1. Suppose a random sample of 36 items is selected from a population. The population standard deviation is known to be 10. The standard error of the mean would be:
(a) 1.333
(b) 1.667
(c) 3.667
(d) 2.333
(e) 1.875
2. From 100 homes of similar sizes, a sample of 25 homes is selected to study the average home heating cost during the winter months. Suppose the heating cost is known to be normally distributed with mean of $220 per month for the four months of winter and standard deviation of $45. Ifthe 100 homes represent the population size, the standard error of the heating cost would be:
(a) 9.00
(b) 8.75
(c) 3.66
(d) 7.83
(e) 1.87
3. Suppose n=64 measurements is selected from a population with mean #220 and standard deviation 0′ =16. The Z-score corresponding to a value of x = 24 would be:
(a) 2.0
(b) 3.0
(c) -2.5
(d) -2.0
(e) 1.5
4. A random sample of n=100 observations is selected from a population with y = 30 and standard deviation 0′ =16 . The probability that p()-c 2 28) is
(a) 0.8236
(b) 0.8936
(c) 0.9036
(d) 0.9983
(e) 0.8944
5. A random sample of n=100 observations is selected from a population with y = 30 and standard deviation 0′ =16 . The probability that p(22.1< ; < 26.8) is
(a) 0.0434
(b) 0.0228
(c) 0.0036
(d) 0.0983
(e) 0.0944
6. A random sample of size 36 is drawn from a population with meany = 278. If 86% of the time the sample mean is less than 281, then the population standard deviation would be:
(a) 16.67
(b) 12.67
(c) 11.12
(d) 13.33
(e) 19.67
7. A random sample of size n=81 is drawn from population with mean equal to 50 and standard deviation 25. The expected value ofthe mean E(xi) [or, y;] and the standard errora;
(a) 50 and 2.95
(b) 50 and 2.78
(c) 28 and 1.72
(d) 50 and 15.00
(e) 80 and 12.0
8. According to a recent news report, the average price of gasoline is $3.80 per gallon (March 2011). This price can be considered as the nationwide population mean price per gallon. Suppose that the standard deviation of the gasoline price per gallon is $0.50. A sample of 49 gas stations in Salt Lake City is taken. The probability that the sample mean price is within $0.10 of the population mean is
(a) 0.9236
(b) 0.8384
(c) 0.9544
(d) 0.9983
(e) 0.8764
9. A finite population is normally distributed with mean #250 and standard deviation 0′ =15. Suppose a sample of size 49 is taken so that the sample mean )-c can be used to estimate the population mean y. The probability that the sample mean is less than or equal to 48 or, p(; S 48) ifthe size ofthe finite population is N=150
(a) 0.1236
(b) 0.2420
(c) 0.1544
(d) 0.1983
(e) 0.1292
10. The average life of a battery used in newly designed electric cars is 150 hours with a standard deviation of 20 hours. Suppose these values are true for all batteries of this type so that these values can be considered true for the population with u = 150 and o
If a sample of size 50 is selected, the probability that the sample mean life is within i 5 of the population mean (between 145 and 155 hours) is
(a) 0.8236
(b) 0.9420
(c) 0.9544
(d) 0.9232
(e) 0.1292
11. The production manager of a bottling plant has acquired new machines to fill beverage cans. These machines are used to fill 16 ounce cans in one of their production lines. If the filling machine is working properly, the mean fill volume should be 16.0 ounces with a standard deviation of 0.30 ounces. If mean fill volume in the cans is over 16.2 ounces or below 15.8 ounces, then an over filling or under filling occurs. To avoid over or under filling the production manager randomly selects a sample of 9 cans periodically and checks the volume. If the average content is less than 15.8 ounces or more than 16.2 ounces, the production manager must stop the line to make adjustments. The probability of stopping the line based on the information above is:
(a) 0.0236
(b) 0.0376
(c) 0.0400
(d) 0.0456
(e) None ofthe above
12. A new Rasmussen Report national telephone survey finds that just 32% of American Adults favor “sin taxes” on soda and junk foods. The survey was based on a sample of 1,000 American Adults. Based on this survey data, the standard deviation of For the standard error of sample proportion would be (considering p = 0.32)
(a) 0.0176
(b) 0.0200
(c) 0.0196
(d) 0.0148
(e) 0.0138
13. Based on a report, 55% of the voters believe that the nation’s current economic problems are the result of recession that started during the Bush administration. A new Rasmussen Reports national telephone survey finds that 51% of likely voters say the nation’s current economic problems are due to the recession which began under the administration of George Bush. This survey was based on 1,000 likely voters and was conducted on March 18-19, 2011 by Rasmussen Reports. Based on this survey report, the probability that the sample proportion is lower than 51% is approximately
(a) 0.0076
(b) 0.0200
(c) 0.0055
(d) 0.0098
(e) 0.0128
14. According to a newly published report approximately 43% of adults say filing their tax paperwork is worse than a trip to the dentist. If a random sample of 200 is chosen, the probability that at least 90 of them share this opinion is
