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Q:
Assuming the same level of significance , as the sample size increases, the value of t/2 approaches the value of z/2.
Q:
The standard error of the sample mean is s/n.
Q:
When the margin of error is added to and subtracted from the sample mean, an interval is formed that will contain with a probability of (1 - ).
Q:
The t distribution always has n degrees of freedom.
Q:
The coffee and soup machine at the local subway station is supposed to fill cups with 6 ounces of soup. Ten cups of soup are bought, with results of a mean of 5.93 ounces and a standard deviation of 0.13 ounces. Construct a 95.44 percent tolerance interval of the machine-fill amounts and a 95 percent confidence interval for the true machine-fill amount.
Q:
When sample size is 16, find t.001.
Q:
When sample size is 11, find t.001.
Q:
When sample size is 20, find t.001.
Q:
When sample size is 16, find t.025.
Q:
When sample size is 11, find t.025.
Q:
When sample size is 20, find t.025.
Q:
When sample size is 16, find t.10.
Q:
When sample size is 11, find t.10.
Q:
When sample size is 20, find t.10.
Q:
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. 225 flight records are randomly selected and the number of unoccupied seats is noted, with a sample mean of 11.6 seats and a standard deviation of 4.1 seats. How many flights should we select if we wish to estimate to within 5 seats and be 95 percent confident?A. 44B. 3C. 2D. 110E. 6
Q:
Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoccupied seats per flight over the past year. 225 flight records are randomly selected and the number of unoccupied seats is noted, with a sample mean of 11.6 seats and a standard deviation of 4.1 seats. Calculate a 95 percent confidence interval for μ, the mean number of unoccupied seats per flight during the past year.
A. [11.06, 12.14]
B. [11.34, 11.86]
C. [10.44, 12.76]
D. [11.15, 12.05]
E. [3.56, 19.64]
Q:
Researchers studied the role that the age of workers has in determining the hours per month spent on personal tasks. A sample of 1,686 adults were observed for one month. The data follow. Construct a 93 percent confidence interval for the mean hours spent on personal tasks for 45- to 64-year-olds.
A. [4.26, 4.36]
B. [4.25, 4.37]
C. [4.27, 4.35]
D. [4.28, 4.34]
E. [2.83, 5.79]
Q:
Researchers studied the role that the age of workers has in determining the hours per month spent on personal tasks. A sample of 1,686 adults were observed for one month. The data follow. Construct an 88 percent confidence interval for the mean hours spent on personal tasks for 18- to 24-year-olds.
A. [4.09, 4.25]
B. [4.11, 4.23]
C. [4.08, 4.26]
D. [4.14, 4.20]
E. [4.15, 4.19]
Q:
A statistical quality control process for cereal production measures the weight of a cereal box. The population standard deviation is known to be .06 ounces. In order to achieve 97 percent confidence with a margin of error of .02 ounces, how large a sample should be used?
A. 32
B. 43
C. 7
D. 664
Q:
A car insurance company would like to determine the proportion of accident claims covered by the company. According to a preliminary estimate, 60 percent of the claims are covered. How large a sample should be taken to estimate the proportion of accident claims covered by the company if we want to be 98 percent confident that the sample percentage is within 3 percent of the actual percentage of the accidents covered by the insurance company?
A. 348
B. 1508
C. 1887
D. 1442
Q:
A computer manufacturing company has sent a mail survey to 2,800 of its randomly selected customers that have purchased a new laptop. The survey asked the customers whether or not they were satisfied with the computer. 800 customers responded to the survey. 640 customers indicated that they were satisfied, while 160 customers indicated they were not satisfied with their new computer. Construct a 96 percent confidence interval estimate of the true proportion of customers satisfied with their new computer.
A. [.771, .829]
B. [.788, .812]
C. [.775, .825]
D. [.785, .816]
Q:
An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered by the company. A random sample of 200 claims shows that the insurance company covered 80 accident claims and did not cover 120 claims. Construct a 90 percent confidence interval estimate of the true proportion of claims covered by the insurance company.
A. [.356, .444]
B. [.360, .440]
C. [.372, .428]
D. [.343, .457]
Q:
An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered by the company. A random sample of 240 claims shows that the insurance company covered 90 accident claims, while 150 claims were not covered. Use a confidence interval of 95 percent and determine the margin of error.
