Explore BrainMass

Explore BrainMass

    Sampling distribution, Normal Distribution, t-distribution

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    Please see attached.
    1. For air travelers, one of the biggest complaints is of the waiting time between when the airplane taxis away from the terminal until the flight takes off. This waiting time is known to have a skewed-right distribution with a mean of 10 minutes and a standard deviation of 8 minutes. Suppose 100 flights have been randomly sampled. Describe the sampling distribution of the mean waiting time between when the airplane taxis away from the terminal until the flight takes off for these 100 flights.

    a. Distribution is skewed-right with mean = 10 minutes and standard error = 0.8 minutes.
    b. Distribution is skewed-right with mean = 10 minutes and standard error = 8 minutes.
    c. Distribution is approximately normal with mean = 10 minutes and standard error = 0.8 minutes.
    d. Distribution is approximately normal with mean = 10 minutes and standard error = 8 minutes.

    2. The standard error of the mean for a sample of 100 is 30. In order to manipulate the standard error of the mean to be 15, we would:
    a. increase the sample size to 200.
    b. increase the sample size to 400.
    c. decrease the sample size to 50.
    d. decrease the sample size to 25.

    3. The use of the finite population correction factor when sampling without replacement from finite populations will:
    a. increase the standard error of the mean.
    b. not affect the standard error of the mean.
    c. reduce the standard error of the mean.
    d. only affect the proportion, not the mean.

    4. Assume that house prices in a neighborhood are normally distributed with standard deviation $20,000. A random sample of 16 observations is taken. What is the probability that the sample mean differs from the population mean by more than $5,000?
    a. 0.3174
    b. 0.1587
    c. 0, because it is assumed that the sample mean is equal to the population mean in a normally distributed population
    d. Cannot be determined from the information given

    5. If the expectation of a sampling distribution is located at the parameter it is estimating, then we call that sampling distribution:
    a. unbiased.
    b. minimum variance.
    c. biased.
    d. random.

    6. The diameter of Ping-Pong balls manufactured at a large factory is expected to be approximately normally distributed with a mean of 1.30 inches and a standard deviation of 0.04 inch. What is the probability that a randomly selected Ping-Pong ball will have a diameter between 1.31 and 1.33 inches?
    a. 0.0987
    b. 0.1747
    c. 0.2734
    d. 0.3721

    7. Approximately 5% of U.S. families have a net worth in excess of $1 million and thus can be called "millionaires." However, a survey in the year 2000 found that 30% of Microsoft's 31,000 employees were millionaires. If random samples of 100 Microsoft employees had been taken that year, what proportion of the samples would have been between 25% and 35% millionaires?
    a. 0.1379
    b. 0.2758
    c. 0.7242
    d. 1.0911

    8. For sample size 16, the sampling distribution of the mean will be approximately normally distributed:
    a. regardless of the shape of the population.
    b. if the shape of the population is symmetrical.
    c. if the sample standard deviation is known.
    d. if the sample is normally distributed.

    9. The fill amount of bottles of soft drink has been found to be normally distributed with a mean of 2.0 liters and a standard deviation of 0.05 liter. If a random sample of bottles is selected, what is the probability that the sample mean will be between 1.99 and 2.0 liters?
    a. 0.4772
    b. 0.5228
    c. 0.9544
    d. Cannot be determined from the information given

    10. If you were constructing a 99% confidence interval of the population mean based on a sample of n = 25 where the standard deviation of the sample s = 0.05, the critical value of t will be:

    a. 2.7969
    b. 2.7874
    c. 2.4922
    d. 2.4851

    11. The width of a confidence interval estimate for a proportion will be:
    a. narrower for 99% confidence than for 95% confidence.
    b. wider for a sample size of 100 than for a sample size of 50.
    c. narrower for 90% confidence than for 95% confidence.
    d. narrower when the sample proportion is 0.50 than when the sample proportion is 0.20.

    12. When determining the sample size for a proportion for a given level of confidence and sampling error, the closer to 0.50 that p is estimated to be, the __________ the sample size required.
    a. smaller
    b. larger
    c. Sample size is not affected.
    d. The effect cannot be determined from the information given.

    13. A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?
    a. 105
    b. 150
    c. 420
    d. 597

    14. A major department store chain is interested in estimating the average amount its credit card customers spent on their first visit to the chain's new store in the mall. Fifteen credit card accounts were randomly sampled and analyzed with the following results: Χ = $50.50 and 2 s = 400. Assuming the distribution of the amount spent on their first visit is approximately normal, what is the shape of the sampling distribution of the sample mean that will be used to create the desired confidence interval for µ?
    a. Approximately normal with a mean of $50.50
    b. A standard normal distribution
    c. A t distribution with 15 degrees of freedom
    d. A t distribution with 14 degrees of freedom

    15. An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What is the upper end point in a 99% confidence interval for the average income?
    a. $15, 052
    b. $15, 141
    c. $15,330
    d. $15,364

    16. A 99% confidence interval estimate can be interpreted to mean that:

    a. if all possible samples are taken and confidence interval estimates are developed, 99% of them would include the true population mean somewhere within their interval.
    b. we have 99% confidence that we have selected a sample whose interval does include the population mean.
    c. Both of the above.
    d. None of the above.

