Statistics for Strategy Exam 2 Fall 2014 11:30 am Friday Oct. 3 100 points 1. Bubble Sheet: (Fill in 2 items only) (a) Write and bubble your Name (Last name, b First name) (b) Write and bubble your 8-digit Student ID Number (Begin in the leftmost box.) 2. Exam Booklet: (Fill in 3 items) Name Student ID Number Section Number (Fill in box, worth 3 points on exam ) 3. Sign Tippie Honor Pledge: \"I have neither given nor received assistance on this exam.\" Signature: This is a closed-book exam. Use a calculator (but no cell phones) and pencils only. You have 50 minutes to complete the exam. There are 25 questions, three tables, and a formula sheet on 21 pages. The exam format is multiple choice. Circle the single best answer on the exam booklet and fill the corresponding bubble on the bubble sheet with a No. 2 pencil. The base score for built-in partial credit is 22 points. In addition, 3 points are earned for the correct section section number and for each question which is answered correctly on the bubble sheet. Check ICON to see your score and a list of exam questions which were answered incorrectly. When finished, please put bubble sheet inside the first page of the exam booklet and deposit into box. 1 Questions 1-5. A corporation is trying to decide which of three makes of electric automobiles to purchase for its fleet: domestic, Japanese, or European. Five cars of each type were ordered and driven for 10,000 miles. The operating costs per hundred miles (in dollars) for these cars are recorded in a table: Domestic 18 18 17 19 16 Japanese 15 17 16 17 17 European 19 18 17 18 20 Define 1 = mean operating cost per hundred miles for all domestic vehicles, in dollars 2 = mean operating cost per hundred miles for all Japanese vehicles, in dollars 3 = mean operating cost per hundred miles for all European vehicles, in dollars Part of the MINITAB ANOVA table for these data is shown below, where some of the numbers have been replaced by question marks. Source Factor Error Total DF 2 12 14 SS 10.13 ? ? MS 5.07 ? F ? P ? 1. Describe the alternative hypothesis for the F test. (a) 1 6= 2 and 1 6= 3 and 2 6= 3 (b) 1 = 2 6= 3 (c) 1 = 3 6= 2 (d) 2 = 3 6= 1 (e) Any of the situations described by answers (a), (b), (c), (d) 2. Find MSE. (a) 1.06 (b) 1.13 (c) 3.17 (d) 3.40 (e) 17.46 (c) 1.60 (d) 4.49 (e) 4.78 3. Find the value of F . (a) 0.29 (b) 1.49 4. Find the P -value. (a) 0.030 (b) 0.035 (c) 0.242 (d) 0.264 (continued) 2 (e) 0.753 5. Which conclusion is implied for the F test, using 5% significance? (a) European cars have the smallest mean operating cost per hundred miles. (b) The mean operating costs per hundred miles from smallest to largest are: European, Japanese, domestic. (c) The mean operating costs per hundred miles from smallest to largest are: Japanese, domestic, European. (d) There's not enough evidence to show differences in mean operating costs per hundred miles between domestic, Japanese, and European cars. (e) None of the answers is correct. The following MINITAB steps produce the output shown below: Calc > Probability Distributions > F > (Numerator degrees of freedom = 2 ) > (Denominator degrees of freedom = 12 ) > (Input constant 4.78) > OK Calc Calc Calc Calc > > > > Probability Probability Probability Probability Distributions Distributions Distributions Distributions > > > > F F F F > > > > (Input (Input (Input (Input constant constant constant constant 0.29) 1.49) 4.49) 1.60) > > > > OK OK OK OK Cumulative Distribution Function F distribution with 2 DF in numerator and 12 DF in denominator x 4.78 P(X<=x) 0.970270 Cumulative Distribution Function F distribution with 2 DF in numerator and 12 DF in denominator x 0.29 P(X<=x) 0.246638 Cumulative Distribution Function F distribution with 2 DF in numerator and 12 DF in denominator x 1.49 P(X<=x) 0.735749 Cumulative Distribution Function F distribution with 2 DF in numerator and 12 DF in denominator x 4.49 P(X<=x) 0.964985 Cumulative Distribution Function F distribution with 2 DF in numerator and 12 DF in denominator x 1.6 P(X<=x) 0.757883 3 Questions 6-11. 