Question 1

Sampling from a population may be preferred, because studying all the items in a population may be too ____________.

Accepted Answer

The answer of Sampling from a population may be preferred,...

Question 2

A __________________ assumes that each item or individual in the population has an equal chance of being included.

Accepted Answer

The answer of A __________________ assumes that each item or...

Question 3

The standard deviation of the sampling distribution of the sample means is called the ______________.

Accepted Answer

The answer of The standard deviation of the sampling distribution...

Question 4

The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. Suppose you bought a set of four tires, what is the likelihood the mean tire life of these four tires is between 57,000 and 63,000 miles?
A)0.4332
B)0.8664
C)1.00
D)Very likely

Accepted Answer

We first need to standardize the values using the formula:
z = (x - μ) / (σ / √n)

Where x = 60,000 miles (mean), μ = 60,000 miles (population mean), σ = 4,000 miles (standard deviation), and n = 4 (sample size).

For the lower bound:
z = (57,000 - 60,000) / (4,000 / √4) = -1.5
For the upper bound:
z = (63,000 - 60,000) / (4,000 / √4) = 1.5

Using a standard normal distribution table or calculator, we can find the area between -1.5 and 1.5:
P(-1.5 < z < 1.5) = 0.8664

Therefore, the likelihood that the mean tire life of these four tires is between 57,000 and 63,000 miles is 0.8664 or approximately 87%.

Question 5

Sampling is required when crash testing cars because the products are __________.

Accepted Answer

The answer of Sampling is required when crash testing cars...

Question 6

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find the mean of this sample is between 95 and 105?
A)0.0182
B)0.9818
C)1.00
D)Very unlikely

Accepted Answer

Using the Central Limit Theorem, we can assume that the sampling distribution of the sample means will also be normally distributed with a mean of 100 and a standard deviation of 15/sqrt(50) = 2.12.

Next, we can find the z-score for the lower and upper bounds:

Lower z-score = (95-100)/2.12 = -2.36 
Upper z-score = (105-100)/2.12 = 2.36

Using a standard normal distribution table, we can find the probability of getting a z-score between -2.36 and 2.36, which is 0.9818 or approximately 98.18%.

Therefore, the probability of selecting a sample of 50 adults and finding the mean of this sample is between 95 and 105 is 0.9818 or approximately 98.18%.

Question 7

The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. Suppose we select a sample of 40 tires and use a simulator to determine the tread life. What is the standard error of the mean?
A)632.46
B)40
C)4000
D)Cannot be determineD.

Accepted Answer

The standard error of the mean (SEM) is calculated by dividing the standard deviation by the square root of the sample size. In this case, SEM = 4000 / sqrt(40) = 632.46.

Question 8

The Office of Student Services at a large western state university maintains information on the study habits of its full-time students. Their studies indicate that the mean amount of time undergraduate students study per week is 20 hours. The hours studied follows the normal distribution with a standard deviation of six hours. Suppose we select a random sample of 144 current students. What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours?
A)0.9544
B)0.0160
C)0.0986
D)0.9104

Accepted Answer

We need to use the central limit theorem since we have a sample size of 144 which is larger than 30. First, we standardize the values using the formula:

z = (x̄ - μ) / (σ / √n)

where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Plugging in the values, we get:

z1 = (19.25 - 20) / (6 / √144) = -1.5
z2 = (21 - 20) / (6 / √144) = 1

Next, we look up the corresponding probabilities from the standard normal table:

P(z < -1.5) = 0.0668
P(z < 1) = 0.8413

To find the probability that the mean is between 19.25 and 21, we subtract the two probabilities:

P(-1.5 < z < 1) = P(z < 1) - P(z < -1.5) = 0.8413 - 0.0668 = 0.7745

Therefore, the probability that the mean of the sample is between 19.25 and 21 hours is 0.7745, which is closest to choice D (0.9104).

Question 9

If a population is normally distributed, then the sampling distribution of the sample means will be __________________________.

Accepted Answer

The answer of If a population is normally distributed, then...

Question 10

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find the mean of this sample is less than 95?
A)0.0091
B)0.9818
C)0.4909
D)0.9544

Accepted Answer

We need to use the Central Limit Theorem since we have a sample size of 50 (n > 30) and we know the population standard deviation. Thus, the sampling distribution of the mean will be approximately normal with a mean of 100 and a standard deviation of 15/sqrt(50) = 2.12. 
We need to find the probability of finding a sample mean less than 95, which is equivalent to finding the z-score of the sample mean:

z = (95-100)/2.12 = -2.36

From standard normal distribution table, the probability of finding a z-score less than -2.36 is 0.0091. Therefore, the answer is A.

Question 11

If the size of a sample equals the size of the population, the sampling error is ________.

Accepted Answer

The answer of If the size of a sample equals...

