MS1023 Business Statistics with Computer Applications

QUESTION

MS1023 Business Statistics with Computer ApplicationsHomework #4MS1023 Business Statistics w/Comp Apps IHomework #4 – Use Red Par Score FormChps. 9 & 10: 50 Questions Only7. The rejection region for a hypothesis testbecomes smaller if the level of significanceis changed from 0.01 to 0.05.1. The first step in testing a hypothesis is toestablish a true null hypothesis and a falsealternative hypothesis.a) Trueb) Falsea) Trueb) False8. Whenever hypotheses are establishedsuch that the alternative hypothesis is "μ>8",where μ is the population mean, thehypothesis test would be a two-tailed test.2. In testing hypotheses, the researcherinitially assumes that the alternativehypothesis is true and uses the sample datato reject it.a) Trueb) Falsea) Trueb) False3. The null and the alternative hypothesesmust be mutually exclusive and collectivelyexhaustive.a) Trueb) False4. Generally speaking, the hypotheses thatbusiness researchers want to prove are statedin the alternative hypothesis.a) Trueb) False5. When a true null hypothesis is rejected,the researcher has made a Type I error.a) Trueb) False6. When a false null hypothesis is rejected,the researcher has made a Type II error.a) Trueb) FalseMaho Sonmez maho.sonmez@utsa.edu9. Whenever hypotheses are establishedsuch that the alternative hypothesis is "μ >8", where μ is the population mean, the pvalue is the probability of observing asample mean greater than the observedsample mean assuming that μ ≥ 8.a) Trueb) False10. If a null hypothesis was not rejected atthe 0.10 level of significance, it will berejected at a 0.05 level of significance basedon the same sample results.a) Trueb) False11. Consider the following null andalternative hypotheses.Ho: 67Ha: > 67These hypotheses _______________.a) indicate a one-tailed test with a rejectionarea in the right tailb) indicate a one-tailed test with a rejectionarea in the left tailc) indicate a two-tailed testd) are established incorrectlye) are not mutually exclusive1MS1023 Business Statistics with Computer ApplicationsHomework #412. Consider the following null andalternative hypotheses.Ho: p .16Ha: p < .16These hypotheses _______________.a) rejectb) not rejectc) redefined) change the alternate hypothesis intoe) restatea) indicate a one-tailed test with a rejectionarea in the right tailb) indicate a one-tailed test with a rejectionarea in the left tailc) indicate a two-tailed testd) are established incorrectlye) are not mutually exclusive16. A researcher is testing a hypothesis of asingle mean. The critical t value for = .05and a two-tailed test is +2.0796. Theobserved t value from sample data is -2.92.The decision made by the researcher basedon this information is to _____ the nullhypothesis.13. Suppose the alternative hypothesis in ahypothesis test is: "the population mean isless than 60". If the sample size is 27, isunknown, and alpha =.05, the critical valueof t is _______.a) rejectb) not rejectc) redefined) change the alternate hypothesis intoe) restatea) 2.0555b) -2.0555c) -1.7056d) -1.7033e) -2.051817. The diameter of 3.5 inch diskettes isnormally distributed. Periodically, qualitycontrol inspectors at Dallas Diskettesrandomly select a sample of 16 diskettes. Ifthe mean diameter of the diskettes is toolarge or too small the diskette punch is shutdown for adjustment; otherwise, thepunching process continues. The nullhypothesis is ______.14. Suppose the alternative hypothesis in ahypothesis test is "the population mean isgreater than 60". If the sample size is 24,is unknown, and alpha = .01, the criticalvalue of t is _______.a) -2.8073b) -2.7969c) 2.8073d) 2.4999e) 2.492215. A researcher is testing a hypothesis of asingle mean. The critical t value for = .05and a one-tailed test is 2.0639. The observedt value from sample data is 1.742. Thedecision made by the researcher based onthis information is to ______ the nullhypothesis.Maho Sonmez maho.sonmez@utsa.edua) n 16b) n = 16c) = 3.5d) 3.5e) ≥ 3.518. The diameter of 3.5 inch diskettes isnormally distributed. Periodically, qualitycontrol inspectors at Dallas Diskettesrandomly select a sample of 16 diskettes. Ifthe mean diameter of the diskettes is toolarge or too small the diskette punch is shutdown for adjustment; otherwise, thepunching process continues. The lastsample showed a mean and standarddeviation of 3.55 and 0.08 