190.3 (b) 13.8 (c) 52.8 (d) 11.3 (e) 13.7 3. The sample median is about (a) 69 (b) 72.5 (c) 76.5 (d) 73.5 (e) 73 4. The sample range is about (a) 52 (b) 90 (c) 64 (d) 13 (e) 98 5. The percentile rank of the measurement 89 is about (a) 89 (b) 91 (c) 87 (d) 10 (e) 84 6. The gestation period for most dog breeds is roughly bell-shaped with mean 63 and standard deviation 1.5 days. The proportion of litters that are born between 60 and 66 days of gestation is about (a) 0.95 (b) 0.68 (c) 0.975 (d) 0.50 (e) 0.997 STAT 1220 Page 2 SPRING 2016 Problems 7 and 8 pertain to the information in the following two-way contingency table, relating the result of a test for a disease and the health with respect to that disease of the person tested: Test positive Test negative Test inconclusive diseased 0.043 0.016 0.001 healthy 0.007 0.914 0.019 7. The probability that a randomly selected person is healthy (free of the disease) is about (a) 0.940 (b) 0.050 (c) 0.914 (d) 0.933 (e) 0.007 8. The probability that a randomly selected person is healthy (free of the disease), given that the test for the disease is negative, is about (a) 1.000 (b) 0.914 (c) 0.972 (d) 0.983 (e) 0.815 Problems 9 and 10 are based on the following information: in a certain region the probability that a randomly selected adult has a credit card is 0.75, that he has a debit card is 0.60, and that he has both is 0.45. 9. The probability that a randomly selected adult has either a credit card or a debit card is about (a) 0.15 (b) 0.90 (c) 1.35 (d) 0.45 (e) impossible to tell (insufficient information) 10. The events C: a randomly selected adult has a credit card and D: a randomly selected adult has a debit card are (a) Independent because P (C) P (D) = P (C D) (b) Independent because P (C) P (D) 6= P (C D) (c) Dependent because P (C) P (D) = P (C D) (d) Dependent because P (C) P (D) 6= P (C D) (e) impossible to tell if independent or not (insufficient information) STAT 1220 Page 3 SPRING 2016 Problems 11-13 pertain to the following probability distribution of the number X of defective items in a randomly selected standard container of 5,000 fasteners. (No container ever has more than 5 defective items.) x P (x) 0 p 1 0.27 2 0.20 3 0.06 4 0.02 5 0.01 11. The missing entry p is about (a) 0.20 (b) 0.00 (c) 0.16 (d) 0.44 (e) 0.32 12. The probability that the container will contain at least two defective fasteners is about (a) 0.20 (b) 0.29 (c) 0.14 (d) 0.71 (e) 0.09 13. The average number of defective fasteners per container of 5,000 is about (a) 2.5 (b) 1.2 (c) 0.11 (d) 0.98 (e) 1.5 Problems 14 and 15 pertain to the following situation: a woman buys 20 one-dollar lottery tickets per month. The probability of any ticket being a winning ticket is 0.10 or 10%. 14. The probability that in any one month at least three of the tickets that the woman buys are winning tickets is about (a) 0.2000 (b) 0.8670 (c) 0.3231 (d) 0.3000 (e) 0.6769 15. The average number of winning tickets that the woman buys each month is about (a) 0.2 (b) 1.3 (c) 2.1 (d) 1.5 (e) 2.0 16. A researcher wishes to estimate the mean level of student loan debt of bachelor's degree students at graduation, with 95% confidence and to within $500. Assuming that the standard deviation of student loan debt to be about $3,000, the estimated minimum sample size needed is about (a) 98 (b) 139 (c) 10 (d) 67 (e) 195 STAT 1220 Page 4 SPRING 2016 17. To estimate the effectiveness of a tutoring center its director chose 20 students at random to compare their test scores before using the center with their test scores after using the center. The correct formula to use to construct a confidence interval for the mean difference in scores is q (n 1)s21 +(n2 1)s22 (a) x t/2 ( sn ) (b) d t/2 sdn (c) ( x1 x 2 ) t/2 s2p ( n11 + n12 ), s2p = 1 n1 +n 2 2 q 2 q s2 s (e) (b p1 pb2 ) z/2 pbn1 q1b1 + pbn2 q2b2 (d) ( x1 x 2 ) z/2 n11 + n22 18. The yield of wheat on unirrigated land owned by a commercial grower is normally distributed with mean 42.3 