Question 12

The tread life of tires mounted on light duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. Suppose you bought a set of four tires, what is the likelihood the mean tire life of these four tires is more than 66,000 miles?
A)0.0013
B)0.9987
C)0.4987
D)0.9544

Accepted Answer

To find the probability that the mean tire life of these four tires is more than 66,000 miles, we use the formula for the z-score: $z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Here, $\bar{x} = 66,000$, $\mu = 60,000$, $\sigma = 4,000$, and $n = 4$. Plugging these values into the formula gives $z = \frac{66,000 - 60,000}{4,000/\sqrt{4}} = \frac{6,000}{2,000} = 3$. Looking up a z-score of 3 in the standard normal distribution table or using a calculator gives a probability of 0.0013 that the mean tire life of these four tires is more than 66,000 miles.

Question 13

In the sampling distribution of the sample means, the standard error of the mean will vary according to the size of the sample. As the sample size n gets larger, the variability of the sampling distribution of the means gets _____________.

Accepted Answer

The answer of In the sampling distribution of the sample...

Question 14

To collect a sample, a population is first divided into subgroups and then a sample is selected from each subgroup. This is called _________ sampling.

Accepted Answer

The answer of To collect a sample, a population is...

Question 15

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find that the mean of this sample is between 98 and 103?
A)0.3264
B)0.9428
C)0.4702
D)0.7471

Accepted Answer

The answer of The Intelligence Quotient (IQ) test scores for...

Question 16

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find that the mean of this sample exceeds 104?
A)0.0294
B)0.9412
C)Approximately one
D)0.9706

Accepted Answer

To find the probability that the mean of a sample exceeds a certain value, we use the formula for the standard error of the mean, which is the standard deviation divided by the square root of the sample size (n). Here, the standard deviation (σ) is 15, and the sample size (n) is 50. The standard error (SE) = σ / sqrt(n) = 15 / sqrt(50). We then find the Z-score for the sample mean of 104, which is (104 - 100) / SE. Using the Z-score, we can find the probability from the standard normal distribution table. The calculation shows that the probability the sample mean exceeds 104 is approximately 0.0294, indicating a relatively small chance.

Question 17

A total of _______ different samples of size 3 can be collected from a population of 10 items.

Accepted Answer

The answer of A total of _______ different samples of...

Question 18

Auditors may select every 20^th file starting with the 5^th file in the top drawer. Then, file numbers 25, 45, 65, 85, . . . are selected. This type of sampling is called ___________.

Accepted Answer

The answer of Auditors may select every 20^th file starting...

Question 19

The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. Suppose we select a sample of 40 tires and use a simulator to determine the tread life. What is the likelihood of finding that the sample mean is between 59,050 and 60,950?
A)0.4332
B)0.8664
C)1.00
D)Very likely

Accepted Answer

We can use the central limit theorem to approximate the distribution of sample means as normal with mean 60,000 and standard deviation 4,000/sqrt(40) = 632.45. Then, we standardize the sample mean as follows:

z = (x̄ - μ) / (σ/√n) = (59050 - 60000) / (4000/√40) = -2.5
z = (x̄ - μ) / (σ/√n) = (60950 - 60000) / (4000/√40) = 2.5

Using a standard normal table, we find the probability of z being between -2.5 and 2.5 is approximately 0.8664. Therefore, the likelihood of finding that the sample mean is between 59,050 and 60,950 is 0.8664 or 86.64%. The best choice is (B).

Question 20

Sampling from a population may be preferred because, contacting or listing the entire population would be too ___________ consuming.

Accepted Answer

The answer of Sampling from a population may be preferred...

Quiz 8: Sampling Methods and the Central Limit Theorem

Sampling from a population may be preferred, because studying all the items in a population may be too ____________.

A __________________ assumes that each item or individual in the population has an equal chance of being included.

The standard deviation of the sampling distribution of the sample means is called the ______________.

Sampling is required when crash testing cars because the products are __________.

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find the mean of this sample is between 95 and 105?

The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. Suppose we select a sample of 40 tires and use a simulator to determine the tread life. What is the standard error of the mean?

If a population is normally distributed, then the sampling distribution of the sample means will be __________________________.

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find the mean of this sample is less than 95?

If the size of a sample equals the size of the population, the sampling error is ________.

The tread life of tires mounted on light duty trucks follows the normal probability distribution with a mean of 60,000 miles and a standard deviation of 4,000 miles. Suppose you bought a set of four tires, what is the likelihood the mean tire life of these four tires is more than 66,000 miles?

In the sampling distribution of the sample means, the standard error of the mean will vary according to the size of the sample. As the sample size n gets larger, the variability of the sampling distribution of the means gets _____________.

To collect a sample, a population is first divided into subgroups and then a sample is selected from each subgroup. This is called _________ sampling.

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find that the mean of this sample is between 98 and 103?

The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find that the mean of this sample exceeds 104?

A total of _______ different samples of size 3 can be collected from a population of 10 items.

Auditors may select every 20th file starting with the 5th file in the top drawer. Then, file numbers 25, 45, 65, 85, . . . are selected. This type of sampling is called ___________.

Sampling from a population may be preferred because, contacting or listing the entire population would be too ___________ consuming.

Auditors may select every 20^th file starting with the 5^th file in the top drawer. Then, file numbers 25, 45, 65, 85, . . . are selected. This type of sampling is called ___________.