inches,2MS1023 Business Statistics with Computer ApplicationsHomework #4respectively. Using = 0.05, theappropriate decision is _______.21. Using = 0.01, the appropriate decisionis to _______ the null hypothesis.a) reject the null hypothesis and shut downthe punchb) reject the null hypothesis and do not shutdown the punchc) do not reject the null hypothesis and shutdown the punchd) do not reject the null hypothesis and donot shut down the punche) do nothinga) rejectb) not rejectc) redefined) change the alternate hypothesis intoe) restateThe following data (in pounds), which wereselected randomly from a normallydistributed population of values, representmeasurements of a machine part that issupposed to weigh, on average, 8.3 pounds.8.18.48.28.58.48.38.38.78.38.48.18.28.28.38.58.88.48.68.28.5Use these data and alpha = 0.01 to test thehypothesis that the parts average 8.3 pounds.Answer the questions 19-21 based on theabove information.19. The null hypothesis is ______.a) n 20b) n = 20c) = 8.3d) 8.3e) ≥ 8.3Downtime in manufacturing is costly andcan result in late deliveries, backlogs, failureto meet orders, and even loss of marketshare. Suppose a manufacturing plant hasbeen averaging 23 minutes of downtime perday for the past several years, but during thepast year, there has been a significant effortby both management and productionworkers to reduce downtime. In an effort todetermine if downtime has beensignificantly reduced, company productivityresearchers have randomly sampled 31 daysover the past several months from companyrecords and have recorded the dailydowntimes shown below in minutes. Usethese data and an alpha of .01 to test todetermine if downtime has beensignificantly reduced. Assume that dailydowntimes are normally distributed in thepopulation.1916232623242218272219172728192619171915213224171816241918252120. The critical value is ______.a) -2.8609b) 2.8609c) ±2.8453d) ±2.5395e) ±2.8609Maho Sonmez maho.sonmez@utsa.eduAnswer the questions 22-25 based on theabove information.22. The alternative hypothesis is ______.a) = 233MS1023 Business Statistics with Computer Applicationsb) > 23c) ≤ 23d) 23e) < 2323. The critical value is ______.a) -2.4528b) 2.4528c) 2.4573d) -2.4573e) ±2.750024. Using = 0.01, the appropriate decisionis to _______ the null hypothesisa) rejectb) not rejectc) redefined) change the alternate hypothesis intoe) restate25. Using = 0.05, the appropriate decisionis to _______ the null hypothesisa) rejectb) not rejectc) redefined) change the alternate hypothesis intoe) restate26. Ophelia O’Brien, VP of ConsumerCredit of American First Banks (AFB),monitors the default rate on personal loansat the AFB member banks. One of herstandards is "no more than 5% of personalloans should be in default." On each Friday,the default rate is calculated for a sample of500 personal loans. Last Friday’s samplecontained 30 defaulted loans. The nullhypothesis is _______.a) p ≥ 0.05b) p ≤ 0.05c) p > 0.05Maho Sonmez maho.sonmez@utsa.eduHomework #4d) p < 0.05e) p = 0.0527. Ophelia O’Brien, VP of ConsumerCredit of American First Banks (AFB),monitors the default rate on personal loansat the AFB member banks. One of herstandards is "no more than 5% of personalloans should be in default." On each Friday,the default rate is calculated for a sample of500 personal loans. Last Friday’s samplecontained 30 defaulted loans. Using =0.10, the critical z value is _______.a) 1.645b) -1.645c) 1.28d) -1.28e) 2.2828. Ophelia O’Brien, VP of ConsumerCredit of American First Banks (AFB),monitors the default rate on personal loansat the AFB member banks. One of herstandards is "no more than 5% of personalloans should be in default." On each Friday,the default rate is calculated for a sample of500 personal loans. Last Friday’s samplecontained 30 defaulted loans. Using =0.10, the observed z value is _______.a) 1.03b) -1.03c) 0.046d) -0.046e) 1.3329. Ophelia O’Brien, VP of ConsumerCredit of American First Banks (AFB),monitors the default rate on personal loansat the AFB member banks. One of herstandards is "no more than 5% of personalloans should be in default." On each Friday,the default rate is calculated for a sample of500 personal loans. Last Friday’s sample4MS1023 Business Statistics with Computer Applicationscontained 30 defaulted loans. Using =0.10, the appropriate decision is _______.a) reduce the sample sizeb) increase the sample sizec) reject the