A. .01513
B. .03125
C. .06125
D. .02965
Q:
The quality control manager of a tire company wishes to estimate the tensile strength of a standard size of rubber used to make a class of radial tires. A random sample of 61 pieces of rubber from different production batches is subjected to a stress test. The test measures the force (in pounds) needed to break the rubber. According to the sample results, the average pressure is 238.4 pounds with a population standard deviation of 35 pounds. Determine the 98 percent confidence interval.
A. [227.96, 248.84]
B. [236.64, 240.16]
C. [233.63, 243.17]
D. [229.21, 247.59]
Q:
A local country club golf tournament organizer is attempting to estimate the average number of strokes for the 13th hole. On a particular day, 64 players completed the play on the 13th hole, with an average of 4.25 strokes and a population standard deviation of 1.6 strokes. Determine the 95 percent confidence interval for the average number of strokes.
A. [3.94, 4.56]
B. [3.86, 4.64]
C. [4.05, 4.45]
D. [3.92, 4.58]
Q:
If the 95 percent confidence interval for a mean is from .771 to 1.629, can we conclude that = .5, using a 95 percent confidence interval?A. YesB. No
Q:
A random sample n = 15 is taken from a population assumed to be normal, and = 1.2 and the sample variance is .36. Calculate a 99 percent confidence interval for .A. [.923, 1.477]B. [.794, 1.607]C. [.801, 1.600]D. [.739, 1.661]
Q:
A random sample n= 15 is taken from a population assumed to be normal, and = 1.2 and s = .6. Calculate a 98 percent confidence interval for A. [.675, 1.725]B. [.794, 1.606]C. [.876, 1.524]D. [.882, 1.518]
Q:
A sample of 2,000 people yielded =.52. Calculate a 99 percent confidence interval for p.
A. [.515, .525]
B. [.494, .546]
C. [.506, .534]
D. [.491, .549]
Q:
A sample of 2,000 people yielded =.52. Calculate a 95 percent confidence interval for p.
A. [.509, .531]
B. [.494, .546]
C. [.498, .542]
D. [.502, .538]
Q:
A sample of 2,000 people yielded =.52. Calculate a 90 percent confidence interval for p.
A. [.506, .534]
B. [.468, .572]
C. [.511, .529]
D. [.502, .538]
Q:
A sample of 2,000 people yielded =.52. What is the variance of the population proportion?A. .0056B. .0112C. .00012D. .0161
Q:
The 99 percent confidence interval for the average weight of a product is from 71.36 lb to 78.64 lb. Can we conclude that μ is equal to 71, using a 99 percent confidence interval?A. YesB. No
Q:
The 95 percent confidence interval for the average weight of a product is from 72.23 lb to 77.77 lb. Can we conclude that μ = 77, using a 95 percent confidence interval?A. YesB. No
Q:
The weight of a product is measured in pounds. A sample of 50 units is taken from a recent production. The sample yielded = 75 lb, and we know that 2 = 100 lb. Calculate a 99 percent confidence interval for .A. [71.36, 78.64]B. [74.25, 75.75]C. [38.58, 111.42]D. [71.71, 78.29]
Q:
The weight of a product is measured in pounds. A sample of 50 units is taken from a recent production. The sample yielded = 75 lb, and we know that 2 = 100 lb. Calculate a 95 percent confidence interval for .A. [71.25, 78.75]B. [72.23, 77.77]C. [47.28, 102.72]D. [72.67, 77.33]
Q:
The weight of a product is measured in pounds. A sample of 50 units is taken from a recent production. The sample yielded = 75 lb, and we know that 2 = 100 lb. Calculate a 90 percent confidence interval for .A. [73.19, 76.81]B. [51.74, 98.26]C. [72.67, 77.33]D. [67.50, 82.50]
Q:
The customer service manager for the XYZ Fastener Manufacturing Company examined 60 invoices and found 9 that contained errors. Find a 98 percent confidence interval for the proportion of invoices with errors.
A. [.147, .153]
B. [.080, .220]
C. [.043, .257]
D. [.112, .188]
Q:
You want to estimate the proportion of customers who are satisfied with their experiences at a local supermarket at = .10 and within .025 of the true value. It has been estimated that p =.85. How large of a sample is needed?A. 1083B. 553C. 71D. 336
Q:
In a randomly selected group of 650 automobile deaths, 180 were alcohol related. Construct a 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol.