(a) 0.2076
(b) 0.3200
(c) 0.0155
(d) 0.2843
(e) 0.3128
15. A study about the students graduating within 4 years of their entrance to the universities indicated that 62% of the students do not graduate within 4 years. Suppose a random sample of 500 students was taken. The sample considered the students after 4 years of their college entrance. The probability that fewer than 285 graduated within 4 years is
(a) 0.4893
(b) 0.9893
(c) 0.0107
(d) 0.0584
(e) 0.0628
16. Historically, a production line produces 6% defective items. The production supervisor takes a sample of 100 items frequently and if he finds 8 or more defective products, he stops the line to make adjustments. The probability that a random sample of 100 would lead to the stoppage ofthe production line is:
(a) 0.2995
(b) 0.3893
(c) 0.2005
(d) 0.4584
(e) 0.7995
17. In simple random sampling
(a) Every sample has equal probability being selected
(b) Every sample is drawn at a pre-specified time
(c) Every item in the sample has equal probability
(d) None ofthe above is correct
(e) Only (a) and (c) are correct
18. From a population of size 8 (N=8), all possible samples of size 3 (n=3) that can be drawn are:
(a) 24
(b) 15
(c) 10
(d) 56
(e) 86
19. A confidence interval for the mean is determined using the following formula Ei1.28[ij J; The confidence level being used in the above interval is
(a) 95%
(b) 98%
(c) 99%
(d) 67%
(e) 80%
20. In a confidence interval, increasing the confidence level while keeping the sample size fixed
(a) increases the width of the confidence interval.
(b) leaves the confidence interval unchanged.
(c) makes the confidence interval estimate more precise.
(d) makes the confidence interval estimate more reliable.
(e) decreases the width of the confidence interval.
21. The general form of a confidence interval is
(a) Point estimate i standard error.
(b) Mean i standard error ofthe mean.
(c) Mean i the margin of error.
(d) Point estimate i the margin of error.
(e) Estimate i the margin of error.
22. A random sample of n measurements is selected from a population with unknown mean y and known standard deviation 0′. A 95% confidence interval for y when n = 200,} 2102,0′ = 4.69 would be
(a) 104.35 and 15.65.
(b) 103.76 and 134.24
(c) 101.35 and 102.65
(d) 108.56 and 120.44
(e) 118.25 and 123.75
23. A random sample of size 20 produced a sample mean $232.8 and a standard deviation 5 = 3.6. A 80% confidence interval using a t-distribution is to be constructed. The t- value for the interval would be
(a) 1.711
(b) 2.064
(c) 2.038
(d) 1.328
(e) 2.485
24. In constructing a confidence interval with known population standard deviation 0 the sample size is increased from n=36 to n=144 while the confidence level is held fixed at 95%. This will
(a) increase the width of the confidence interval making the estimate less precise.
(b) decrease the width ofthe confidence interval making the estimate less precise.
(c) leave the width ofthe confidence interval unchanged.
(d) double the width of the confidence interval.
(e) decrease the width of the confidence interval making the estimate more precise.
25. In constructing a confidence interval with known population standard deviation (0′ = 8), the sample size is increased from n=64 to n=256 while the confidence level is held fixed at 95%. This will
(a) decrease the width of the confidence interval by 75%.
(b) leave the width of the confidence interval unchanged.
(c) double the width of the confidence interval.
(d) reduce the width of the interval to one-half.
(e) Increase the width ofthe interval by 100%
27. The length of time that a space rocket component functions is approximately normally distributed. A sample of 20 of these components showed a mean of $2900 hours with a standard deviation 5 = 87 hours. A 95% confidence interval for the mean time that the component will function is to be constructed. The margin of error would be
(a) 42.9
(b) 40.7
(c) 81.4
(d) 34.2
(e) Can’t be determined
28. The length of time that a space rocket component functions is approximately normally distributed. A sample of 20 of these components showed a mean of $2900 hours with a standard deviation 5 = 87 hours. A 95% confidence interval for the mean time that the component will function
(a) 824.94 to 927.06
(b) 724.94 to 927.46
(c) 823.45 to 928.55
(d) 859.3 to 940.7
(e) 824.00 to 927.06
29. The standard deviation of the test scores on a certain college placement test is known to be 12.5. A random sample of 81 students had a mean score of 86.8. A 90% confidence interval estimate for the average score of all students is
(a) 82.94 to 92.06
(b) 84.52 to 89.08
(c) 82.45 to 92.55
(d) 85.32 to 94.68
(e) 84.00 to 92.70
30. Which of the following statement is true for constructing the confidence interval estimate for the population mean
(a) higher is the confidence interval, wider is the confidence interval for a fixed samplesize.