bushels and standard deviation 2.4 bushels per acre. The probability that a randomly selected tract of unirrigated land will produce at least 40.5 bushels per acre is about (b) 0.2266 (a) 0.8361 (c) 0.6628 (d) 0.2266 (e) 0.7734 19. Letter grades will be assigned in a multi-section course according to the principle that the top and bottom 5% of final scores will be awarded A's and F's, respectively, the next 10% will be awarded B's and D's, and the remaining middle 70% will be awarded C's. If final scores this semester are normally distributed with mean 68.3 and standard deviation 9.9 then the range of scores that will be awarded C's (that is, the interval that is symmetric about 68.3 and has area 0.70 above it) is about (a) [61.9, 74.7] (b) [63.2, 73.4] (c) [58.0, 78.6] (d) [56.5, 80.1] (e) [60.0, 76.6] 20. Household incomes in a certain state have mean $53,550 and standard deviation $3,750. The probability that the mean of a random sample of 200 households will be within $500 of the true population mean is about (a) 0.10 (b) 0.94 (c) 0.88 (d) 0.34 (e) 0.65 21. Based on past experience an airline food caterer knows that when presented with the choice between meal A and meal B, 58% will select meal A. On a flight seating 175 passengers, the probability that at least 65% will select meal A is about (a) 0.97 (b) 0.43 (c) 0.15 (d) 0.08 (e) 0.03 STAT 1220 Page 5 SPRING 2016 22. In a sample of 100 individuals who had retired in 2015, the number of years worked had mean 45.6 and standard deviation 3.8 years. A 90% confidence interval for the mean number of years worked by all individuals who had retired in 2015 is about (a) [45.0, 46.2] (b) [45.4, 45.8] (c) [45.1, 46.1] (d) [44.9, 46.3] (e) [44.6, 46.6] 23. In a sample of 335 small businesses (those employing at most 50 workers), owners of 81 of them indicated that they wanted to expand their business. A 98% confidence interval for the proportion of all small businesses that intend to expand is about (a) 0.2420.060 (b) 0.2420.046 (c) 0.2420.038 (d) 0.2420.054 (e) 0.2420.073 24. The estimated ages (in thousands of years) obtained by radiocarbon dating of five artifacts from an archaeological site were: 16.82, 16.30, 15.30, 15.88, 16.54 Assuming that the population of ages of all artifacts is normally distributed, an 90% confidence interval for the mean age of all artifacts from this site is about (a) 16.17 0.52 (b) 16.17 0.39 (c) 16.17 0.54 (d) 16.17 0.41 (e) 16.17 0.57 25. A pollster wishes to estimate, with 95% confidence and to within three percentage points, the proportion of all registered voters who favor a flat income tax. Using the fact that one year ago the proportion was 22% or 0.22, the minimum estimated sample size needed to meet these objectives is about (a) 1,214 (b) 752 (c) 1,068 (d) 516 (e) 733 26. In a test of hypotheses of the form H0 : = 0 versus Ha : < 0 using = 0.01, when the sample size is 16 and the population is normally distributed but of unknown standard deviation the rejection region will be the interval (a) (, 2.947] (b) (, 2.583] (c) (, 2.921] (d) (, 2.602] (e) (, 2.326] STAT 1220 Page 6 SPRING 2016 27. In a test of hypotheses H0 : = 27 versus Ha : 6= 27 in a normally distributed population, the rejection region is the union of intervals (, 2.120] [2.120, ), the value of the sample mean computed from a sample of size 17 is x = 28.5, and the value of the test statistic is t = 2.811. The correct decision and justification are: (a) Do not reject H0 because the sample is small. (b) Do not reject H0 because 28.5 lies in the rejection region. (c) Reject H0 because x = 28.5 6= 27 = 0 . (d) Reject H0 because the test statistic is negative. (e) Reject H0 because 2.811 < 2.120. 