1. Ginger root is utilized by many as a dietary complement. A producer ofsupplements produces capsules which could be marketed to incorporate a minimal of 500 mg.of ground ginger root. A shopper advocacy group doubts this declare and teststhe hypothesesH0: ? = 500 Ha: ? < 500based on measuring the amount of ginger root in a SRS of 100 capsules. Supposethe results of the test fail to reject H0 when, in fact, the alternative hypothesis is true.In this case the consumer advocacy group will havea. committed a Type I error.b. committed a Type II error.c. no power to detect a mean of 500.2. A researcher reports that a test is "significant at 5%." This test will bea. Significant at 1%.b. Not significant at 1%.c. Significant at 10%.3. Suppose the average Math SAT score for all students taking the exam this yearis 480 with standard deviation 100. Assume the distribution of scores is normal.The senator of a particular state notices that the mean score for students in hisstate who took the Math SAT is 500. His state recently adopted a newmathematics curriculum and he wonders if the improved scores are evidencethat the new curriculum has been successful. Since over 10,000 students in hisstate took the Math SAT, he can show that the P-value for testing whether themean score in his state is more than the national average of 480 is less than0.0001. We may correctly conclude thata. there is strong statistical evidence that the new curriculum has improvedMath SAT scores in his state.b. although the results are statistically significant, they are not practicallysignificant, since an increase of 20 points is fairly small.c. these results are not good evidence that the new curriculum has improvedMath SAT scores.4. I want to construct a 92% confidence interval. The correct z* to use isa. 1.75b. 1.41c. 1.6455. The teacher of a class of 40 high school seniors is curious whether the meanMath SAT score ? for the population of all 40 students in his class is greater than500 or not. To investigate this, he decides to test the hypothesesH0: ? = 500 Ha: ? > 500at diploma ? = Zero.05. To take motion, he computes that frequent Math SAT ranking of all thestudents in his class and constructs a 95% confidence interval for the populationmean. The indicate Math SAT ranking of all the students was 502 and, assuming thestandard deviation of the scores is ? = 100, he finds the 95% confidence interval is502 ? 31. He may concludea. H0 cannot be rejected at diploma ? = Zero.05 because of 500 is inside confidenceinterval.b. H0 cannot be rejected at diploma ? = Zero.05, nevertheless this ought to be determined bycarrying out the hypothesis verify considerably than using the boldness interval.c. We would make sure that H0 is simply not true.6. I wish to uncover a 95% confidence interval for the indicate number of cases menchange channels with a distant administration all through a industrial. Primarily based totally on apreliminary analysis, I estimate ? = 15. What variety of commercials’ worth of information doI should have a margin of error no more than three?a. 10b. 97c. 967. Experiments on finding out in animals sometimes measure how prolonged a laboratoryrat takes to go looking out its strategy by the use of a maze. Suppose for one specific maze, themean time is regarded as 20 seconds with an abnormal deviation of = 2seconds. Suppose moreover that cases for laboratory rats are often distributed. Aresearcher decides to verify whether or not or not rats uncovered to cigarette smoke take longeron frequent to complete the maze. She exposes 25 rats to cigarette smoke for 15minutes after which info how prolonged each takes to complete the maze. The meantime for these rats is 20.6 seconds. Are these outcomes important on the =Zero.05 diploma? Assume the researcher’s rats is perhaps thought-about a SRS from thepopulation of all laboratory rats.a. Certain.b. No.c. The question cannot be answered as a result of the outcomes shouldn’t practicallysignificant.eight. The cases for untrained rats to run an abnormal maze has a N (65, 15) distributionwhere the cases are measured in seconds. The researchers hope to level out thattraining improves the cases. The selection hypothesis isa. Ha: Âµ > 65.b. Ha: > 65.c. Ha: Âµ < 65.9. Does taking garlic tablets twice a day provide significant health benefits? Toinvestigate this issue, a researcher conducted a study of 50 adult subjects whotook garlic tablets twice a day for a period of six months. At the end of the study,100 variables related to the health of the subjects were measured on eachsubject and the means com-pared to known means for these variables in thepopulation of all adults. Four of these variables were significantly better (in thesense of statistical significance) at the 5% level for the group taking the garlictablets as compared to the population as a whole, and one variable wassignificantly better at the 1% level for the group taking the garlic tablets ascompared to the population as a whole. It would be correct to concludea. there is good statistical evidence that taking garlic tablets twice a dayprovides some health benefits.b. there is good statistical evidence that taking garlic tablets twice a dayprovides benefits for the variable that was significant at the 1% level. We shouldbe somewhat cautious about making claims