(b) larger is the sample standard deviation, wider is the confidence interval when the sample size and confidence level are fixed
(c) in cases where the population standard deviation 0 is known, the appropriate distribution to use to construct the interval is the normal distribution.
(d) In a confidence interval, the margin of error gets larger as the sample size isincreased.
(e) all of the above statements are true.
31. The average life of a sample of 30 car tires was found to be 60,000 miles. It is known that the lifetimes of such tires are normally distributed with a standard deviation of 7,500 miles. A 95% confidence interval estimate of the mean life of all such tires was calculated. The width of this confidence interval is
(a) 5937
(b) 5368
(c) 6357
(d) 5731
(e) 5846
32. The Nielsen Company reported that as of the third quarter of 2010, 28 percent of U.S. mobile subscribers now have smart phones, cell phones with operating systems resembling those of computers. The growing popularity of smart phones like Apple’s iPhone, RIM’s Blackberry devices and a variety of Google Android-based models on the market, has accelerated the adoption rate. Among those who acquired a new cell phone in the past six months, 41 percent opted for a Smartphone over a standard feature phone. A sample of 850 high school students was asked if they had a smart phone. An overwhelming 578 indicated that they had a smart phone. The margin of error at a 95% confidence would be
(a) 0.02
(b) 0.05
(c) 0.03
(d) 0.09
(e) 0.06
33. Approximately 51 percent of the U.S. population has at least two credit cards. (Source: Experian national score index study, February 2007). Suppose a study of 500 consumers showed that 315 carried three or more credit cards. A 95% confidence interval for the proportion of U.S. population who carried three or more credit cards would be
(a) 0.588 to 0.734
(b) 0.598 to 0.689
(c) 0.633 to 0.732
(d) 0.588 to 0.672
(e) 0.355 to 0.896
34. A quality engineer is interested in estimating the mean time required to assemble a bar code scanner. If the engineer wishes to be 95% confident that the error in estimating the mean time is less than 0.25 minutes, and she knows from the past experience that the standard deviation of the assembly time is 0.75 minutes; the sample size she would need is
(a) 20
(b) 24
(c) 25
(d) 35
(e) 30
35. The required sample size to estimate the mean for an a particular study resulted into significantly large sample size. The analyst believes that he does not have the time or the resources to collect such a large sample. Which of the following actions would lead to a reduced sample size?
(a) the margin of error required in the actual study should be increased.
(b) the confidence level should be reduces.
(c) the variation in the population should be reduced.
(d) all of the above will lead to a reduced sample size.
36. A large hospital wants to estimate the mean time before the patients are attended upon arrival to the hospital emergency room. From a past study, it is known that the standard deviation of the waiting time is 12 minutes. If the hospital administration wishes to estimate the mean time within 3 minutes with a 95% confidence, the sample size needed will be
(a) 107
(b) 190
(c) 500
(d) 62
(e) 290
37. A manufacturer of plasma TVs has problems with excessive customer complaints and consequent return of the product for repair or replacement. The quality control department wants to determine the magnitude of the problem so that it can estimateits warranty liability. The number of plasma TVs the quality engineer should sampleand inspect in order to estimate the fraction defective, p within 2% with a 95% confidence would be
(a) 2500
(b) 2401
(c) 3000
(d) 2000
(e) 5300
38. A manufacturer of electronic chip is interested in estimating the fraction defective chips produced in one of their plants. A random sample of 200 chips produced 12 defectives. The point estimate and a 95% confidence interval on the fraction defective would be
(a) E : 0.12 and 95% confidence interval 0.03 ≤ p ≤ 0.09
(b) E : 0.06 and 95% confidence interval 0.03 ≤ p ≤ 0.09
(c) E : 0.06 and 95% confidence interval 0.02 ≤ p ≤ 0.07
(d) E : 0.06 and 95% confidence interval 0.06 ≤ p ≤ 0.09
(e) None ofthe above.