Statistics for Management And Economics Abbreviated 10th Edition By Gerald Keller – Test Bank

Statistics for Management And Economics Abbreviated 10th Edition By Gerald Keller – Test Bank

CHAPTER 9: SAMPLING DISTRIBUTIONS

TRUE/FALSE

1. The Central Limit Theorem permits us to draw conclusions about a population based on a sample alone, without having any knowledge about the distribution of that population. And this works no matter what the sample size is.

ANS: F PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

2. When a great many simple random samples of size n are drawn from a population that is normally distributed, the sampling distribution of the sample mean is normal regardless of the sample size n.

ANS: T PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

3. Consider an infinite population with a mean of 100 and a standard deviation of 20. A random sample of size 64 is taken from this population. The standard deviation of the sample mean equals 2.50.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

4. If all possible samples of size n are drawn from an infinite population with standard deviation 8, then the standard error of the sample mean equals 1.0 if the sample size is 64.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

5. A sample of size n is selected at random from an infinite population. As n increases, the standard error of the sample mean increases.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

6. A sample of size 25 is selected from a population of size 500. The finite population correction is needed to find the standard error of .

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

7. A sample of size 25 is selected from a population of size 75. The finite population correction needed to find the standard error of .

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

8. The amount of time it takes to complete a final examination is negatively skewed distribution with a mean of 70 minutes and a standard deviation of 8 minutes. If 64 students were randomly sampled, the probability that the sample mean of the sampled students exceeds 73.5 minutes is approximately 0.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

9. If all possible samples of size n are drawn from a normal population, the probability distribution of the sample mean is an normal distribution.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

10. If the sample size increases, the standard error of the mean also increases.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

11. The amount of bleach a machine pours into bottles has a mean of 50 ounces with a standard deviation of 0.25 ounces. We take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean has a mean of 50 ounces.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

12. If the population distribution is skewed, in most cases the sampling distribution of the sample mean can be approximated by the normal distribution if the samples contain at least 30 observations.

ANS: T PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

13. A sampling distribution is a probability distribution for a statistic, not a parameter.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

14. A sampling distribution is defined as the probability distribution of means from all possible sample sizes that are taken from a given population.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

15. If the population distribution is unknown, in most cases the sampling distribution of the mean can be approximated by the normal distribution if the samples contain at least 30 observations.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

16. The amount of bleach a machine pours into bottles has a mean of 50 ounces with a standard deviation of 0.25 ounces. Suppose we take a random sample of 36 bottles filled by this machine. The sampling distribution of the sample mean has a standard error of 0.25 ounces.

ANS: F PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

17. As the size of the sample is increased, the standard error of decreases.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

18. As the size of the sample is increased, the mean of increases.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

19. In inferential statistics, the standard error of the sample mean assesses the uncertainty or error of estimation.

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NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

MULTIPLE CHOICE

1. The standard deviation of the sampling distribution of is also called the:

a.

central limit theorem.

c.

finite population correction factor.

b.

population standard deviation.

d.

standard error of the sample mean.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

2. Random samples of size 49 are taken from an infinite population whose mean is 300 and standard deviation is 21. The mean and standard error of the sample mean, respectively, are:

a.

300 and 21

c.

300 and 0.43

b.

300 and 3

d.

None of these choices.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

3. Given an infinite population with a mean of 75 and a standard deviation of 12, the probability that the mean of a sample of 36 observations, taken at random from this population, is less than 78 is:

a.

0.9332

c.

1.5000

b.

0.5987

d.

None of these choices.

ANS: A PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

4. An infinite population has a mean of 40 and a standard deviation of 15. A sample of size 100 is taken at random from this population. The standard error of the sample mean equals:

a.

15

c.

15/100

b.

15/Ö100

d.

None of these choices.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

5. If all possible samples of size n are drawn from an infinite population with a mean of 15 and a standard deviation of 5, then the standard error of the sample mean equals 1.0 for samples of size:

a.

5

c.

25

b.

15

d.

None of these choices.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

6. As a general rule in computing the standard error of the sample mean, the finite population correction factor is used only if the:

a.

sample size is smaller than 5% of the population size.

b.

sample size is greater than 5% of the sample size.

c.

sample size is more than half of the population size.

d.

None of these choices.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

7. Consider an infinite population with a mean of 160 and a standard deviation of 25. A random sample of size 64 is taken from this population. The standard deviation of the sample mean equals:

a.