350 business executives from the Des Moines area were randomly surveyed for their annual incomes and whether or not they have earned a Masters in Business Administration (MBA) degree. The survey includes 110 executives who earn less than $80,000 annually, 140 executives who earn between $80,000 and $120,000 annually, and 100 executives who earn more than $120,000 annually. The numbers of executives in each income category who have earned an MBA are shown below: Annual Income Less Than $80,000 Between $80,000 and $120,000 More Than $120,000 Number who have Earned an MBA 30 60 60 Number of Des Moines Executives Surveyed 110 140 100 (This page and the next are blank. Questions begin on page 6.) 4 (blank page) 5 6. Which of the following statements about Des Moines executives contradicts the null hypothesis of the Chi-Square (2 ) test? (a) The percentage of MBA recipients who earn less than $80,000 annually differs from the percentage of non-MBA recipients who earn less than $80,000 annually. (b) The percentage of MBA recipients who earn less than $80,000 annually is the same as the percentage of non-MBA recipients who earn less than $80,000 annually. 7. Find the critical value of the Chi-Square test, at 5% significance. (a) 3.84 (b) 5.99 (c) 7.81 (d) 9.49 (e) None of the answers is correct to the second decimal place. 8. Calculate the value of the 2 statistic. (Carry calculations to at least one decimal place.) (a) 4.1 (b) 9.2 (c) 22.9 (d) 25.6 (e) 292.4 9. Provide an English interpretation of the Chi-Square test. (a) There is sufficient evidence to show that the percentage of MBA recipients who earn less than $80,000 annually differs from the percentage of non-MBA recipients who earn less than $80,000 annually. (b) There is insufficient evidence to show that the percentage of MBA recipients who earn less than $80,000 annually differs from the percentage of non-MBA recipients who earn less than $80,000 annually. (c) Neither of the above is a correct interpretation. 10. The MINITAB commands and corresponding output for three separate statistical comparisons are shown on the next page. Which comparisons reinforce the conclusion from the Chi-Square test? (a) Comparison 1 only (b) Comparison 2 only (c) Comparison 3 only (d) Comparisons 1 and 3 (e) None of the comparisons reinforce the conclusion (continued on next two pages) 6 11. With 95% confidence, which of the following is a valid conclusion about Des Moines executives? (a) Those without an MBA have between 10.4% and 29.6% greater chance of earning more than $120,000 annually than those with an MBA. (b) Those without an MBA have between 10.4% and 29.6% less chance of earning more than $120,000 annually than those with an MBA. (c) Those with an MBA have between 10.4% less and 10.4% more chance of earning more than $120,000 annually than those without an MBA. (d) There is not a significant difference in the percentage earning more than $120,000 annually between those with an MBA and those without an MBA. (e) None of the answers is correct to the first decimal place. (See MINITAB steps and resulting output next page) 7 Comparison 1 Stat > Basic Statistics > 2 Proportions > (Choose \"Summarized data\") Number of events: Number of trials: Sample 1 80 200 Sample 2 30 150 > OK Test and CI for Two Proportions Sample 1 2 X 80 30 N 200 150 Sample p 0.400000 0.200000 Difference = p (1) - p (2) Estimate for difference: 0.2 95% CI for difference: (0.106687, 