AP Review Questions 262-322

Question 1

262. Suppose X is a geometric random variable that counts the number of trials necessary to obtain the first successful trial. If the probability of failure for any given trial is 0.42, what is the probability that the first success will occur on the first or second trial?
     A) (0.58)(0.42)^0 + (0.58)(0.42)^1
     B) (0.42)(0.58)^0 + (0.42)(0.58)^1
     C) (26 over 1)(0.58)^1(0.42)^25 + (26 over 2)(0.58)^2(0.42)^24
     D) (26 over 1)(0.42)^1(0.58)^25 + (26 over 2)(0.42)^2(0.58)^24
     E) (0.42)(0.58)^1 + (0.42)(0.58)^2

Answer 1

(A) The probability of failure on any trial is 0.42, therefore the probability of success on any trial is 1-0.42=0.58. Since x is geometric, the probability of the first success occurring on the first or second trial is P(X=1) + p(X=2) = 0.58(0.42)^0 + 0.58(0.42)^1.

Question 2

263. A coin is weighted so that the probability of heads if 0.64. On average how many coin flips will it take for tails to first appear?
     A. 1.56
     B. 2
     C. 2.78
     D. 6.4
     E. This cannot be determined since the total number of flips is unknown

Answer 2

A. For a geometric random variable , the average time to the first success is 1/p = 1/0.64 = 1.56

Question 3

264. In a binomial experiment with 45 trials the probablility of more than 25 successes can be approximated by P(z>(25-27)/3.29). What is the probability of successes of a single trial of this experiment?
A. .07
B. .56
C. .6
D. .61
E. .79

Answer 3

C The binomial distribution can be approximated by the normal distribution with u= np and o= sqrt(np(1-p)) when n is large enough. In this case, P(z>(25-27)/3.29) = P(z>(x-u)/o)) If np= 27 and n=45, p=0.6

Question 4

265. Let X represent the number of successes in 30 trials of a binomial experiment. If P(X=2)=(30 over 2)(0.45)^2(0.55)^28, what is the approximate probability of fewer than 10 successes in the 30 trials?
     A) P(z

Answer 4

(E) Since x is binomial, P(X=x) = (n over x)p^x(1-p)^n-x. Therefore p=0.45 and np=13.5, giving us the approximation P[z

Question 5

267. The distribution of the binomial random variable that counts the number of successful trials in 70, each of which have 40% probability of failure, can be approximated by which of the following distributions?
     A) A slightly skewed right distribution with a mean of 28 and a standard deviation of 16.8.
     B) A slightly skewed right distribution with a mean of 42 and a standard deviation of 4.1.
     C) An aproximately normal distribution with a mean of 28 and a standard deviation of 16.8.
     D) An approxiamtely normal distribution with a mean of 42 and a standard deviation of 4.1.
     E) The number of trials is not large enough to make a determination.

Answer 5

(D) If the probability of failure is 40% then the probability of success is 60%. Further, since np greater than or equal to 5, the distribution can be approximated by a normal distribution with u=np=42 and o=sqrt[np(1-p)] = 4.1

Question 6

269. An experiment consists of 31 independent trials. The probability that any one trial is a failure is 45%. Which is the best estimate of the probability that more than 20 trials will be failures?
A) 0.01
B) 0.02
C) 0.11
D) 0.89
E) 0.99

Answer 6

(A) The described situation is binomcdf with n=31 and p=1-0.45=0.55 . The probability of more than 20 failures os equivalent to 10 or fewer successes, which is P(X less than or equal to 10) symbolically. This is best found using the calaculator: binomcdf (31, 0.55, 10).

Question 7

266. The distribution of the binomial random variable that counts the number of successful trials in 65, each of which have a 30% probabiility of success can be approximated by which of the following distributions?
     A) A slightly skewed left distribution with a mean of 65 and a standard deviation of 19.5.
     B) Slightly skewed eft distribution with a mean of 19.5 and a standard deviation of 3.7.
     C) An approximately normal distribution with a mean of 65 and a standard deviation of 19.5.
     D) An approximately normal distribution with a mean of 19.5 and a standard deviation of 3.7.
     E) The number of trials is not large enough to make a determination.