A. [.243, .311]
B. [.262, .292]
C. [.259, .294]
D. [.263, .291]
Q:
An insurance company estimates 45 percent of its claims have errors. The insurance company wants to estimate with 99 percent confidence the proportion of claims with errors. What sample size is needed if they wish to be within 5 percent of the actual?
A. 664
B. 657
C. 163
D. 1084
Q:
A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour. An approximation is 20 percent. They want the estimate to be at the 90 percent confidence level and within 2 percent of the actual proportion. What sample size is needed?
A. 22
B. 1692
C. 1537
D. 1083
Q:
We want to estimate with 99 percent confidence the percentage of automobile purchasers who are under 30 years of age. A margin of error of 5 percentage points is desired. What sample size is needed? In an earlier sample, we found a 99 percent confidence interval of purchasers under 30 years of age to be [.18, .27].
A. 104
B. 664
C. 392
D. 523
Q:
Find a 98 percent confidence interval for p when = .25 and n = 400.
A. [.206, .294]
B. [.231, .269]
C. [.228, .272]
D. [.200, .300]
Q:
Given the following test scores, find a 95 percent confidence interval for the population mean: 148, 154, 158, 160, 161, 162, 166, 170, 182, 195, 236. Assume population normality.
A. [155.24, 188.76]
B. [168.64, 175.36]
C. [157.25, 186.75]
D. [116.41, 227.59]
Q:
In a study of 265 engineering students, the average score on the final exam in the statistics course was 63.8 and s = 3.08. What is a 95 percent confidence for ?A. [63.59, 64.01]B. [57.76, 69.84]C. [63.43, 64.17]D. [63.56, 64.04]
Q:
What is a 95 percent confidence interval for when n = 10, =35.6, and s = 13.0? Assume population normality.A. [26.44, 44.76]B. [26.30, 44.90]C. [33.02, 38.18]D. [27.54, 43.66]
Q:
What sample size is needed to estimate with 95 percent confidence the mean intake of daily calcium within 20 units of the true mean if the intake is normally distributed with a variance of 1900 units?
A. 34,671
B. 187
C. 32
D. 19
Q:
What sample size is needed to estimate the proportion of highway speeders within 5 percent using a 90 percent confidence level?
A. 385
B. 68
C. 271
D. 165
Q:
What sample size is needed to obtain a 95 percent confidence interval for the proportion of fat in meat that is within 3 percent of the true value?
A. 267
B. 1068
C. 17
D. 545
Q:
What sample size is needed to obtain a 90 percent confidence interval for the mean protein content of meat if the estimate is to be within 2 pounds of the true mean value? Assume that the variance is 49 pounds.
A. 34
B. 1625
C. 21
D. 987
Q:
The success rate of an experimental medical procedure is 37 per 120 cases in a sample. Find a 95 percent confidence interval for the actual success proportion of the procedure.
A. [.2975, .4425]
B. [.2389, .3776]
C. [.2836, .4564]
D. [.2250, .3910]
Q:
In a survey of 1,000 people, 420 are opposed to an income tax increase. Construct a 95 percent confidence interval for the proportion of people in the population opposed to this tax increase.
A. [.394, .446]
B. [.389, .451]
C. [.380, .460]
D. [.399, .441]
Q:
A sample of 100 items has a population standard deviation of 5.1 and a mean of 21.6. Construct a 95 percent confidence interval for A. [11.60, 31.60]B. [21.16, 22.04]C. [20.60, 22.60]D. [20.76, 22.43]
Q:
A sample of 12 items yields = 48.5 grams and s = 1.5 grams. Assuming a normal distribution, construct a 90 percent confidence interval for the population mean weight.
A. [47.722, 49.278]
B. [47.788, 49.212]
C. [45.806, 51.194]
D. [47.865, 49.135]
Q:
A sample set of weights in pounds are 1.01, .95, 1.03, 1.04, .97, .97, .99, 1.01, and 1.03. Assume the population of weights is normally distributed. Find a 99 percent confidence interval for the mean population weight.
A. [.965, 1.035]
B. [.969, 1.031]
C. [.973, 1.027]
D. [.941, 1.059]
Q:
A random sample of size 30 from a normal population yields = 32.8 with a population standard deviation of 4.51. Construct a 95 percent confidence interval for .A. [23.96, 41.64]B. [32.04, 33.56]C. [31.45, 34.15]D. [31.19, 34.41]
Q:
In 2005, 13,000 Internet users were surveyed and asked about their willingness to pay fees for access to websites. Of these, 2,938 were definitely not willing to pay such fees. How large a sample is necessary to estimate the proportion of interest to within 2 percent in a 95 percent confidence interval?