for the variables that weresignificant at the 5% level.c. None of the above.10. To assess the accuracy of a kitchen scale a standard weight known to weigh 1gram is weighed a total of n times and the mean, , of the weighings iscomputed. Suppose the scale readings are normally distributed with unknownmean, Âµ , and standard deviation = 0.01 g. How large should n be so that a90% confidence interval for Âµ has a margin of error of Â± 0.0001?a. 165b. 27061c. 3841611.Suppose that you are a student worker in the statistics department and agree tobe paid by the Random Pay system. Each week the Chair flips a coin. If the coincomes up heads, your pay for the week is $80; if it comes up tails, your pay forthe week is $40. You work for the department for 100 weeks (at which point youhave learned enough probability to know the system is not to your advantage).The probability that , your average earnings in the first two weeks, is greaterthan $65 isa. 0.2500.b. 0.3333.c. 0.5000.12.A set of ten cards consists of five red cards and five black cards. The cards areshuffled thoroughly and I am given the first four cards. I count the number of redcards X in these four cards. The random variable X has which of the followingprobability distributions?a. The binomial distribution with parameters n = 10 and p = 0.5.b. The binomial distribution with parameters n = 4 and p = 0.5.c. None of the above.13. In a pre-election poll, 400 of the 500 probable voters polled favored theincumbent. In this poll, the sample proportion, , of those favoring thechallenger isa. 0.80b. 0.20c. 0.5014. Incomes in a certain town are strongly right skewed with mean $36000 andstandard deviation $7000. A random sample of 10 households is taken. What isthe probability the average of the sample is more than $38000?a. 0.3875b. 0.1831.c. Cannot say.15. The scores of individual students on the American College Testing (ACT)Program composite college entrance examination have a normal distributionwith mean that varies slightly from year to year and standard deviation 6.0. Youplan to take an SRS of size n of the students who took the ACT exam this yearand compute the mean score of the students in your sample. You will use thisto estimate the mean score of all students this year. In order for the standarddeviation of to be no more than 0.1, how large should n be?a. At least 60.b. At least 3600.c. This cannot be determined because we do not know the true mean of thepopulation.16. We want to take a sample of 100 items out of a large batch for quality controlpurposes. Based on past history, the proportion of defective items is 4%. Can weuse the normal approximation to the binomial distribution to find the probabilityof finding more than 5 defective items in the sample of 100?a. Yes, because n is large.b. Noc. We do not have enough information.17. The SAT scores of entering freshmen at University X have a N (1200, 90)distribution and the SAT scores of entering freshmen at University Y have a N(1215,110) distribution. A random sample of 100 freshmen is sampled fromeach University, with the sample mean of the 100 scores from University Xand the sample mean of the 100 scores from University Y. The probability thatis greater than Âµ Y, the population mean for University Y, isa. 0.0475.b. 0.0869.c. 0.4325.Quizzes prepared by Dr. Patricia Humphrey, Georgia Southern University18.Suppose that you are a student worker in the Statistics Department and theyagree to pay you using the Random Pay system. Each week the Chair flips a coin.If it comes up heads, your pay for the week is $80, and if it comes up tails, yourpay for the week is $40. You work for the department for two weeks. Let bethe average of the pay you receive for the first and second week. The samplingdistribution of is: $40 $60 $80a. given by the following probability distribution.Probability: 0.25 0.50 0.25b. approximately normal with mean $60 and standard deviation 14.14.c. exactly normal with mean $60 and standard deviation 14.14.19. From previous polls, it is believed that 66% of likely voters prefer the incumbent.A new poll of 500 likely voters will be conducted. In the new poll, if theproportion favoring the incumbent has not changed, what is the mean andstandard deviation of the number preferring the incumbent?a. ? = 330, ? = 10.59b. ? = 0.66, ? = 0.021c. ? = 330, ? = 18.1720. From previous polls, it is believed that 66% of likely voters prefer the incumbent.A new poll of 500 likely voters will be conducted. In the new poll, if theproportion favoring the incumbent has not changed, what is the probability thatmore than 68% will favor the incumbent?a. 32%.b. 82.7%c. 17.1%21.A local farmer is interested in comparing the yields of two varieties of tomatoes.In an experimental field, she selects 40 locations and assigns 20 plants fromeach variety at random to the locations. She determines the average per plant(in pounds). She computes a 95% confidence interval for the difference in meanyields between the two varieties using the two-sample t procedures with theresulting interval (2.13, 6.41). For testing using thetwo-sample t procedures we can say thata. the P-value could be greater than 0.05.b. the P-value must be less than 0.05.c. no information about the P-value can be obtained without the test statistic.22.As the degrees of freedom