25

c.

25/Ö64

b.

25/64

d.

None of these choices.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

8. The Central Limit Theorem states that, if a random sample of size n is drawn from a population, then the sampling distribution of the sample mean :

a.

is approximately normal if n < 30.

b.

is approximately normal if n > 30.

c.

is approximately normal if the underlying population is normal.

d.

None of these choices.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

9. A sample of size 40 is taken from an infinite population whose mean and standard deviation are 68 and 12, respectively. The probability that the sample mean is larger than 70 equals

a.

P(Z > 70)

c.

P(Z > 0.17)

b.

P(Z > 2)

d.

P(Z > 1.05)

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

10. The finite population correction factor should be used:

a.

whenever we are sampling from an infinite population.

b.

whenever we are sampling from a finite population.

c.

whenever the sample size is large compared to the population size.

d.

whenever the sample size is small compared to the population size.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

11. Random samples of size 81 are taken from an infinite population whose mean and standard deviation are 45 and 9, respectively. The mean and standard error of the sampling distribution of the sample mean are:

a.

45 and 9

c.

45 and 9/Ö81

b.

45/81 and 9/81

d.

45/Ö81 and 9/Ö81

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

12. A sample of size 25 is selected at random from a finite population. If the finite population correction factor is 0.63, then the population size is:

a.

25

c.

41

b.

66

d.

None of these choices.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

13. If all possible samples of size n are drawn from an infinite population with a mean of m and a standard deviation of s, then the standard error of the sample mean is inversely proportional to:

a.

m

c.

n

b.

s

d.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

14. A sample of size n is selected at random from an infinite population. As n increases, which of the following statements is true?

a.

The population standard deviation decreases.

b.

The standard error of the sample mean decreases.

c.

The population standard deviation increases.

d.

The standard error of the sample mean increases.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

15. The expected value of the sampling distribution of the sample mean equals the population mean m :

a.

only when the population is normally distributed.

b.

only when the sample size is large.

c.

only when the population is infinite.

d.

for all populations.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

16. If a random sample of size n is drawn from a normal population, then the sampling distribution of the sample mean will be:

a.

normal for all values of n.

c.

approximately normal for all values of n.

b.

normal only for n > 30.

d.

approximately normal only for n > 30.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

17. If all possible samples of size n are drawn from a population, the probability distribution of the sample mean is called the:

a.

standard error of .

c.

sampling distribution of .

b.

expected value of .

d.

normal distribution.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

18. Sampling distributions describe the distributions of:

a.

sample statistics.

c.

both parameters and statistics

b.

population parameters.

d.

None of these choices.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

19. Suppose X has a distribution that is not normal. The Central Limit Theorem is important in this case because:

a.

it says the sampling distribution of is approximately normal for any sample size.

b.

it says the sampling distribution of is approximately normal if n is large enough.

c.

it says the sampling distribution of is exactly normal, for any sample size.

d.

None of these choices.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

20. Which of the following statements about the sampling distribution of is NOT true?

a.

It is generated by taking all possible samples of size n and computing their sample means.

b.

Its mean is equal to the population mean m.

c.

Its standard deviation is equal to the population standard deviation s.

d.

All of these choices are true.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

21. The standard error of the mean:

a.

is never larger than the standard deviation of the population.

b.

decreases as the sample size increases.

c.

measures the variability of the mean from sample to sample.

d.

All of these choices are true.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

22. Which of the following is true about the sampling distribution of the sample mean?

a.

Its mean is always equal to

b.

Its standard error is always equal to the population standard deviation s.

c.

Its shape is exactly normal if n is large enough.

d.

None of these choices.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

23. The owner of a meat market has an assistant who has determined that the weights of roasts are normally distributed, with a mean of 3.2 pounds and standard deviation of 0.8 pounds. If a sample of 25 roasts yields a mean of 3.6 pounds, what is the Z-score for this sample mean?

a.

-2.50

c.

-0.50

b.

2.50

d.

None of these choices.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

24. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with a mean of 3.2 pounds and standard deviation of 0.84 pounds. If a sample of 16 fish is taken, what is the standard error of the mean weight?

a.

0.840

c.

0.210

b.

0.053

d.

None of these choices.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

25. Suppose the ages of students in your program follow a positively skewed distribution with mean of 24 years and a standard deviation of 4 years. If we randomly sampled 100 students, which of the following statements about the sampling distribution of the sample mean age is NOT true?

a.