Answer 7

D. Since np > 5, the distribution can be approximated by a normal distribution with u= np = 19.5 and o = sqrt(np(1-p)) = 3.7

Question 8

268. The probability that any one trial out of 23 independent trials will be successful is 17%. Find the probability that fewer than three trials will be successful. 
A) 0.43
B) 0.15
C) 0.22
D) 0.56
E) 0.78

Answer 8

C The described situation is binomcdf with n = 23 and p = 0.17. The probability that fewer than three trials will be successful is P(X

Question 9

270. In a large population, it is known that 55% of items in the population have a certain attribute. Whether an item has the attribute or not is independent of whether any other item has the attribute. If a sample of 95 is drawn from this population, what is the probability that more than 45% of the items in the sample will have the attribute?
A) P(Z>0.55-0.45/0.051)
B) P(Z>0.45-0.55/0.051) 
C) P(Z>0.55-0.45/9.33)
D) P(Z>0.45-0.55/9.33)
E) P(Z>42.75-52.25/9.33)

Answer 9

B Using the normal Appromixation to the sampling distribution of p, this question is asking for P(phat>.45). Since the mean is p and the standard distribution is sqrt(p(1-p)/n) this is equivalent to P( Z> .45-.55/sqrt(.55(.45)/95)

Question 10

272. Suppose that X is a binomial random variable that counts the number of successes in 12 trials. Find P(X is greater than or equal to 3) if the probability of success for a single trial is 0.29.
A) 0.1807
B) 0.2460
C) 0.2775
D) 0.4765
E) 0.5235

Answer 10

E This probability can be found on a TI83 or 84 using binomial (12, .29, 3)

Question 11

274. The random variable X represents the number of successes in 10 independent trials. Which of the following represents the probability of fewer than two failures?
     A) P(X greater than or equal 2)
     B) P(X8)
     D) P(X greater than or equal to 8)
     E) The probability of success for an individual trial must be known to determine which of these represents the given probability.

Answer 11

C If there are fewer than 2 failures, there must be either one or no failures in other words either 9 or 10 successes

Question 12

273. Let X represent the number of successes in n independent trials. If the probability of success for any given trials is p. which of the following is true?
     I. The mean number of successes in n trials is sqrt[np(1-p)].
     II. The probability of failure for any given trial is 1-p.
     III. The probability that there will be exactly two successful trials is 2/n.

Answer 12

B The described random variable is binomial, therefore the mean number of successes would be np and the probability of two successes would be found using the binomial formula.

Question 13

275. A binomial random variable X counts the number of successes in three trials. Let the probability of success for any given trial be 0.4.
     a) Find the probability distribution of X in table form.
     b) Explain how the probability of two or more successes can be found using your table from a).

Answer 13

A each individual probability is found using the formula for exactly that number of successes For instance p(x=0) is the value .216

Question 14

276. A small candy company specializes in creating various candies, all of which contain peanuts. Their latest product is the “Chocolate and Peanut Bonanza Bag,” a 1-pound (16 ounce) bag of chocolate peanut clusters. The company claims that each bag contains an everage of 10 ounces of peanuts and distribution of the weight of peanuts in this product is approximately normal.
     a) Suppose that the standard deviation of the weight of peanuts contained in the new product is 1.1 ounces. If the company’s claim is true, what is the probability that a smaple of 40 Chocolate and Peanut Bonanza bags will contain an average less than 9.5 ounces of peanuts.
     b) If the distribution of the weights of peanuts in the new product was not normal but had the same mean and standard deviation, would your calculations in part a) still be appropriate? Explain.

Answer 14

A Since the sample size is 40, the sampling distribution of the sample mean will be approximately normal with a mean of 10 and a standard deviation of 1.1/sqrt40= .1739. Therefore the probability a sample would have a mean less that 9.5 is .002

BYes, as long as the sample size is larger than 30, the distribution will be approximately normal regardless of the distribution of the population, however, the standard deviation depends on the sample size. it decreases as the sample size increases.

