1. Ginger root is used by many as a dietary supplement. A manufacturer ofsupplements produces capsules that are advertised to contain at least 500 mg.of ground ginger root. A consumer advocacy group doubts this claim 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 level ? = 0.05. To do so, he computes that average Math SAT score of all thestudents in his class and constructs a 95% confidence interval for the populationmean. The mean Math SAT score 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 level ? = 0.05 because 500 is within confidenceinterval.b. H0 cannot be rejected at level ? = 0.05, but this must be determined bycarrying out the hypothesis test rather than using the confidence interval.c. We can be certain that H0 is not true.6. I wish to find a 95% confidence interval for the mean number of times menchange channels with a remote control during a commercial. Based on apreliminary study, I estimate ? = 15. How many commercials’ worth of data doI need to have a margin of error no more than 3?a. 10b. 97c. 967. Experiments on learning in animals sometimes measure how long a laboratoryrat takes to find its way through a maze. Suppose for one particular maze, themean time is known to be 20 seconds with a standard deviation of = 2seconds. Suppose also that times for laboratory rats are normally distributed. Aresearcher decides to test whether rats exposed to cigarette smoke take longeron average to complete the maze. She exposes 25 rats to cigarette smoke for 15minutes and then records how long each takes to complete the maze. The meantime for these rats is 20.6 seconds. Are these results significant at the =0.05 level? Assume the researcher’s rats can be considered a SRS from thepopulation of all laboratory rats.a. Yes.b. No.c. The question cannot be answered since the results are not practicallysignificant.8. The times for untrained rats to run a standard maze has a N (65, 15) distributionwhere the times are measured in seconds. The researchers hope to show thattraining improves the times. The alternative 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 bank is investigating ways to entice customers to charge more on their creditcards. (Banks earn a fee from the merchant on each purchase, and hope tocollect interest from the customers as well). A bank selects a random group ofcustomers who are told their “cash back” will increase from 1% to 2% for allcharges above a certain dollar amount each month. Of the 500 customers whowere told the increase applied to charges above $1000 each month, the averageincrease in spending was $527 with standard deviation $225. Of the 500customers who were told the increase applied to charges above $2000 eachmonth, the average increase in spending was $439 with standard deviation$189. When testing whether or not the increases in spending are different, thetest is significant ata. 5%b. 1%c. 0.5%29.A researcher wished to compare the average amount of time spent inextracurricular activities by high school students in a suburban school districtwith that in a school district of a large city. The researcher obtained an SRS of 60high school students in a large suburban school district and found the mean timespent in extracurricular activities per week to be = 6 hours with a standarddeviation s1 = 3 hours. The researcher also obtained an independent SRS of 40high school students in a large city school district and found the mean time spentin extracurricular activities per week to be = 4 hours with a standarddeviation s2 = 2 hours. Let Âµ 1 and Âµ 2 represent the mean amount of time spentin extracurricular activities per week by the populations of all high schoolstudents in the suburban and city school districts, respectively. If the researcherused the more accurate software approximation to the degrees of freedom, hewould have used which of the following for the number of degrees 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 selected at random. The dial oneach is set to 300Â° F and after one hour, the actual temperature of each ismeasured. The temperatures measured are 305Â°, 310Â°, 300Â°, and 305Â°.Assuming that the actual temperatures for this model when the dial is set to300Â° are normally distributed with mean Âµ , we test whether the oven is properlycalibrated by testing the hypothesesH 0: Âµ = 300, H a: Âµ ? 300.Based on the data, the P -value for this test isa. between 0.10 and 0.05.b. between 0.05 and 0.025.c. between 0.025 and 0.01.31.An inspector inspects large truckloads of potatoes to determine the proportion pwith major defects prior to using the potatoes to be made into potato chips. Sheintends to compute a 95% confidence interval for p. To do so, she selects an SRSof 50 potatoes from the over 2000 potatoes on the truck. Suppose that only 2 ofthe potatoes sampled are found to have major defects. Which of the followingassumptions for inference about a proportion using a confidence interval areviolated?a. The population is at least 10 times as large as the sample.b. n is so large that both the count of successes n and the count of failuresn (1 – ) are 10 or more.c. There appear to be no violations.32.A manufacturer 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 2,respectively, that are defective. A 98% confidence interval for p1 – p2, thedifference in the two proportions isa. -.05 ? 0.033.b. -.05 ? 0.068.c. – .05 ? 0.074.33. I want to estimate the proportion of individuals in my area who think the publicschool system needs major overhauling. If I believe the proportion will be about35%, how many individuals will I need to sample if I want a 95% margin of errorto be no more than 3%?a. At least 30.b. At least 1068.c. At least 972.34.A poll finds that 54% of the 600 people polled favor the incumbent. Shortly afterthe poll is taken, it is disclosed that he had an extramarital affair. A new pollfinds that 50% of the 1030 polled now favor the incumbent. The standard errorfor a confidence interval for the candidate’s latest support level isa. 0.016b. 0.020c. 0.02535.A student believes that 20% of all students think pepperoni is their favorite pizza.He performs a test of hypothesis, H0: p = 0.2, having taken a sample of 200students and finding that 52 think pepperoni is their favorite. He finds a p-valueof 0.0338, so rejects the null at ? = 0.05. He then computes a 95% confidenceinterval for the true proportion and finds it is (0.199, 0.321). He is confused.20% is in the interval! What is the difference?a. He made a mistake in one of his calculations.b. The two use different values of p in computing the standard deviation.c. He should have used the plus four confidence interval.36.A poll finds that 54% of the 600 people polled favor the incumbent. Shortly afterthe poll is taken, it is disclosed that he had an extramarital affair. A new pollfinds that 50% of the 1030 polled now favor the incumbent. We want to know ifhis support has decreased. The test statistic isa. z = 1.56b. z = -2.57c. z = -1.5537. I want to know which of two manufacturing methods will be better. I create 10prototypes using the first process, and 10 using the second. There were 3defectives in the first batch and 5 in the second. Find a 95% confidence intervalfor the difference in the proportion of defectives.a. (-0.62, 0.22)b. (-0.56, 0.22)c. (-0.493, 0.160)38.A sample of 75 students found that 55 of them had cell phones. The margin oferror for a 95% confidence interval estimate for the proportion of all studentswith cell phones isa. 0.084b. (0.633, 0.833)c. 0.10039.A poll finds that 54% of the 600 people polled favor the incumbent. Shortly afterthe poll is taken, it is disclosed that he had an extramarital affair. A new pollfinds that 50% of the 1030 polled now favor the incumbent. We want to know ifhis support has decreased. In computing a test of hypothesis with ,what is the estimate of the overall proportion, ?a. 52%b. 52.5%c. 51.5%40. 100 rats whose mothers were exposed to high levels of tobacco smoke duringpregnancy were put through a simple maze. The maze required the rats to makea choice between going left or going right at the outset. 80 of the rats went rightwhen running the maze for the first time. Assume that the 100 rats can beconsidered an SRS from the population of all rats born to mothers exposed tohigh levels of tobacco smoke during pregnancy (note that this assumption mayor may not be reasonable, but researchers often assume lab rats arerepresentative of large populations since they are often bred to have uniformcharacteristics). Let p be the proportion of rats in this population that would goright when running the maze for the first time. A 90% confidence interval for pisa. 0.8 Â± .040.b. 0.8 Â± .066.c. 0.8 Â± .078.