The mean of the sampling distribution of sample mean is equal to 24 years.

b.

The standard deviation of the sampling distribution of sample mean is equal to 4 years.

c.

The shape of the sampling distribution of sample mean is approximately normal.

d.

All of these choices are true.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

26. Suppose that 100 items are drawn from a population of manufactured products and the weight, X, of each item is recorded. Prior experience has shown that the weight has a non-normal probability distribution with m = 8 ounces and s = 3 ounces. Which of the following is true about the sampling of ?

a.

Its mean is 8 ounces.

c.

Its shape is approximately normal.

b.

Its standard error is 0.3 ounces.

d.

All of these choices are true.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

27. The standard error of the mean for a sample of 100 is 25. In order to cut the standard error of the mean in half (to 12.5) we must:

a.

increase the sample size to 200.

b.

decrease the sample size to 50.

c.

keep the sample size at 100 and change something else.

d.

None of these choices.

ANS: D PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

28. Which of the following is true regarding the sampling distribution of the mean for a large sample size? Assume the population distribution is not normal.

a.

It has the same shape, mean and standard deviation as the population.

b.

It has the same mean as the population, but a different shape and standard deviation.

c.

It the same mean and standard deviation as the population, but a different shape.

d.

It has the same shape and mean as the population, but a different standard deviation.

ANS: B PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

29. For sample sizes greater than 30, the sampling distribution of the mean is approximately normally distributed:

a.

regardless of the shape of the population.

b.

only if the shape of the population is symmetric.

c.

only if the population is normally distributed.

d.

None of these choices.

ANS: A PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

30. For a sample size of 1, the sampling distribution of the mean is normally distributed:

a.

regardless of the shape of the population.

b.

only if the population values are larger than 30.

c.

only if the population is normally distributed.

d.

None of these choices.

ANS: C PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

COMPLETION

1. Because the value of the ____________________ varies randomly from sample to sample, we can regard it as a new random variable created by sampling.

ANS: sample mean

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

2. The variance of the sampling distribution of is ____________________ the variance of the population we’re sampling from for all sample sizes.

ANS: less than

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

3. A randomly selected value of is likely to be ____________________ to(than) the mean of the population than is a randomly selected value of X.

ANS: closer

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

4. As the number of throws of a fair die increases, the probability that the sample mean is close to (the number) ____________________ increases.

ANS: 3.5

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

5. The width of the sampling distribution of gets ____________________ as the sample size increases.

ANS:

narrower

more narrow

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

6. As n gets ____________________, the shape of the sampling distribution of becomes increasingly bell shaped.

ANS: larger

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

7. As n gets larger, the sampling distribution of becomes increasingly bell shaped. This phenomenon is due to the ____________________.

ANS: Central Limit Theorem

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

8. The accuracy of the approximation of with a normal distribution depends on the probability distribution of the ____________________ and on the sample ____________________.

ANS: population; size

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

9. If the population is normal, then is normally distributed for __________________ values of n.

ANS: all

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

10. The finite population correction is not needed if the population size is large relative to the sample size. As a rule of thumb, we will treat any population that is at least ____________________ times larger than the sample size as large.

ANS:

20

twenty

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Knowledge

SHORT ANSWER

Heights of Fences

Heights of fences are normally distributed with a mean of 52 inches and a standard deviation of 4 inches.

1. {Heights of Fences Narrative} Find the probability that one randomly selected fence is under 54 inches.

ANS:

0.6915

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

2. {Heights of Fence Narrative} Find the probability that two randomly selected fences are both under 54 inches.

ANS:

(0.6915)(0.6915) = 0.4782

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

3. {Heights of Fence Narrative} Find the probability that the mean height of 4 randomly selected fences is under 54 inches.

ANS:

0.8413

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

Average Annual Revenue

Suppose that the average annual revenue of a small business is $150,000 with a standard deviation of $40,000. Assume that the revenue distribution is normal.

4. {Average Annual Revenue Narrative} What is the probability that one business selected at random makes less than $120,000?

ANS:

0.2266

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

5. {Average Annual Revenue Narrative} What is the probability that the average annual revenue of a random sample of 4 businesses is less than $120,000?

ANS:

0.0668

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

6. {Average Annual Revenue Narrative} Why are your answers to the previous two questions different?