Question 15

277. A large population has a skewed right distribution with a mean of 46.1 and a standard deviation of 5.3.
     a) Describe the shape of the sampling distribution of the mean if all possible samples of size 10 are taken.
     b) Describe the shape of the sampling distribution of the mean if all possible samples of size 50 are taken.
     c) Will the mean and the standard deviation be the same in a) and b)? Is this true in general or only for this particular problem? Explain.

Answer 15

.A distribution will be skewed right with a mean of 46.1 and a standard deviation of 5.3/sqrt10= 1.67

B The distribution will be approximately normal with a mean of 46.1 and a standard deviation of .75

C in both cases the mean will be the same. the standard deviation decreases as the size of the sample increases.

Question 16

271. Suppose that X is a binomial random variable that counts the number of successes in 30 trials. Find P(X>1) if the probability of success for a single trial is 0.14.
A) 0.0108
B) 0.0529
C) 0.0638
D) 0.9362
E) 0.9892

Answer 16

D using the TI83 or 84 this can be found using the binomial (30, .14, 1).

Question 17

278. Previous studies have suggested that the average weight of a particular animal species is 121.3 pounds with a standard deviation of 20.9 pounds. The distribution of the weights has not been determined.
     a) A sample of 90 of these animals had an average weight of 115 pounds. What is the probability that a sample of this size would have a mean of 115 pounds or fewer?
     b) Based on your answer in a), it is likely that the true mean weight of these animals is 121.3 pounds as has been suggested? Justify your answer.

Answer 17

A since the sample is large the sampling distribution is approximately unimodal and P=.002
b it is unlikely to get a sample with a mean of 115 or less if the true mean is 121.3, However, such a sample was found. Therefore unlikely the true me is 121.3

Question 18

279. A survey of students at a large state university found that 82% had purchased textbooks from an off-campus vendor at least once during their college career.
     a) If 45 students are sampled, how many students would you expect to have purchased textbooks from an off-campus vendor at least once during their college career?
     b) If 150 students are sampled, what is the probability that more than 130 of them have purchased textbooks from an off-campus vendor at least once during their college career?

Answer 18

a This is a binomial experiment, the expected value of x is 36.9. Therefore you would expect about 37 students to have brought textbooks off campus.
B The probability that more that 130 students would have done this is binomcdf (150,.82, 130) = .051 p=.82 s=150

Question 19

280. Quality control technicians at a large manufacturing firm have found that approximately 1.2% of parts from the main production line are defective. It is believed that the defects are independent of one another.
     a) An outside consultant is brought in to test parts from the main production line. In a random sample of 150 parts, he finds that four are defective. What is the probability that the consultant would get a sample size of this with four or more defects if the quality control technicians’ estimate is correct? Justify your answer.
     b) Assuming the original estimate of 1.2% was correct, if the consultant decides to sample another 150 parts and the first 145 are not defective, what is the probability the 146th part will be defective? Justify your answer.

Answer 19

.A Since it is thought that the defects are independent, this can be thought of as a binomial with x=150 and p=.012. Therefore 1- binomcdf (150,.12,3) =.1075

Question 20

281. You want to create a 95% confidence interval with a margin of error of no more than 0.05 for a population proportion. The historical data indicate that the population has remained constant at about 0.55. What is the minimum size random sample you need to construct this interval?
A) 378
B) 380
C) 223
D) 324
E) None of the above

Answer 20

B P= .55, =1.96, M=.05

Question 21

282. A random sample of 30 households was selected as part of a study on electricity usage, and the number of kWh was recorded for each household in the sample for the March quarter of 2011. The average usage was found to be 450kWh. In a very large study in the March quarter of the previous year it was found that the standard deviation of the usage was 81 kWh. Assuming the standard deviation is unchanged and that the usage is normally distributed, provide an expression for calculating a 99% confidence interval for the mean usage in the March quarter of 2011.
A) 450±2.756(81/√30)
B) 450±2.575(9/√30)
C) 450±2.33(81/√30)
D) 450±2.575(81/√30)
E) None of the above

Answer 21

D since o is assumed known, we use an interval where o=81, x= 30.