0.293313) Test for difference = 0 (vs not = 0): Z = 4.20 P-Value = 0.000 Comparison 2 Stat > Basic Statistics > 2 Proportions > (Choose \"Summarized data\") Number of events: Number of trials: Sample 1 80 200 Sample 2 60 150 > OK Test and CI for Two Proportions Sample 1 2 X 80 60 N 200 150 Sample p 0.400000 0.400000 Difference = p (1) - p (2) Estimate for difference: 0 95% CI for difference: (-0.103712, 0.103712) Test for difference = 0 (vs not = 0): Z = 0.00 P-Value = 1.000 Comparison 3 Stat > Basic Statistics > 2 Proportions > (Choose \"Summarized data\") Number of events: Number of trials: Sample 1 40 200 Sample 2 60 150 > OK Test and CI for Two Proportions Sample 1 2 X 40 60 N 200 150 Sample p 0.200000 0.400000 Difference = p (1) - p (2) Estimate for difference: -0.2 95% CI for difference: (-0.296018, -0.103982) Test for difference = 0 (vs not = 0): Z = -4.08 8 P-Value = 0.000 Questions 12-15. Bob works for Target Stores, Inc. as a data analyst. Target marketing executives have assigned Bob the following project: Determine whether men spend more than women per trip to Target, on average. Bob interviews 70 randomly-selected husband-wife couples while they shop at Target and records the dollar amounts spent by both the husband and the wife during those shopping trips. (This results in 70 men and 70 women in Bob's survey.) 12. Which of the following could be Bob's alternative hypothesis? (a) 1 6= 2 (b) x1 6= x2 (c) p1 > p2 (d) 1 < 2 (e) x1 > x2 13. Is there a pairing mechanism in Bob's analysis? (a) Yes, the data are paired by gender. (b) Yes, the data are paired by store. (c) Yes, the data are paired by Target. (d) Yes, the data are paired by couple. (e) No 14. Which analysis should Bob perform? (a) two-sided one-sample t test (b) one-sided paired two-sample t test (c) one-sided independent two-sample t test (d) two-sided paired two-sample t test (e) two-sided independent two-sample t test 15. Now suppose that Bob wishes to answer a different question: Do three groups of Target customers those younger than 25, those between 25 and 50 years old, and those older than 50 spend different dollar amounts, on average, per trip to Target? What is the first type of test which Bob should apply? (a) Z test (b) F test (c) one-sided t test (d) two-sided t test (e) 2 test 9 Questions 16-17. An experiment was performed in which 78 children were divided into four groups: 34 children received piano lessons, 10 received singing lessons, 20 received computer instruction lessons, and 14 received no lessons. Each child was tested twice for skills in spatial reasoning: on a pretest before any lessons, and again on a posttest after any lessons. (Higher scores indicate greater skill.) A child's improvement in spatial reasoning from receiving lessons is defined as Improvement = Posttest Score Pretest Score Refer to MINITAB output on page 11. 16. We worked with clusters of means in Homework 4. Using a 5% significance level and 95% confidence, identify the distinct clusters. (a) There is one cluster which contains all four groups of children. (b) There are three clusters: Cluster 1 = singing Cluster 2 = piano, computer Cluster 3 = none (c) There are three clusters: Cluster 1 = piano Cluster 2 = singing Cluster 3 = computer, none (d) There are two clusters: Cluster 1 = singing, computer, none Cluster 2 = piano (e) None of the answers is correct since some of the clusters are not distinct. 