A. 18
B. 307
C. 1680
D. 2000
E. 2965
Q:
In 2005, 13,000 Internet users were surveyed and asked about their willingness to pay fees for access to websites. Of these, 2,938 were definitely not willing to pay such fees. Construct a 95 percent confidence interval for the proportion definitely unwilling to pay fees.
A. [0.286, 0.302]
B. [0.219, 0.233]
C. [0.220, 0.232]
D. [0.212, 0.241]
E. [0.214, 0.245]
Q:
The state agriculture department tested 203 fuel samples across the state in 2009 for accuracy of the reported octane level. For premium grade, 14 out of 105 samples failed. How many samples would be needed to create a 99 percent confidence interval that is within 0.02 of the true proportion of premium grade fuel-quality failures?
A. 4148
B. 2838
C. 1913
D. 744
E. 54
Q:
The state agriculture department tested 203 fuel samples across the state in 2009 for accuracy of the reported octane level. For premium grade, 14 out of 105 samples failed. Find a 99 percent confidence interval for the true population proportion of premium grade fuel-quality failures.
A. [.045, .221]
B. [.068, .198]
C. [.023, .115]
D. [.048, .219]
E. [.100, .276]
Q:
Recently, a case of food poisoning was traced to a particular restaurant chain. The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur. Despite the response from the restaurant chain, many consumers refused to visit the restaurant for some time after the event. A survey was conducted three months after the food poisoning occurred, with a sample of 319 former customers contacted. Of the 319 contacted, 29 indicated that they would not go back to the restaurant because of the potential for food poisoning. What sample size would be needed in order to be 99 percent confident that the sample proportion is within .02 of Ï, the true proportion of customers who refuse to go back to the restaurant?
A. 14
B. 38
C. 129
D. 1,371
E. 1,777
Q:
Recently, a case of food poisoning was traced to a particular restaurant chain. The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur. Despite the response from the restaurant chain, many consumers refused to visit the restaurant for some time after the event. A survey was conducted three months after the food poisoning occurred, with a sample of 319 former customers contacted. Of the 319 contacted, 29 indicated that they would not go back to the restaurant because of the potential for food poisoning. Construct a 95 percent confidence interval for the true proportion of the market who still refuse to visit any of the restaurants in the chain three months after the event.
A. [.059, .122]
B. [.090, .091]
C. [.000, .196]
D. [.240, .339]
E. [.118, .244]
Q:
A manufacturer of NFL regulation footballs uses a machine to inflate its new balls to a pressure of 13.5 pounds (σ = 0.1). When the machine is properly calibrated, the mean inflation pressure is 13.5 pounds, but uncontrollable factors can cause the pressure of individual footballs to vary. For quality control purposes, the manufacturer wishes to estimate the mean inflation pressure to within 0.025 pounds of its true value with 99 percent confidence. What sample size should be used?
A. 677
B. 107
C. 35
D. 27
E. 11
Q:
On a standard IQ test, the standard deviation is 15. How many random IQ scores must be obtained if we want to find the true population mean (with an allowable error of 0.5) and we want 97 percent confidence in the results?
A. 4,239
B. 283
C. 212
D. 131
E. 66
Q:
A botanist measures the heights of 16 seedlings and obtains a mean and standard deviation of 72.5 cm and 4.5 cm, respectively. Find the 95 percent confidence interval for the mean height of seedlings in the population from which the sample was selected.
A. [62.91, 82.09]
B. [70.30, 74.70]
C. [70.10, 74.90]
D. [70.54, 74.46]
E. [71.90, 73.10]
Q:
An experiment was conducted to determine the effectiveness of a method designed to remove oil wastes found in soil. Three contaminated soil samples were treated. After 95 days, the percentage of contamination removed from each soil sample was measured, with a mean of 49.3 percent and a standard deviation of 1.5 percent. If we wished to narrow the boundary around μ for a 98 percent confidence interval to within 0.5 percent, how many soil samples should be in our experiment.
A. 437
B. 33
C. 9
D. 6
E. 3
Q:
An experiment was conducted to determine the effectiveness of a method designed to remove oil wastes found in soil. Three contaminated soil samples were treated. After 95 days, the percentage of contamination removed from each soil sample was measured, with a mean of 49.3 percent and a standard deviation of 1.5 percent. Estimate the mean percentage of contamination removed at a 98 percent confidence level.