28. In a test of hypotheses H0 : p = 0.38 vs. Ha : p 6= 0.38, at the 5% level of significance, a sample of size 782 produced the test statistic z = 1.483. The p-value (the observed significance) of the test is about (a) 0.050 (b) 0.1388 (c) 1.483 (d) 0.025 (e) 0.0694 29. The National Association of Colleges and Employers conducted a survey to test whether mean starting salaries of graduates from Region I differed from those of Region II. To do so they recorded the starting salaries of 145 randomly selected graduates from Region I and of 145 randomly selected graduates of Region II. The correct null and alternative hypotheses for this test are (a) H0 : x 1 x 2 = 0 vs. Ha : x 1 x 2 6= 0 (b) H0 : d = 0 vs. Ha : d 6= 0 (c) H0 : p1 p2 = 0 vs. Ha : p1 p2 6= 0 (d) H0 : 1 2 = 0 vs. Ha : 1 2 6= 0 (e) H0 : 1 2 = 0 vs. Ha : 1 2 > 0 30. Analysis of data gathered by the U. S. Census Bureau relating level of education x (in years) and unemployment rate y (in percentage points) in a recent year yielded r = 0.826, se = 1.224, and the regression equation y = 1.17x + 26.2. For each additional year of education the unemployment rate (a) decreases by about 1.17 percentage points (b) decreases by about 26.2 percentage points (c) decreases by about 0.83 percentage points (d) decreases by about 1.22 percentage points (e) changes by an amount that cannot be determined from the information given STAT 1220 Page 7 SPRING 2016 Part II 1. Historically the mean nurse-to-patient ratio at top rated cancer care facilities has been 1.6. To test whether it is different now with changes that have occurred in the health care system, the ratios at nine facilities were measured, yielding x = 1.93 and s = 0.50. Perform the relevant test at the 1% level of significance, in the following sequence of steps. Assume that the population of nurse-to-patient ratios is normally distributed. (a) State the null and alternative hypotheses for the test. [2 points] (b) State the formula for the test statistic and compute its value. Justify your answer. [4 points] (c) Construct the rejection region and make a decision. [4 points] (d) State a conclusion about the mean nurse-to-patient ratio at top rated cancer care facilities, based on the test you performed. [2 points] STAT 1220 Page 8 SPRING 2016 2. In order to test whether the presence of fetal stem cells that have crossed the placenta and remained in the woman's body have afforded some level of protection against breast cancer, 54 women with breast cancer and 45 who were cancer-free were examined. Fetal cells were present in 14 of the women with breast cancer and in 25 of the women without. Test the hypothesis that the proportion p1 of women with breast cancer whose bodies contain fetal stem cells is less than the proportion p2 of women without breast cancer whose bodies contain fetal stem cells, at the 1% level of significance, in the following series of steps. (The samples are sufficiently large to perform the test.) (a) State the null and alternative hypotheses for the test. [2 points] (b) State the formula for the test statistic and compute its value. Justify your answer. [4 points] (c) Construct the rejection region and make a decision. [4 points] (d) State a conclusion about the two population proportions based on the test you performed. [2 points] (e) Compute the p-value (the observed significance) of the test. [2 points] STAT 1220 Page 9 SPRING 2016 3. The total assets x and shareholders' equity y for a large corporation were recorded in each of six randomly selected recent years, in units of billions of dollars. Summary information is: 11.4 6 x 6 37.5 5.4 6 y 6 14.8 SSxx = 509.7283 x = 24.016 SSxy = 178.3283 y = 9.916 SSyy = 62.5283 (a) Find the proportion of the variability in shareholders' equity that is accounted for by total assets. [4 points] (b) Find the regression line for predicting y from x. [4 points] (c) Describe how shareholder equity responds to a $1 billion increase in total assets. [2 points] (d) For any year in which total assets is $25 billion find the shareholders' equity predicted by the regression equation found in part (b). [2 points] (e) State whether or not the same computation as in part (d) but for a year in which total assets is $40 billion is valid, and why. [2 points]