    17. An internal control policy for an online fashion accessories store requires a quality assurance check before a shipment is made. The tolerable exception rate for this internal control is 0.05. During an audit, 500 shipping records were sampled from a population of 5,000 shipping records and 12 were found that violated the control. What is the upper bound for a 95% one-sided confidence interval estimate for the rate of noncompliance?
    a. 0.0060
    b. 0.0067
    c. 0.0353
    d. 0.0374

    © BrainMass Inc. brainmass.com June 3, 2020, 5:52 pm ad1c9bdddf
    https://brainmass.com/statistics/confidence-interval/sampling-distribution-normal-distribution-t-distribution-40684

    Attachments

    Solution Preview

    See attached excel file where the formatting is conserved

    1. For air travelers, one of the biggest complaints is of the waiting time
    between when the airplane taxis away from the terminal until the
    flight takes off. This waiting time is known to have a skewed-right
    distribution with a mean of 10 minutes and a standard deviation of
    8 minutes. Suppose 100 flights have been randomly sampled.
    Describe the sampling distribution of the mean waiting time
    between when the airplane taxis away from the terminal until the
    flight takes off for these 100 flights.

    a. Distribution is skewed-right with mean = 10 minutes and
    standard error = 0.8 minutes.
    b. Distribution is skewed-right with mean = 10 minutes and
    standard error = 8 minutes.
    c. Distribution is approximately normal with mean = 10 minutes
    and standard error = 0.8 minutes.
    d. Distribution is approximately normal with mean = 10 minutes
    and standard error = 8 minutes.

    Answer:
    c. Distribution is approximately normal with mean = 10 minutes
    and standard error = 0.8 minutes.

    The sampling distribution of the mean approaches a normal distribution as n increases, even if the underlying population distribution is far from normal.

    Standard deviation =σ= 8 min
    sample size=n= 100
    σx=standard error of mean=σ/√n= 0.8 = ( 8 /√ 100)

    2. The standard error of the mean for a sample of 100 is 30. In order
    to manipulate the standard error of the mean to be 15, we would:
    a. increase the sample size to 200.
    b. increase the sample size to 400.
    c. decrease the sample size to 50.
    d. decrease the sample size to 25.

    Answer: b. increase the sample size to 400.

    σx=standard error of mean=σ/√n= 30
    sample size=n= 100
    Standard deviation =σ= 300 = ( 30 *√ 100)

    to manipulate the standard error of the mean to be 15
    or σx = 15
    σx=standard error of mean=σ/√n
    σ= 300
    or n=(σ^2)/(σx^2)= 400 =300^2/15^2

    sample size = 400

    3. The use of the finite population correction factor when sampling
    without replacement from finite populations will:
    a. increase the standard error of the mean.
    b. not affect the standard error of the mean.
    c. reduce the standard error of the mean.
    d. only affect the proportion, not the mean.

    Answer: c. reduce the standard error of the mean.

    The correction factor is √(N-n)/(N-1)
    where N= population size
    n= sample size
    N-n < N-1
    Therefore the correction factor &#8730;(N-n)/(N-1) is < 1
    We multiply the standard error of mean by the correction factor
    So the standard error term is reduced.

    4. Assume that house prices in a neighborhood are normally
    distributed with standard deviation $20,000. A random sample of 16
    observations is taken. What is the probability that the sample mean
    differs from the population mean by more than $5,000?
    a. 0.3174
    b. 0.1587
    c. 0, because it is assumed that the sample mean is equal to the
    population mean in a normally distributed population
    d. Cannot be determined from the information given

    Answer: b. 0.1587

    Standard deviation =&#963;= 20000
    sample size=n= 16
    &#963;x=standard error of mean=&#963;/&#8730;n= 5000 = ( 20000 /&#8730; 16)

    z=(sample mean-&#956;)/&#963;x= 1 =5000/5000
    Cumulative Probability corresponding to ...

    Solution Summary

    Answers to 17 Multiple choice questions on Sampling distribution, Normal Distribution, t-distribution, confidence interval estimate, standard error of the mean, finite population correction factor, sample size, degrees of freedom, one-sided confidence interval estimate

    $2.19

    ADVERTISEMENT