A. [48.36, 50.24]
B. [43.27, 55.33]
C. [47.29, 51.31]
D. [47.88, 50.72]
E. [46.47, 52.13]
Q:
A federal bank examiner is interested in estimating the mean outstanding defaulted loans balance of all defaulted loans over the last three years. A random sample of 20 defaulted loans yielded a mean of $67,918 with a standard deviation of $16,552.40. Calculate a 90 percent confidence interval for the mean balance of defaulted loans over the past three years.
A. [66,487, 69,349]
B. [39,299, 96,537]
C. [57,329, 78,507]
D. [61,829, 74,007]
E. [61,519, 74,317]
Q:
There is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area, with a standard deviation of 18.52. Construct a 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers.
A. [11.014, 33.896]
B. [11.275, 33.635]
C. [12.284, 32.626]
D. [12.513, 32.397]
E. [20.072, 24.838]
Q:
In a manufacturing process, a machine produces bolts that have an average length of 3 inches with a variance of .03. If we randomly select three bolts from this process, what is the probability the mean length of the bolt is at most 3.1 inches?
A. 84.13%
B. 100%
C. 71.57%
D. 28.43%
E. 15.87%
Q:
In a manufacturing process, a machine produces bolts that have an average length of 3 inches with a variance of .03. If we randomly select three bolts from this process, What is the probability the mean length of the bolt is at least 3.16 inches?
A. 97.72%
B. 5.48%
C. 94.52%
D. 44.52%
E. 2.28%
Q:
In a manufacturing process, a machine produces bolts that have an average length of 3 inches with a variance of .03. If we randomly select three bolts from this process, what is the standard deviation of the sampling distribution of the sample mean?
A. .03
B. .01
C. .1732
D. .0577
E. .10
Q:
According to a hospital administrator, historical records over the past 10 years have shown that 20 percent of the major surgery patients are dissatisfied with after-surgery care in the hospital. A scientific poll based on 400 hospital patients has just been conducted. Sixty-four patients indicated that they were dissatisfied with the after-surgery care. What are the mean and the standard deviation of the sampling distribution of ?
A. 16% and .034%
B. 20% and 1.83%
C. 20% and 2%
D. 20% and .034%
E. 20% and 16%
Q:
According to a hospital administrator, historical records over the past 10 years have shown that 20 percent of the major surgery patients are dissatisfied with after-surgery care in the hospital. A scientific poll based on 400 hospital patients has just been conducted. What is the probability that at least 70 patients will not be satisfied with the after-surgery care?
A. 89.44%
B. 39.44%
C. 10.56%
D. 78.88%
E. 84.49%
Q:
According to a hospital administrator, historical records over the past 10 years have shown that 20 percent of the major surgery patients are dissatisfied with after-surgery care in the hospital. A scientific poll based on 400 hospital patients has just been conducted. What is the probability that fewer than 64 patients will not be satisfied with the after-surgery care?
A. 47.72%
B. 2.28%
C. 97.72%
D. 95.44%
E. 4.56%
Q:
The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are more than 90 days overdue (delinquent). The historical records of the company show that over the past 8 years, the average has been that 14 percent of the accounts have been delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What are the mean and the standard deviation of the sampling distribution of ?
A. 14 and 250
B. .14 and .1204
C. 35 and 5.486
D. .14 and .0219
E. 35 and 30.1
Q:
The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are more than 90 days overdue (delinquent). The historical records of the company show that over the past 8 years, the average has been that 14 percent of the accounts have been delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What is the probability that at least 30 accounts will be classified as delinquent?
A. 31.86%
B. 18.14%
C. 81.86%
D. 63.72%
E. 75.84%
Q:
The internal auditing staff of a local manufacturing company performs a sample audit each quarter to estimate the proportion of accounts that are more than 90 days overdue (delinquent). The historical records of the company show that over the past 8 years, the average has been that 13 percent of the accounts have been delinquent. For this quarter, the auditing staff randomly selected 250 customer accounts. What is the probability that no more than 40 accounts will be classified as delinquent?
A. 42.07%
B. 92.07%
C. 7.93%
D. 40.15%
E. 90.15%
Q:
If we have a sample size of 100 and the estimate of the population proportion is .10, what is the standard deviation of the sampling distribution of the sample proportion?
A. .0009
B. .03
C. 3
D. .01
E. .10