Question 22

283. If our sample size from the previous question were tripled, how would that change the confidence interval size?
A) Divides the interval size by 3
B) Multiplies the interval size by 1.732
C) Divides the interval size by 1.732
D) Triples the interval size
E) None of the above

Answer 22

C Increasing the sample size by a multiple of d divides the internal by sqrt(d)

Question 23

284. Other things being equal, which of the following actions will reduce the power of a hypothesis test?
I. Increasing sample size
II. Increasing significance level 
III. Increasing beta, the probability of a Type II error
A) I only
B) II only
C) III only
D) All of the above
E) None of the above

Answer 23

C More likely to reject the null hypothesis. It is in fact false. Increasing the significance level reduces the region of acceptance, which makes the hypothesis test more likely to reject the null.

Question 24

285. While conducting an experiment to test a hypothesis, we found that our sample size tripled. Which of the following will increase?
     I. The power of the hypothesis test
     II. The effect size of the hypothesis test
     III. The probability of making a type II error
     A) I only
     B) II only
     C) III only
     D) All of the above
     E) None of the above

Answer 24

A Increasing sample size makes the hypothesis test more sensitive more likely to reject null hypothesis. the probability of making a type II error gets smaller, not bigger, as sample size increases.

Question 25

286. What is the critical t-value of a 95% confidence interval estimate for a sample size of 25?
A) 2.06
B) 1.711
C) 2.064
D) 1.708
E) None of the above

Answer 25

C

Question 26

287. Suppose you construct a 95% confidence interval from a random sample of size n=20 with a sample mean 100 taken from a population with unknown u and a known o=10, and the interval is fairly wide. Which of the following conditions would not lead to a narrower confidence interval?
     A) If you decreased your confidence interval
     B) If you increased your sample size
     C) If the sample size was smaller
     D) If the population standard deviation was smaller
     E) All of the above would lead to a narrower confidence interval

Answer 26

C

Question 27

288. Consider the following data:
1, 7, 3, 3, 6, 4
Assuming these data are drawn from a normal population with mean=u and variance =6, the 95% confidence interval for the appropriate z-score is \_\_\_\_\_\_\_ standard units long.
A) 5.14
B) 4.19
C) 4.04
D) 3.92
E) None of the above

Answer 27

DZ(o=.02/2)=1.96

2x1.96=3.92

Question 28

289. In a middle school with 500 students, it was found that the true proprtion of brunettes is 17.8%. If you were to construct a sonfidence interval estimate of the proportion of brunettes with sample size n=50, which of the following statements would be true?
I. The interval contains 17.8%
II. A 95% confidence interval estimate contains 17.8%
III. The center of the interval is 17.8%
A) I and II 
B) II and III
C) I and III
D) All of the above
E) None of the above

Answer 28

There is no guarantee it is anywhere close to the interval

Question 29

290. A random sample of the actual weight of 5-lb bags of mulch produces a mean of 4.963 lb and a standard deviation of 0.067 lb. If n=50, which of the following will give a 95% confidence interval for the mean weight (in pounds) of the mulch produced by this company?
A) 4.963±0.016
B) 4.963±0.019
C) 4.963±0.067
D) 4.963±0.009
E) None of the above

Answer 29

The margin of error is .019, the solution is 4.963+-.019

Question 30

291. A confidence interval estimate is determined from the SAT scores of an of n students. All other things being equal, which of he following will result in a small margin of error?
I. Smaller standard deviation
II. Smaller sample size
III. Smaller confidence interval
A) I and II
B) I and III
C) II and III
D) I, II, and III
E) None of the above

Answer 30

B The margin of error varies directly with the critical z value and directly with the standard deviation of the sample, but inversely with the square root of the sample size.