ANS:

The revenue of one business chosen from the population has more variability than the average revenue of 4 businesses. That means that a revenue lower than 120,000 is harder to achieve with an average of 4 businesses, compared to one business.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

7. {Average Annual Revenue Narrative} Could you have used a normal distribution to find an (approximate) probability for the average of 4 revenues if the population of revenues did not have a normal distribution?

ANS:

No, because the sample size of 4 is too small for the Central Limit Theorem to be used.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

Mean Salary

In order to estimate the mean salary for a population of 500 employees, the president of a certain company selected at random a sample of 40 employees.

8. {Mean Salary Narrative} Would you use the finite population correction factor in calculating the standard error of the sample mean in this case? Explain.

ANS:

Since the population size is 12.5 times as large as the sample size (500/40 = 12.5), the finite population correction factor is necessary. (It doesn’t have to be used if the population is at least 20 times as large as the sample.)

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

9. {Mean Salary Narrative} If the population standard deviation is $800, compute the standard error both with and without using the finite population correction factor.

ANS:

= $121.448 and $126.491 with and without the finite population correction factor, respectively. The finite population correction factor does make a difference.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

Retirees

A sample of 50 retirees is drawn at random from a normal population whose mean age and standard deviation are 75 and 6 years, respectively.

10. (Retirees Narrative} Describe the shape of the sampling distribution of the sample mean in this case.

ANS:

is normal because the original population is normal. This is true for any sample size.

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Comprehension

11. (Retirees Narrative} Find the mean and standard error of the sampling distribution of the sample mean.

ANS:

= 75 and = .8485

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

12. {Retirees Narrative} What is the probability that the mean age exceeds 73 years?

ANS:

P( > 73) = 0.9909

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

13. {Retirees Narrative} What is the probability that the mean age is at most 73 years?

ANS:

P( £ 73) = 0.0091

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

14. {Retirees Narrative} What is the probability that two randomly selected retirees are over 73 years of age?

ANS:

0.9909 ´ 0.9909 = 0.9819

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

Heights of Men

The heights of men in the USA are normally distributed with a mean of 68 inches and a standard deviation of 4 inches.

15. {Heights of Men Narrative} What is the probability that a randomly selected man is taller than 70 inches?

ANS:

0.3085

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

16. {Heights of Men Narrative} A random sample of five men is selected. What is the probability that the sample mean is greater than 70 inches?

ANS:

0.1314

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

17. {Heights of Men Narrative} What is the probability that the mean height of a random sample of 36 men is greater than 70 inches?

ANS:

0.0013

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

18. {Heights of Men Narrative} If the population of men’s heights is not normally distributed, which, if any, of the previous questions can you answer?

ANS:

If heights were not normal, we could only answer the question about the mean height of 36 men.

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

Poker Time

The amount of time spent by American adults playing Poker per week is normally distributed with a mean of 4 hours and standard deviation of 1.25 hours.

19. {Poker Time Narrative} Find the probability that a randomly selected American adult plays Poker for more than 5 hours per week.

ANS:

0.2119

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

20. {Poker Time Narrative} Find the probability that if four American adults are randomly selected, their average number of hours spent playing Poker is more than 5 hours per week.

ANS:

0.0548

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

21. {Poker Time Narrative} Find the probability that if four American adults are randomly selected, all four play Poker for more than 5 hours per week.

ANS:

(0.2119)⁴ = 0.0020

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

Number of Fish Tanks

The following data give the number of fish tanks owned for a population of 4 families.

Family

A

B

C

D

Number of Fish Tanks Owned

2

1

4

3

22. {Number of Fish Tanks Narrative} Find the mean and the standard deviation for the population.

ANS:

m = 2.5 fish tanks and s = 1.12 fish tanks

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

23. {Number of Fish Tanks Narrative} A sample of size 2 is drawn at random from the population. Use the formulas and to calculate the mean and the standard deviation of the sampling distribution of the sample means.

ANS:

and

PTS: 1 DIF: Moderate OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application

24. {Number of Fish Tanks Narrative} List all possible samples of 2 families that can be selected without replacement from this population, and compute the sample mean for each sample.

ANS:

Sample

A, B

A, C

A, D

B, C

B, D

C, D

1.5

3.0

2.5

2.5

2.0

3.5

PTS: 1 DIF: Easy OBJ: SFME.KELL.15.09.01

NAT: BUSPROG.SFME.KELL.15.03 STA: DISC.SFME.KELL.15.04

KEY: Bloom’s: Application