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UNT DSCI 3710 - Exam 1-DSCI 3710-Spring 2015 Version A KEY

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C. Ha: the mean time to deliver orders for each driver is different from the others.D. Ha: the mean time to deliver orders is the same for drivers 1 and 2.E. Ha: the mean time to deliver orders is the same among order destinations.B. 0.05 < p < 0.10*C. 0.025 < p < 0.05D. 0.01 <p < 0.025E. p < 0.01COURSE: DSCI 3710 Print Name: Exam 1 Version A Signature: Spring 2015 Student ID#: INSTRUCTIONS:- Please print your name and student ID number on this exam. Also, put your signature on this exam.- On your scantron PRINT your name and exam version. - This exam has 25 questions. You have 75 minutes to complete this exam. The exam is open book and open notes. You may use a laptop computer or any type ofhand calculator but please show all your work on the exam and mark all answers on the scantron. Usage of cell phones, digital cameras and other communication devices is prohibited.- Please DO NOT pull this exam apart. When you have completed the exam, pleaseturn your scantron and exam booklet into your instructor, at the front desk.- Good luck and we wish you well on the exam.Note: Whenever question(s) are connected you may be asked to assume a result (givena value) as an answer for the previous question but this result (value) may or may not be correct. The procedure is set in place to prevent you from losing points on a subsequent question because you made a mistake on some previous question/s.Use the information given in the following paragraph to answer the first four questions.260 people are randomly chosen at a shopping mall to taste-test a new brand of fruit drink. They are asked to rate the drink on a scale from 1 to 7, with 1 being very bad and 7 being very good. The results of the survey reveal that the average rating is 5.22 with a standard deviation of 2.31. The marketing division of the fruit drink distributor is only interested in selling this drink if the true mean rating is more than 4.75. 1. What is the alternative hypothesis for testing whether the fruit drink distributor should sell this drink?A. Ha: µ ≠ 4.75B. Ha: µ = 2.25C. Ha: µ > 4.75*D. Ha: µ < 4.75E. Ha: µ > 2.252. What is the calculated value of the test statistic?A. 3.28*B. 2.86C. 4.24D. 2.86E. 1.653. What is the rejection region for testing at the 0.10 level of significance whether the fruit drink distributor should sell this drink?A. Z > 1.645B. Z > 1.96C. Z < -1.28D. Z < -1.96E. Z > 1.28*4. Assuming the calculated value of the test statistic is 1.78, what is the conclusion of testing at the .10 level of significance whether the fruit drink distributor should sell this drink?A. Based on the sample data, there isn't sufficient evidence to conclude that the average rating is more than 4.75.B. Based on the sample data, there isn't sufficient evidence to conclude that the average rating is no more than 4.75.C. Based on the sample data, there is sufficient evidence to conclude that the average rating is no more than 4.75.D. Based on the sample data, there is sufficient evidence to conclude that the average rating is more than 4.75.*E. None of the above.Use the information given in the next paragraph to answer the next four questions.The table below shows the test scores of 10 employees who participated in a new corporate training program that lasted one week. The scores shown are those before and after completing the program, for each employee. Test whether there is significant improvement in the scores afterthe training. Excel analysis at the 5% significance level is shown.EmployeeBefore(1)After(2) t-Test: Paired Two Sample for Means1 98 952 75 81 Before After3 64 75 Mean 69.7 75.64 86 89 Variance 215.57121.825 xx yy Observations 10 106 52 59 Pearson Correlation 0.9317 61 68Hypothesized Mean Difference 08 74 79 df 99 63 64 t Stat -3.124610 51 70 P(T<=t) one-tail 0.0061t Critical one-tail 1.8331P(T<=t) two-tail 0.0122t Critical two-tail 2.2622 5. What is the alternative hypothesis for testing the belief that there was an increase in the mean test scores after the training program? A. Ha: µ2 > 0B. Ha: µ1 > µ2C. Ha: µ2 > µ1 *D. Ha: µ2 = µ1E. Ha: µ1 ≠ µ26. What is the table value (at the 5 % level) of the appropriate critical test statistic to test the belief that there is an increase in the mean test scores of employees who go through the training program?A. 1.833* B. 0.000 C. -3.1246 D. 2.2622 E. 0.01227. What is the p-value for testing the hypothesis that there is an increase in the mean test scoresof employees who participated in the training program?A. -3.1246 B. 0.0061* C. 0.000 D. 2.2622 E. 0.01228. Based on the excel output and the significance level of 5%. What would be your conclusion about whether the training program increased the mean test scores?A. Conclude that there is insufficient evidence that the training program increased the mean test scores since p is less than the significance levelB. Conclude that there is sufficient evidence that the training program increased the mean test scores since p-value is less than the significance level.*C. Conclude that there is insufficient evidence that the training program increased the mean test scores since p is more than the significance level.D. Conclude that there is sufficient evidence that the training program increased the mean test scores since p-value is more than the significance level.E. none of the above Please use the information given in the paragraph below to answer the four questions that follow. 1A university librarian at a university with a large multinational student body is interested in the proportion of students who would like the library search functions available in a language other than English. If over half of the student population favors the Multilanguage search option, the university administration will approve the cost of the new service. A random sample of 350 students was obtained and 186 students indicated they would like the new services. A hypothesis test was conducted using the KPK macro-based MS Excel output provided below to determine if a majority of students favor the new hours. A significance level of 5% was used.Z Test for One Proportion Sample Proportion 0.531429 Number of Observations 350 Ho:XXXX Ha:XXXX Z* 1.175949 P[Z  Z*] XXXX Z Critical,  = 0.05 XXXX 95% CI for Pop. Proportion 0.479075 to 0.5837829. What are the null and alternative hypotheses for this test?A. Ho: p = 0.5 Ha: p ≠ 0.5B. Ho: p < 0.5 Ha: p > 0.5


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