null hypothesisd) do not reject the null hypothesise) do nothing30. Ophelia O’Brien, VP of ConsumerCredit of American First Banks (AFB),monitors the default rate on personal loansat the AFB member banks. One of herstandards is "no more than 5% of personalloans should be in default." On each Friday,the default rate is calculated for a sample of500 personal loans. Last Friday’s samplecontained 38 defaulted loans. Using =0.10, the appropriate decision is _______.a) reduce the sample sizeb) increase the sample sizec) reject the null hypothesisd) do not reject the null hypothesise) do nothing31. The executives of CareFree Insurance,Inc. feel that "a majority of our employeesperceive a participatory management style atCareFree." A random sample of 200CareFree employees is selected to test thishypothesis at the 0.05 level of significance.Eighty employees rate the management asparticipatory. The null hypothesis is__________.a) p ≥ 40b) p < 40c) p ≥ 0.50d) p < 0.50e) n > 20032. A study by Hewitt Associates showedthat 79% of companies offer employeesflexible scheduling. Suppose a researcherbelieves that in accounting firms this figureMaho Sonmez maho.sonmez@utsa.eduHomework #4is lower. The researcher randomly selects415 accounting firms and through interviewsdetermines that 303 of these firms haveflexible scheduling. With a 1% level ofsignificance, does the test show enoughevidence to conclude that a significantlylower proportion of accounting firms offeremployees flexible scheduling?a) Reject the null hypothesis, concludingthat the lower proportion of accountingfirms offer employees flexible scheduling.b) Do not reject the null hypothesis,concluding that the lower proportion ofaccounting firms offer employees flexiblescheduling.c) Reject the null hypothesis, concludingthat the higher proportion of accountingfirms offer employees flexible scheduling.d) Do not reject the null hypothesis,concluding that the higher proportion ofaccounting firms offer employees flexiblescheduling.e) The test is inconclusive.Suppose a Realtor is interested in comparingthe asking prices of midrange homes inPeoria, Illinois, and Evansville, Indiana. TheRealtor conducts a small telephone survey inthe two cities, asking the prices of midrangehomes. A random sample of 21 listings inPeoria resulted in a sample average price of$116,900, with a standard deviation of$2,300. A random sample of 26 listings inEvansville resulted in a sample averageprice of $114,000, with a standard deviationof $1,750. The Realtor assumes prices ofmidrange homes are normally distributedand the variance in prices in the two cities isabout the same. The researcher wishes totest whether there is any difference in themean prices of midrange homes of the twocities for alpha = .01.5MS1023 Business Statistics with Computer ApplicationsAnswer the questions 33-36 based on theabove information.33. The null hypothesis for this problem is______.a) 1 – 2 < 0b) 1 – 2 > 0c) 1 – 2 = 1d) 1 – 2 0e) 1 – 2 = 034. The degrees of freedom for this problemare _______.a) 20b) 25c) 46d) 45e) 4335. The critical t value from table is ____.a) ±2.6870b) ±2.6896c) ±2.4121d) ±2.0141e) ±1.679436. The appropriate decision for thisproblem is to _____________.Homework #4nine car rental companies in Dallas results ina sample mean of $44 and a standarddeviation of $3. Use alpha = 0.05 to test todetermine whether the average daily carrental rates in Boston are significantly higherthan those in Dallas. Assume car rental ratesare normally distributed and the populationvariances are equal.Answer the questions 37-40 based on theabove information.37. The null hypothesis for this problem is______.a) 1 – 2 < 0b) 1 – 2 > 0c) 1 – 2 = 1d) 1 – 2 0e) 1 – 2 = 038. The degrees of freedom for this problemare _______.a) 7b) 8c) 9d) 15e) 16a) not reject the null hypothesis, 1 – 2 = 0b) reject the null hypothesis, 1 – 2 ≥ 0c) reject the null hypothesis 1 – 2 = 0d) not reject the null hypothesis, 1 – 2 ≤ 0e) reject the null hypothesis 1 – 2 = 139. The critical t value from table is ____.Based on an indication that mean daily carrental rates may be higher for Boston thanfor Dallas, a survey of eight car rentalcompanies in Boston is taken and the samplemean car rental rate is $47, with a standarddeviation of $3. Further, suppose a survey of40. The appropriate decision for thisproblem is to _____________.Maho Sonmez maho.sonmez@utsa.edua) -1.7531b) 