become larger, the difference between the t and zdistributions becomesa. Narrowerb. Stays the samec. Wider23.Some researchers have conjectured that stem-pitting disease in peach-treeseedlings might be controlled through weed and soil treatment. An experimentwas conducted to compare peach-tree seedling growth with soil and weedstreated with one of two herbicides.In a field containing 10 seedlings, five were randomly selected throughout thefield and assigned to receive Herbicide A. The remainder received Herbicide B.Soil and weeds for each seedling were treated with the appropriate herbicide,and at the end of the study period the height in (cm) was recorded for eachseedling. The following results were obtained:Herbicide A 87 80 80 76 73Herbicide B 78 77 74 68 62A 90% confidence interval for the difference in mean heights for the two herbicidesis (0.2, 14.6). Which statement is correct?a. The P -value for a test of the null hypothesis of equal means and thealternative of different means would be greater than 10% since the interval doesn'tinclude 0.b. A 95% confidence could not include zero either, since we would be even moreconfident of a difference in the groups.c. Both (a) and (b) are incorrect.24. I did an eggsperiment with a fellow instructor one day. His calculus class wasstudying volumes of solids by rotating curves around the x-axis. We modeledthe volume of an egg as an ellipsoid, and measured eggs with calipers, using thecalculus formula. Each egg was also measured for volume using a waterdisplacement method. We wanted to know if the two methods agreed or not. Thedata wereA 95% confidence interval for the average difference (calculus - water) isa. (-1.69, 5.04)b. (-7.54, 10.90)c. (-2.29, 5.64)25.Suppose that a random sample of 41 state college students is asked to measurethe length of their right foot in centimeters. A 95% confidence interval for themean foot length for students at this university turns out to be (21.709,25.091).Which of the following is true?a. The sample mean was 23.4 cm.b. The margin of error is 3.382c. If the confidence level is changed to 90% we will get a wider interval.Quizzes prepared by Dr. Patricia Humphrey, Georgia Southern University26.A local farmer is interested in comparing the yields of two varieties of tomatoes.In an experimental field, he selects 20 locations and assigns 10 plants from eachvariety at random to the locations. He determines the yield per plant (in pounds).The mean yield for plants of variety 1 was = 16.3 pounds with a standarddeviation = 3 pounds. The mean yield for plants of variety 2 was =18.4pounds with a standard deviation = 4 pounds. The standard error of thedifference in sample means isa. 2.10 pounds.b. 2.50 pounds.c. 1.58 pounds.27. The manager of an automobile dealership is considering a new bonus plan toincrease sales. Currently, the mean sales rate per salesperson is fiveautomobiles per month. The correct set of hypotheses to test the effect of thebonus plan isa. H0: ? = 5, HA: ? > 5b. H0: ? > 5, HA: ? = 5c. H0: = 5, HA: > 528.A monetary establishment is investigating strategies to entice shoppers to price additional on their creditcards. (Banks earn a cost from the service supplier on each purchase, and hope tocollect curiosity from the consumers as successfully). A monetary establishment selects a random group ofcustomers who’re suggested their “cash once more” will improve from 1% to 2% for allcharges above a certain dollar amount each month. Of the 500 shoppers whowere suggested the rise utilized to prices above $1000 each month, the averageincrease in spending was $527 with customary deviation $225. Of the 500shoppers who’ve been suggested the rise utilized to prices above $2000 eachmonth, the everyday improve in spending was $439 with customary deviation$189. When testing whether or not or not or not the desire improve in spending are completely completely different, thetest is necessary ata. 5percentb. 1percentc. Zero.5%29.A researcher wished to verify the everyday time period spent inextracurricular actions by highschool school college students in a suburban school districtwith that in a school district of an enormous metropolis. The researcher obtained an SRS of 60highschool school college students in an enormous suburban school district and positioned the indicate timespent in extracurricular actions per week to be = 6 hours with a standarddeviation s1 = three hours. The researcher moreover obtained an unbiased SRS of 40highschool school college students in an enormous metropolis school district and positioned the indicate time spentin extracurricular actions per week to be = 4 hours with a standarddeviation s2 = 2 hours. Let Âµ 1 and Âµ 2 characterize the indicate time period spentin extracurricular actions per week by the populations of all extreme schoolstudents inside the suburban and metropolis school districts, respectively. If the researcherused the additional appropriate software program program approximation to the degrees of freedom, hewould have used which of the subsequent for the number of ranges of freedom forthe two-sample t procedures?a. 39.b. 59.c. 98.30. We wish to see if the dial temperature for a certain model oven is properlycalibrated. four ovens of a certain model are chosen at random. The dial oneach is able to 300Â° F and after one hour, the exact temperature of each ismeasured. The temperatures measured are 305Â°, 310Â°, 300Â°, and 305Â°.Assuming that the exact temperatures for this model when the dial is able to300Â° are often distributed with indicate Âµ , we verify whether or not or not the oven is properlycalibrated by testing the hypothesesH Zero: Âµ = 300, H a: Âµ ? 