Question 31

292. In a test to check pH level, 49 samples showed its level to be 2.4 and with a standard deviation of 0.35. Find the 90% confidence interval estimate for the mean pH level.
A) 2.4±0.01
B) 2.4±0.08
C) 2.4±0.32
D) 2.4±0.35
E) None of the above

Answer 31

B Using z scores df=48 2.4+- 1.677(.35/sqrt49)

Question 32

293. In a recent online survey of 1,700 adults it was tevealed that only 17% felt the internet was secure. With what degree of confidence can you say that 17%±2% of adults believe that online security is secure?
A) 72.9%
B) 90%
C) 95%
D) 97.2%
E) 98.6%

Answer 32

D =.0091
z=2.2
0.9861-.0139=97.2%

Question 33

294. One of the students in the school council election wants to make healthy lunch menus her pitch to gain votes. What size voter sample would she need to obtain a 90% confidence that her support level margin of error is no more than 4%?
A) 25 
B) 600
C) 423
D) 1691
E) None of the above

Answer 33

C n=423

Question 34

295. Biology professor Doug Miller wants to reduce the width of the confidence interval around the proportion of people who don’t wash their hands and therefore carry a certain type of bacteria. What can he do to accomplish this?
     A) Increase the sample size
     B) Decrease the sample size
     C) Change the estimate of the true proportion of people who wash their hands
     D) Increase the confidence interval
     E) None of the above

Answer 34

A Professor Miller can increase the sample size to accomplish his aims.

Question 35

296. What is t(sub 0.01) (20)?
A) 2.845
B) 2.528
C) 2.197
D) 1.325
E) None of the above

Answer 35

B.

Question 36

297. You want to create a 95% confidence interval of a population proprtion with a margin of error of no more than 0.05. How large of a sample would you need?
A) 93
B) 200
C) 325
D) 385
E) 1093

Answer 36

D Because there is no indication what to expect for the population proportion we will use p=.5
n=(1.96/.05)^2= 384.16.

Question 37

298. What is the critical value of t for a 99% confidence interval based on a sample size of 30?
A) 2.750
B) 2.756
C) 2.457
D) 2.462
E) None of the above

Answer 37

B z*= 2.756 with df=29.

Question 38

299. When comparing a 95% confidence interval with a confidence interval of 99% created from the same data, how will the intervals differ?
     A) The sample must be known to determine the difference.
     B) The mean of the sample must be known to determine the difference.
     C) The 95% interval will be wider than the 99% interval.
     D) The 95% interval will be narrower than the 99% interval.
     E) The use of the t-distribution or the z-distribution will determine how the two intervals differ.

Answer 38

D.

Question 39

300. Which of these is true about standard error?
     A) It is another wat of expressing standard deviation.
     B) It is an estimator of standard deviation.
     C) It is another way of expressing the mean.
     D) It is an estimator of the mean.
     E) It is an estimator of confidence interval.

Answer 39

B When we estimate a standard deviation from data, we call the estimator the standard error.

Question 40

301. What is the effect of sample size on the width of a confidence interval?
     A) The smaller the sample size, the smaller the width of the confidence interval.
     B) The larger the sample size, the smaller the width of the confidence interval.
     C) The smaller the sample size, the larger the width of the confidence interval.
     D) The larger the sample size, the larger the width of the confidence interval.
     E) None of the above.

Answer 40

B.

Question 41

302. In a particular neighborhood a survey was conducted to determine the percentage of people who did their own gardening. A random sample of 100 householders showed that 63 of them did gardening themselves. An actual 90% confidence interval for householders doing their own gardening is:
A) 0.63+- 1.960xsqrt.[(0.63)(0.37)/100]
B) 0.63+-2.560xsqrt.[(0.63)(0.37)/100]
C) 0.63+-1.645xsqrt.[(0.63)(0.37)/100]
D) 0.63+-1.960xsqrt.[(0.63)(0.37)/100]
E) 0.63+-1.645xsqrt.[(0.63)(0.37)/100]

Answer 41

C 90% confident there are only 2 that have 1.645 as critical values. the sample size is 100. .

