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SCENARIO 8-11 A poll was conducted by the marketing department of a video game company to determine the popularity of a new game that was targeted to be launched in three months.Telephone interviews with 1,500 young adults were conducted which revealed that 49% said they would purchase the new game.The margin of error was ±3 percentage points. -Referring to Scenario 8-11, you are 99% confidence that the percentage of the targeted young adults who will purchase the new game is somewhere between 46% and 52%.

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SCENARIO 8-6 After an extensive advertising campaign, the manager of a company wants to estimate the proportion of potential customers that recognize a new product.She samples 120 potential consumers and finds that 54 recognize this product.She uses this sample information to obtain a 95% confidence interval that goes from 0.36 to 0.54. -Referring to Scenario 8-6, the parameter of interest is 54/120 = 0.45.

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SCENARIO 8-2 A quality control engineer is interested in the mean length of sheet insulation being cut automatically by machine.The desired mean length of the insulation is 12 feet.It is known that the standard deviation in the cutting length is 0.15 feet.A sample of 70 cut sheets yields a mean length of 12.14 feet.This sample will be used to obtain a 99% confidence interval for the mean length cut by machine. -Referring to Scenario 8-2, suppose the engineer had decided to estimate the mean length to within0.03 with 99% confidence.Then the sample size would be _.

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165.8724 r...

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The sample mean is a point estimate of the population mean.

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A prison official wants to estimate the proportion of cases of recidivism.Examining the records of250 convicts, the official determines that there are 65 cases of recidivism.A 99% confidence interval for the proportion of cases of recidivism would go from to .

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SCENARIO 8-11 A poll was conducted by the marketing department of a video game company to determine the popularity of a new game that was targeted to be launched in three months.Telephone interviews with 1,500 young adults were conducted which revealed that 49% said they would purchase the new game.The margin of error was ±3 percentage points. -Referring to Scenario 8-11, what is the needed sample size to obtain a 90% confidence interval estimate of the percentage of the targeted young adults who will purchase the new game by allowing the same level of margin of error?

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For a given data set, the confidence interval will be wider for 95% confidence than for90% confidence.

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SCENARIO 8-11 A poll was conducted by the marketing department of a video game company to determine the popularity of a new game that was targeted to be launched in three months.Telephone interviews with 1,500 young adults were conducted which revealed that 49% said they would purchase the new game.The margin of error was ±3 percentage points. -Referring to Scenario 8-11, what is the needed sample size to obtain a 95% confidence interval in estimating the percentage of the targeted young adults who will purchase the new game to within±5%?

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SCENARIO 8-8 The president of a university would like to estimate the proportion of the student population that owns a personal computer.In a sample of 500 students, 417 own a personal computer. -Referring to Scenario 8-8, we are 99% confident that the mean numbers of student population who own a personal computer is between 0.7911 and 0.8769.

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The t distribution approaches the standardized normal distribution when the number of degrees of freedom increases.

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Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses.Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund.A recent survey of 8 private colleges in the United States revealed the following endowments (in millions of dollars) : 60.2, 47.0, 235.1, 490.0, 122.6, 177.5,95.4, and 220.0.Summary statistics yield Xˉ=180.975\bar { X } = 180.975 and S= 143.042 .Calculate a 95% confidence interval for the mean endowment of all the private colleges in the United States assuming a normal distribution for the endowments.


A) $180.975 ±\pm $94.066
B) $180.975 ±\pm $99.123
C) $180.975 ±\pm $116.621
D) $180.975 ±\pm $119.586

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It is desired to estimate the mean total compensation of CEOs in the Service industry.Data were randomly collected from 18 CEOs and the 95% confidence interval was calculated to be ($2,181,260,$5,836,180) .Which of the following interpretations is correct?


A) 95% of the sampled total compensation values fell between $2,181,260 and $5,836,180.
B) We are 95% confident that the mean of the sampled CEOs falls in the interval $2,181,260 to $5,836,180.
C) In the population of Service industry CEOs, 95% of them will have total compensations that fall in the interval $2,181,260 to $5,836,180.
D) We are 95% confident that the mean total compensation of all CEOs in the Service industry falls in the interval $2,181,260 to $5,836,180.

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SCENARIO 8-6 After an extensive advertising campaign, the manager of a company wants to estimate the proportion of potential customers that recognize a new product.She samples 120 potential consumers and finds that 54 recognize this product.She uses this sample information to obtain a 95% confidence interval that goes from 0.36 to 0.54. -Referring to Scenario 8-6, it is possible that the true proportion of people that recognize the product is not between 0.36 and 0.54.

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SCENARIO 8-8 The president of a university would like to estimate the proportion of the student population that owns a personal computer.In a sample of 500 students, 417 own a personal computer. -Referring to Scenario 8-8, a 99% confidence interval for the proportion of the student population who own a personal computer is from to .

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Which of the following statements is(are) correct concerning a 95% confidence interval?


A) There is a 0.95 probability that the population parameter of interest is contained within the interval.
B) In computing confidence intervals from 20 samples, 19 of the intervals would contain the population mean.
C) If you were to produce all possible confidence intervals using the sample mean of each sample of a given size from the population, then 95% of the intervals would contain the population mean.
D) There is a 0.05 probability the population parameter of interest is not contained within theinterval.

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An economist is interested in studying the incomes of consumers in a country.The population standard deviation is known to be $1,000.A random sample of 50 individuals resulted in a mean income of $15,000.What is the upper end in a 99% confidence interval for the average income?


A) $15,052
B) $15,141
C) $15,330
D) $15,364

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SCENARIO 8-8 The president of a university would like to estimate the proportion of the student population that owns a personal computer.In a sample of 500 students, 417 own a personal computer. -Referring to Scenario 8-8, it is possible that the 99% confidence interval calculated from the data will not contain the sample proportion of students who own a personal computer.

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The standardized normal distribution is used to develop a confidence interval estimate of the population proportion when the sample size is sufficiently large.

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SCENARIO 8-4 The actual voltages of power packs labeled as 12 volts are as follows: 11.77, 11.90, 11.64, 11.84, 12.13, 11.99, and 11.77. -Referring to Scenario 8-4, it is possible that the 99% confidence interval calculated from the data will not contain the mean voltage for the entire population.

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SCENARIO 8-12 The Three Brothers Energy Drink Company bottles and distributes a popular drink for athletes and exercise enthusiasts.Because of its marketing successes the company has installed an additional filling machine and the managers are eager to use it in daily operations.The machine is set to fill bottles at 16 oz. However, we know there is inherent machine variability and quality control has determined through testing a mean of 16.2 oz.and a standard deviation of 0.3 oz.using a 100 bottle sample. -Refer to Scenario 8-12 It is possible that a 90% confidence interval will not include the mean fill volume.

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