1.7531c) -2.1314d) 2.1314e) ±1.6794a) not reject the null hypothesis, 1 – 2 ≤ 0b) reject the null hypothesis, 1 – 2 ≥ 0c) reject the null hypothesis 1 – 2 = 06MS1023 Business Statistics with Computer Applicationsd) reject the null hypothesis, 1 – 2 ≤ 0e) not reject the null hypothesis 1 – 2 ≥ 041. A researcher wants to conduct abefore/after study on 11 subjects todetermine if a treatment results in anydifference in scores. The null hypothesis isthat the average difference is zero while thealternative hypothesis is that the averagedifference is not zero. Scores are obtainedon the subjects both before and after thetreatment. After subtracting the after scoresfrom the before scores, the averagedifference is computed to be -2.40 with asample standard deviation of 1.21. Assumethat the differences are normally distributedin the population. The degrees of freedomfor this test are _______.a) 11b) 2c) 9d) 20e) 1042. A researcher wants to conduct abefore/after study on 11 subjects todetermine if a treatment results in anydifference in scores. The null hypothesis isthat the average difference is zero while thealternative hypothesis is that the averagedifference is not zero. Scores are obtainedon the subjects both before and after thetreatment. After subtracting the after scoresfrom the before scores, the averagedifference is computed to be -2.40 with asample standard deviation of 1.21. Assumethat the differences are normally distributedin the population. The observed t value forthis test is _______.a) -6.58b) -21.82c) -2.4d) 1.98e) 2.33Maho Sonmez maho.sonmez@utsa.eduHomework #443. A researcher wants to conduct abefore/after study on 11 subjects todetermine if a treatment results in anydifference in scores. The null hypothesis isthat the average difference is zero while thealternative hypothesis is that the averagedifference is not zero. Scores are obtainedon the subjects both before and after thetreatment. After subtracting the after scoresfrom the before scores, the averagedifference is computed to be -2.40 with asample standard deviation of 1.21. A 0.05level of significance is selected. Assumethat the differences are normally distributedin the population. The table t value for thistest is _______.a) ±1.8125b) ±2.2622c) ±1.7959d) ±2.2281e) ±3.1693One of the new thrusts of quality controlmanagement is to examine the process bywhich a product is produced. This approachalso applies to paperwork. In industrieswhere large long-term projects areundertaken, days and even weeks mayelapse as a change order makes its waythrough a maze of approvals beforereceiving final approval. This process canresult in long delays and stretch schedules tothe breaking point. Suppose a quality controlconsulting group claims that it cansignificantly reduce the number of daysrequired for such paperwork to receiveapproval. In an attempt to “prove” its case,the group selects five jobs for which itrevises the paperwork system. The followingdata show the number of days required for achange order to be approved before thegroup intervened and the number of daysrequired for a change order to be approved7MS1023 Business Statistics with Computer Applicationsafter the group instituted a new paperworksystem.Before After1287310816985Use alpha = 0.05 to determine whether therewas a significant drop in the number of daysrequired to process paperwork to approvechange orders. Assume that the differencesin days are normally distributed.Homework #447. What is the appropriate critical value forthis problem for alpha = 0.01?a) 4.6041b) 4.0321c) 3.3649d) 4.5407e) 3.746948. Mean sample difference (̅ ) is _______.a) 10.6b) 6.6c) 4d) -4e) 17.2Answer the questions 44-50 based on theabove information.49. The observed value of t for this problemis __________.44. The null hypothesis for this problem is_____.a) 4.78b) 2.05c) 4.87d) -4.78e) -2.05a) D = 0b) D ≥ 0c) D ≤ 0d) D ≠ 0e) D ≥ 145. The degrees of freedom in this problemare _______.a) 3b) 8c) 5d) 9e) 450. The appropriate conclusion for thisproblem is ___________.a) not reject the null hypothesis, D ≤ 0b) reject the null hypothesis, D ≥ 0c) reject the null hypothesis D = 0d) reject the null hypothesis, D ≤ 0e) not reject the null hypothesis D ≥ 046. The sample standard deviation (sd) of thedifferences is _______.a) 1.6733b) 1.6373c) 1.8807d) 1.8708e) 1.8078Maho Sonmez maho.sonmez@utsa.edu8

ANSWER:

REQUEST HELP FROM A TUTOR