Question 42

303. A recent election poll found that 79% of the population favored the president. If the president’s opponents believed this percentage was lower, the null and the alternative hypoithesis would be:
     A) Ho: p=0.79; Ha: p0.79
     C) Ho: p=0.21; Ha: p

Answer 42

A the null is a statement of no difference, Ho= p=.79.

Ha=p

Question 43

304. In a recent health survey it was found that average beef consumption in the US per person was 90 pounds per year. A research firm wishes to find whether beef consumption has decreased and does a study to test the null and the alternative hypothesis. The test statistic is t= -2.147 with 30 df. Compute the p-value.
A) 0.98
B) 0.02
C) 2.147
D) 0.2
E) None of the above

Answer 43

B df=30 t=2.147, p value=.02

Question 44

305. If the confidence interval for a population proportion changes from 90% to 98% with all else being equal, then ____________.
     A) the size of the interval increases by 41%
     B) the size of the interval decreases by 9%
     C) the size of the interval increases by 9%
     D) the size of the interval decreases by 41%
     E) None of the above

Answer 44

A with values ranging from +- 1.645 to +- 2.326

2.326/1.645= an increase of 41%

Question 45

306. Which of the following margins of error in a confidence interval results using z-scores?
     I. In obtaining the sample surveys errors occurred due to no response
     II. Random sample surveys
     III. Errors occurred due to using sample standard deviations as estimates for population standard deviations
     A) I only
     B) II only
     C) III only
     D) I and III
     E) I and III

Answer 45

.B the margin of error has to do with measuring variation that occurs due to chance but has nothing to do with defective survey designs

Question 46

307. A gallon of gas in 30 test automobiles resulted in xbar=28 and s=1.2. Determine a 85% confidence interval to estimate the mean mileage. 
A) 28+-2.045(1.2/sqrt(30)
B) 28+-1.960(1.2/sqrt(30)
C) 28+-2.326(1.2/sqrt(30)
D) 28+-2.042(1.2/sqrt(30)
E) 28+-2.750(1.2/sqrt(30)

Answer 46

A 28+-2.045(1.2/sqrt(30).

Question 47

308. An Australian hospital's administrators wish to learn the average length of stay of all patients. The Australian Bureau of Statistics determines that for a 95% confidence interval the avg. length of stay needs to be within +-0.5 days for 50 patients. How many records would need to be looked at to obtain the same confidence interval to be within +-0.25 days?
A) 25
B) 50
C) 100
D) 200
E) 300

Answer 47

D 4(50)=200.

Question 48

309. What is the critical value z of a 99% confidence interval for a known standard deviation?
A) 1.645
B) 2.33
C) 2.345
D) 2.56
E) 2.75

Answer 48

C the nearest table entry shows .9905 as between .9904 and .9906, which corresponds to 2.345.

Question 49

310. While weighing five infants in the hospital we found their weights in kg to be 3.1, 3.5, 2.8, 3.2, and 3.4. What is a 90% confidence interval estimate of the infants' weights? 
A) 3.2+-0.202
B) 3.2+-0.247
C) 3.2+-0.261
D) 4.0+-0.202
E) 4.0+-0.261

Answer 49

C

=3.2+-0.261

Question 50

311. A poll identifies the proportions of high-income and low-oncome voters who support a decrease in taxes. If we need to know the answer to be within =-0.02 at the 95% confidence interval, what sample size should be taken?
A) 3383
B) 4803
C) 7503
D) 9453
E) 4503

Answer 50

B

n>or equal to 69.3^2= 4802.5, hence the pool uses a sample of 4803.