300.Primarily based totally on the data, the P -value for this verify isa. between Zero.10 and Zero.05.b. between Zero.05 and Zero.Zero25.c. between Zero.Zero25 and Zero.01.31.An inspector inspects large truckloads of potatoes to seek out out the proportion pwith predominant defects earlier to using the potatoes to be made into potato chips. Sheintends to compute a 95% confidence interval for p. To take motion, she selects an SRSof 50 potatoes from the over 2000 potatoes on the truck. Suppose that solely 2 ofthe potatoes sampled are found to have predominant defects. Which of the followingassumptions for inference a few proportion using a confidence interval areviolated?a. The inhabitants is a minimal of 10 cases as large as a result of the sample.b. n is so large that every the rely of successes n and the rely of failuresn (1 – ) are 10 or additional.c. There appear like no violations.32.A producer receives parts from two suppliers. An SRS of 400 parts fromsupplier 1 finds 20 defective. An SRS of 100 parts from supplier 2 finds 10defective. Let p1 and p2 be the proportion of all parts from suppliers 1 and a pair of,respectively, which could be defective. A 98% confidence interval for p1 – p2, thedifference inside the two proportions isa. -.05 ? Zero.033.b. -.05 ? Zero.Zero68.c. – .05 ? Zero.Zero74.33. I want to estimate the proportion of individuals in my area who suppose the publicschool system needs predominant overhauling. If I think about the proportion will possible be about35%, how many people will I need to sample if I want a 95% margin of errorto be no more than three%?a. At least 30.b. At least 1068.c. At least 972.34.A poll finds that 54% of the 600 of us polled favor the incumbent. Shortly afterthe poll is taken, it is disclosed that he had an extramarital affair. A model new pollfinds that 50% of the 1030 polled now favor the incumbent. The same old errorfor a confidence interval for the candidate’s latest assist diploma isa. Zero.016b. Zero.Zero20c. Zero.02535.A pupil believes that 20% of all school college students suppose pepperoni is their favorite pizza.He performs a verify of hypothesis, H0: p = Zero.2, having taken a sample of 200school college students and discovering that 52 suppose pepperoni is their favorite. He finds a p-valueof Zero.0338, so rejects the null at ? = Zero.05. He then computes a 95% confidenceinterval for the true proportion and finds it is (Zero.199, Zero.321). He is confused.20% is inside the interval! What is the distinction?a. He made a mistake in a single amongst his calculations.b. The two use completely completely different values of p in computing the same old deviation.c. He must have used the plus four confidence interval.36.A poll finds that 54% of the 600 of us polled favor the incumbent. Shortly afterthe poll is taken, it is disclosed that he had an extramarital affair. A model new pollfinds that 50% of the 1030 polled now favor the incumbent. We want to know ifhis assist has decreased. The verify statistic isa. z = 1.56b. z = -2.57c. z = -1.5537. I want to know which of two manufacturing methods will possible be larger. I create 10prototypes using the first course of, and 10 using the second. There have been 3defectives inside the first batch and 5 inside the second. Uncover a 95% confidence intervalfor the excellence inside the proportion of defectives.a. (-Zero.62, Zero.22)b. (-Zero.56, Zero.22)c. (-Zero.493, Zero.160)38.A sample of 75 school college students found that 55 of them had cell telephones. The margin oferror for a 95% confidence interval estimate for the proportion of all studentswith cell telephones isa. Zero.084b. (Zero.633, Zero.833)c. Zero.10039.A poll finds that 54% of the 600 of us polled favor the incumbent. Shortly afterthe poll is taken, it is disclosed that he had an extramarital affair. A model new pollfinds that 50% of the 1030 polled now favor the incumbent. We want to know ifhis assist has decreased. In computing a verify of hypothesis with ,what is the estimate of the overall proportion, ?a. 52%b. 52.5percentc. 51.5%40. 100 rats whose mothers have been uncovered to extreme ranges of tobacco smoke duringpregnancy have been put by the use of a straightforward maze. The maze required the rats to makea various between going left or going correct on the outset. 80 of the rats went rightwhen working the maze for the first time. Assume that the 100 rats can beconsidered an SRS from the inhabitants of all rats born to mothers uncovered tohigh ranges of tobacco smoke all through being pregnant (phrase that this assumption mayor might be not low-cost, nevertheless researchers often assume lab rats arerepresentative of giant populations since they’re often bred to have uniformcharacteristics). Let p be the proportion of rats on this inhabitants that may goright when working the maze for the first time. A 90% confidence interval for pisa. Zero.eight Â± .Zero40.b. Zero.eight Â± .Zero66.c. Zero.eight Â± .Zero78.