17. With 95% confidence, which of the following conclusions is correct? (a) All three types of lessons piano, singing, and computer improve childrens' spatial reasoning, on average. (b) None of three types of lessons piano, singing, and computer improve childrens' spatial reasoning, on average. (c) Singing lessons are associated with an actual decline in childrens' spatial reasoning, on average, while piano and computer lessons improve childrens' spatial reasoning, on average. (d) Piano lessons improve childrens' spatial reasoning, on average, while computer and singing lessons are not shown to do so. (e) Piano lessons improved childrens' spatial reasoning scores in the sample, on average, but not by a statistically significant amount. 10 One-way ANOVA: Score versus Lesson Method Null hypothesis Alternative hypothesis Significance level All means are equal At least one mean is different alpha = 0.05 Equal variances were assumed for the analysis. Factor Information Factor Lesson Levels 4 Values Computer, None, Piano, Singing Analysis of Variance Source Lesson Error Total DF 3 74 77 Adj SS 207.3 553.4 760.7 Adj MS 69.094 7.479 F-Value 9.24 P-Value 0.000 Model Summary S 2.73475 R-sq 27.25% R-sq(adj) 24.30% R-sq(pred) 20.04% Means Lesson Computer None Piano Singing N 20 14 34 10 Mean 0.450 0.786 3.618 -0.300 StDev 2.212 3.191 3.055 1.494 95% (-0.768, (-0.671, ( 2.683, (-2.023, CI 1.668) 2.242) 4.552) 1.423) Pooled StDev = 2.73475 Tukey Pairwise Comparisons Tukey Simultaneous Tests for Differences of Means Difference of Levels None - Computer Piano - Computer Singing - Computer Piano - None Singing - None Singing - Piano Difference of Means 0.336 3.168 -0.75 2.832 -1.09 -3.918 SE of Difference 0.953 0.771 1.06 0.868 1.13 0.984 Individual confidence level = 98.96% 11 95% CI (-2.171, 2.842) ( 1.140, 5.195) ( -3.54, 2.04) ( 0.548, 5.116) ( -4.06, 1.89) (-6.505, -1.330) T-Value 0.35 4.11 -0.71 3.26 -0.96 -3.98 Adjusted P-Value 0.985 0.001 0.894 0.009 0.773 0.001 Questions 18-21. A group of individuals who live either in Cedar Rapids or Iowa City were cross-classified according to annual income and city of residence, as shown below: Annual Income Under $10,000 $10,000 to $19,999 $20,000 to $24,999 $25,000 to $34,999 $35,000 and more Cedar Rapids 70 52 69 22 28 Iowa City 62 63 50 19 24 (This page and the next are blank. Questions begin on page 14.) 12 (blank page) 13 18. Find the critical value of the Chi-Square test, at 5% significance. (a) 3.84 (b) 5.99 (c) 7.81 (d) 9.49 (e) None of the answers is correct to the second decimal place. 19. Calculate the value of the 2 statistic. (Carry calculations to at least 3 decimal places.) (a) 0.40 (b) 2.62 (c) 3.96 (d) 12.54 (e) 459.00 20. Provide an English interpretation of the Chi-Square test. (a) There is sufficient evidence to show that the percentage of Iowa City residents who earn at least $35,000 annually differs from the percentage of Cedar Rapids residents who earn less at least $35,000 annually. (b) There is insufficient evidence to show that the percentage of Iowa City residents who earn at least $35,000 annually differs from the percentage of Cedar Rapids residents who earn less at least $35,000 annually. (c) There is insufficient evidence to show that city of residence and annual income are related. (d) There is sufficient evidence to show that city of residence and annual income are not related. (e) None of the above is a correct interpretation. 21. Suppose you know that a particular person lives in either Iowa City or Cedar Rapids, but you don't know which. Further suppose that the person tells you that her annual income is at least $35,000. Has the Chi-Square test shown that this information about her income will help you to guess the city in which she lives? (a) Yes (b) No (c) It is not possible to answer this question without further information. 