Question 51

312. A survey conducted on 1,000 Canadians found that 700 of them refused to receive the H1N1 vaccination. Construct a 95% confidence interval of the estimated proportion of Canadians who refused to receive the vaccination. 
A) (0.723, 0.777)
B) (0.686, 0.714)
C) (0.728, 0.672)
D) (0.986, 0.914)
E) (0.672, 0.728)

Answer 51

D

0.672, 0.728

Question 52

313. For a 95% confidence interval, you want to create a population proportion with a marginal error of no more than 0.025. How large of a sample would you need?
A) 423
B) 1537
C) 385
D) 2237
E) 1082

Answer 52

B

n>or equal to 1537.

Question 53

314. In a random sample of 100 balloons, 30 of them are blue. Construct a 95% confidence  interval for the true proportion of the blue balloon.
A) (0.210, 0.399)
B) (0.193, 0.407) 
C) (0.165, 0.334)
D) (0.200, 0.380)
E) (0.198, 0.378)

Answer 53

A

0.210,0.399

Question 54

315. What is the meaning of null hypothesis?
A) Population parameter
B) Population mean
C) Population proportion
D) Population analogy
E) Population interval

Answer 54

A the null is a statement about a population parameter. It usually predicts no difference or no relationship in population.

Question 55

316. Which of the following statements sounds like typical null or alternative hypotheses?
I. The coin is fair
II. There is no correlation in the population
III. The defendant is guilty. 
A) I only
B) I and II
C) I and III
D) II and III
E) I, II, and III

Answer 55

B We assume that it is fair, when testing significance of a correlation using hypothesis testing, we assume the correlation in the population is zero.

Question 56

317. If a 90% confidence interval is to be constructed, it will be \_\_\_\_\_\_\_\_\_ the 95% confidence interval.
A) narrower than
B) thinner than
C) wider than
D) the same as 
E) None of the above

Answer 56

WHEN YOU REDUCE THE CONFIDENCE INTERVAL, THE INTERVAL GETS SMALLER.

Question 57

318. A consumer survey found that more than 12% of the screws from a supplier are defective. To test this claim, the correct alternative hypothesis is:
A) p=0.12
B) p=/= 0.12
C) p0.12
E) p

Answer 57

D The claim is that the percentage is more than 12% hence p>.12.

Question 58

319. Which of the following statements is true?
     I. If the margin of error is small, then the confidence level is high
     II. A confidence interval is a type of point estimate
     III. A population mean is a point estimate
     IV. If the margin of error is small, then the confidence level is low
     A) I only
     B) II only
     C) III only
     D) IV only
     E) None of the above

Answer 58

E CONFIDENCE LEVEL IS NOT AFFECTED BY MARGIN OF ERROR, it is a type of interval estimate. population is not an example of point estimate: a sample, on the other hand is an example of a point estimate.

Question 59

320. While conducting an experiment to test a hypothesis, a researcher decideds to double her sample size. Which of the following will increase?
I. The effective size of the hypothesis test
II. Type II error probability
III. Power of the hypothesis test
A) I only
B) II only
C) III only 
D) All of the above
E) None of the above

Answer 59

C the effect size is not affected by sample size. type II gets smaller as sample gets bigger.

Question 60

321. A 99% confidence interval for the true mean of a normal population was found to be (123.33, 140.67). If the sample size was 40, find the margin of error for this estimation.
     A) 8.67
     B) 17.34
     C) 264
     D) 132
     E) The sample mean must be known to find the margin of error

Answer 60

A.

Question 61

322. In a sample of 145 cups of coffee, customers rated 37 of them as being served “too hot”. Calculate a 95% confidence interval for the true proportion of cups of coffee that customers will feel are served “too hot.”
     A) 36.8% to 37.2%
     B) 23.5% to 27.5%
     C) 18% to 33%
     D) 16% to 35%
     E) Since the population is not stated to be normal, we cannot use typical techniques to find this interval

Answer 61

C.

Question 62

322. In a sample of 145 cups of coffee, customers rated 37 of them as being served "too hot". Calculate a 95% confidence interval for the true proportion of cups of coffee that customers will feel are served "too hot."
A) 36.8% to 37.2%
B) 23.5% to 27.5%
C) 18% to 33%
D) 16% to 35

Answer 62

C.