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# Normal Distribution, Sampling distribution

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1) If you were constructing a 99% confidence interval of the population mean based on a sample of n=25 where the standard deviation of sample s = 0.50, the critical t will be:
A) 2.7969 (B) 2.7874 (C) 2.4922 (D) 2.4851

2) A statistician wishes to estimate the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and 97% confidence interval was calculated to be (\$2,181,260, \$5,836,180). Which of the following interpretation is correct?
A) 97% of the sampled total compensation value fell between \$2,181,260 and 5,836,180.(B) We are 97% confident that the mean of sampled CEOs fall in the interval \$2,181,260 to \$5,836,180 (C) In the population of service industry CEOs, 97% of them will have total compensation that fall in the interval \$2,181,260 to \$5,836,180.(D) We are 97% confident that average total compensation of all CEO's in the service industry falls in the interval \$2,181,260 to \$ 5,836,180.

3) The personnel of a large corporation wishes to study absenteeism among clerical workers at the corporation's central office during the year. A random sample of 25 clerical workers revealed the following:
Absenteeism: X = 9.7 days and S=4.0 days
12 clerical workers were absent more than 10days

What are the upper and lower limits of the 95% confidence interval estimate of the mean number of absence for clerical workers last year .
A) 8.132 and 11.268 (B) 8.049 and 11.351 (C) 8.052 and 11.348 (D) 8.331 and 11.069

4) 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 population is symmetrical (C) if the sample standard deviation is know (D) if the sample is normally distributed

5) 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 to 25

6) A confidence interval was used to estimate the proportion of statistic student that was a female. A random sample of 72 statistics student generated the following 90% confidence interval( 0.438, 0.642) using the information, what size of 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

7) A stationery store wants to estimate the total retail value of the 300 greeting cards that it has in its inventory. What are the upper and lower limits of the 95% confidence interval estimate of the population total value of all greeting cards that are inventory if a random sample of 20 greeting card indicates an average value of \$1.67 and a standard deviation of \$0.32
A) \$457.52 and 544.48 (B) \$465.08 and \$536.92 (C) \$460.29 and \$541.72 (D) \$457.67 and \$544.33

8) A major department store chain is interested in estimating the average amount its credit card customer spent on their first visit to the chain's new store in the mall. 15 credits accounts were randomly sampled and analyzed with the following results X=\$50.50 s2 and =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 u?
A) Approximately normal with a mean of \$50.50 (B) standard normal distribution (C) A t distribution with 15 degrees of freedom (D) A t distribution with 14 degrees of freedom

9) The standard error of the mean:
A) is never larger than the standard deviation of the population (B) decrease as the sample size increases. (C) measures the variability of the mean from sample to sample (D) All of the above

10) 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 estimate to be, the _____the sample size required.
A) smaller (B) larger (C) sample size is not effected (D) the effect cannot be determined from the information given

11) 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 sample of 100 Microsoft employees had been taken that year, what proportion of the sample would have been between 25% and 35% millionaires ?
A) 0.1379 (B) 0.2758 (C) 0.7242 (D) 1.0911

https://brainmass.com/statistics/confidence-interval/180500

#### Solution Preview

1) If you were constructing a 99% confidence interval of the population mean based on a sample of n=25 where the standard deviation of sample s = 0.50, the critical t will be:
A) 2.7969 (B) 2.7874 (C) 2.4922 (D) 2.4851

99% Confidence limits

sample size=n= 25
Confidence level= 99%
Therefore Significance level=&#945;= 1% =100% -99%
No of tails= 2
This is 2 tailed because we are calculating the confidence interval

Since sample size= 25 < 30
and we are using sample standard deviation to estimate the population standard deviation
use t distribution
t at the 0.01 level of significance and 24 degrees of freedom (=n-1) and 2 tailed test= 2.7969

2) A statistician wishes to estimate the average total compensation of CEOs in the service industry. Data were randomly collected from 18 CEOs and 97% confidence interval was calculated to be (\$2,181,260, \$5,836,180). Which of the following interpretation is correct?
A) 97% of the sampled total compensation value fell between \$2,181,260 and 5,836,180.(B) We are 97% confident that the mean of sampled CEOs fall in the interval \$2,181,260 to \$5,836,180 (C) In the population of service industry CEOs, 97% of them will have total compensation that fall in the interval \$2,181,260 to \$5,836,180.(D) We are 97% confident that average total compensation of all CEO's in the service industry falls in the interval \$2,181,260 to \$ 5,836,180.

(D) We are 97% confident that average total compensation of all CEO's in the service industry falls in the interval \$2,181,260 to \$ 5,836,180.

3) The personnel of a large corporation wishes to study absenteeism among clerical workers at the corporation's central office during the year. A random sample of 25 clerical workers revealed the following:
Absenteeism: X = 9.7 days and S=4.0 days
12 clerical workers were absent more than 10days

What are the upper and lower limits of the 95% confidence interval estimate of the mean number of absence for clerical workers last year .
A) 8.132 and 11.268 (B) 8.049 and 11.351 (C) 8.052 and 11.348 (D) 8.331 and 11.069

95% Confidence limits

Mean=&#956;= 9.7 days
Standard deviation =&#963;= 4.0
sample size=n= 25
&#963;x=standard error of mean=&#963;/&#8730;n= 0.8 = ( 4 /&#8730; 25)
Confidence level= 95%
Therefore Significance level=&#945;= 5% =100% -95%
No of tails= 2
This is a 2 tailed test because we are calculating the confidence interval

Since sample size= 25 < 30
and we are using sample standard deviation to estimate the population standard deviation
use t ...

#### Solution Summary

The solution provides answers and explanations to 11 Multiple choice questions on Normal Distribution, t distribution, confidence interval, sampling distribution of the mean, standard error of the mean etc.

\$2.19

## Basic Business Statistics: 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

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: &#935; = \$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

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