14 Questions 22-25. A study recorded the numbers of houses sold in Iowa City in February, 2014 by nine realtors who were randomly selected from all realtors who worked in the Iowa City market in February, 2014. A second study recorded the numbers of houses sold in Iowa City in February, 2015 by nine realtors who were randomly selected from all realtors who worked in the Iowa City market in February, 2015. Data from the two studies are shown below. Feb. 2014 11 30 6 13 5 4 15 17 12 Feb. 2015 8 19 5 9 3 0 13 11 9 Is the housing market cooling in Iowa City? In particular, has the mean number of houses sold per realtor in the Iowa City market decreased in February 2015 compared to February 2014? Test using 5% significance. The MINITAB commands Stat > Basic Statistics > 2-sample t > (Each sample is in its own column): Sample 1: 'Feb. 2014' Sample 2: 'Feb. 2015' > Options > Alternative hypothesis: Difference > hypothesized difference > OK > OK produce the following output: Two-Sample T-Test and CI: Feb. 2014, Feb. 2015 Two-sample T for Feb. 2014 vs Feb. 2015 Feb. 2014 Feb. 2015 N 9 9 Mean 12.56 8.56 StDev 7.95 5.61 SE Mean 2.7 1.9 Difference = mu (Feb. 2014) - mu (Feb. 2015) Estimate for difference: 4.00 95% lower bound for difference: -1.72 T-Test of difference = 0 (vs >): T-Value = 1.23 The additional MINITAB commands Stat > Basic Statistics > Paired t > (Each sample is in a column): Sample 1: 'Feb. 2014' Sample 2: 'Feb. 2015' 15 P-Value = 0.119 DF = 14 > Options > Alternative hypothesis: Difference > hypothesized difference > OK > OK produce the additional output: Paired T-Test and CI: Feb. 2014, Feb. 2015 Paired T for Feb. 2014 - Feb. 2015 Feb. 2014 Feb. 2015 Difference N 9 9 9 Mean 12.56 8.56 4.00 StDev 7.95 5.61 3.00 SE Mean 2.65 1.87 1.00 95% lower bound for mean difference: 2.14 T-Test of mean difference = 0 (vs > 0): T-Value = 4.00 P-Value = 0.002 22. Let 1 = mean number of houses sold per Iowa City realtor in Feb. 2015 2 = mean number of houses sold per Iowa City realtor in Feb. 2014 Define the hypotheses: (a) HA : 1 > 2 (b) HA : 1 < 2 H0 : 1 2 H0 : 1 2 (c) HA : 1 6= 2 H0 : 1 = 2 (d) HA : (1 2 ) = 0 H0 : (1 2 ) 6= 0 (e) None of the hypotheses is correct. 23. Find the P -value. (a) 0.119 (b) 0.881 (c) 0.002 (d) 0.998 (e) None of the answers is correct to the third decimal place. 24. Make a decision. (a) Reject H0 (b) Fail to Reject H0 (c) Not possible to determine without further information 25. Interpret the decision. (a) There is sufficient evidence to show that the mean number of sales per realtor is less in Feb. 2014 than in Feb. 2015. (b) There is insufficient evidence to show that the mean number of sales per realtor is less in Feb. 2014 than in Feb. 2015. (c) There is sufficient evidence to show that the mean number of sales per realtor is greater in Feb. 2014 than in Feb. 2015. (d) There is insufficient evidence to show that the mean number of sales per realtor is greater in Feb. 2014 than in Feb. 2015. (e) None of the answers is correct. 16 Exam 2 Formulas s x t n r pb z pb (1 pb ) n (b p1 pb2 ) z pb1 bp2 s pb1 bp2 pb1 (1 pb1 ) pb2 (1 pb2 ) + n1 n2 s ( x1 x2 ) t t= s2 s21 + 2 n1 n2 ( x1 x2 ) (1 2 ) q 2 s1 s2 + n22 n1 2 = X (O E)2 E all cells s2p = (average of s21 , s22 , . . . , s2k ) = MSE if n1 = n2 = = nk (sample sizes are all equal) F = DFG = k 1, where k = # of groups MSG MSE DFE = N k, where N = total # of measurements 17 Solution 1. e 2. b 3. d 4. b 5. e The F test can indicate if there is a difference in mean operating costs among the three auto types but cannot determine significant differences between any two particular auto types. (Tukey CI's are needed to do that.) 6. a 7. b 8. c 9. c There is enough evidence to show that income and having an MBA are related. 10. d 11. b 12. d If 1 = average spent per trip to Target by women and 2 = average spent per trip to Target by men 13. d 14. b 15. b 16. d 17. d Compare each of the three groups piano, computer and singing to none 18. d 19. c 20. c 21. b 22. b 23. a 24. b 25. d 22