Question 1

A sample of n = 9 scores is selected from a normal population with a mean of µ = 80 and a standard deviation of σ = 12.The probability that the sample mean will be greater than M = 86 is equal to the probability of obtaining a z-score greater than z = 1.50

Accepted Answer

This statement is true because if we standardize the sample mean using the formula z = (M - µ) / (σ / sqrt(n)), we get z = (86 - 80) / (12 / sqrt(9)) = 1.50. Therefore, the probability of obtaining a sample mean greater than 86 is equal to the probability of obtaining a z-score greater than 1.50, which can be found using a z-table or calculator.

Question 2

A sample is selected from a population with µ = 30 and σ = 10.If the sample mean of M = 34 corresponds to z = +2.00,then the sample size is n = 25.

Accepted Answer

We can use the formula for converting a z-score to a raw score:

z = (X - µ) / (σ / sqrt(n))

Plugging in the given values:

2.00 = (34 - 30) / (10 / sqrt(n))

Solving for n:

n = (10 / 2.00)^2 = 25

Therefore, the sample size is indeed n = 25.

Question 3

If the sample size is equal to the population standard deviation (n = σ),then the standard error for the sample mean is equal to 1.00.

Accepted Answer

The formula for calculating the standard error of the mean is SE = σ/√n. If n = σ, then we have SE = σ/√σ = √σ. Therefore, the standard error would not be equal to 1.00.

Question 4

On average,a sample of n = 16 scores will provide a better estimate of the population mean than you would get with a sample of n = 9 scores from the same population.

Accepted Answer

This is true. As the sample size increases, the standard error of the mean decreases, which means that the sample mean provides a better estimate of the population mean. Therefore, a sample size of 16 will provide a better estimate than a sample size of 9.

Question 5

For a population with µ = 50 and ? = 6,describe the shape,the mean,and the standard deviation for the distribution of sample means for each of the following:
A)Samples of n = 4.
B)Samples of n = 36.

Accepted Answer

The answer of For a population with µ = 50...

Question 6

A population has µ = 60 and σ = 30.For a sample of n = 25 scores from this population,a sample mean of M = 55 would be considered an extreme value.

Accepted Answer

To determine whether a sample mean of M = 55 is considered extreme, we need to compute the z-score for this sample mean using the formula:

z = (M - µ) / (σ / sqrt(n))
z = (55 - 60) / (30 / sqrt(25))
z = -1.67

The z-score tells us how many standard deviations away from the mean the sample mean is. A z-score less than -1.96 or greater than 1.96 would be considered extreme (i.e., falling in the extreme 5% of the distribution). Since the z-score for M = 55 is -1.67, it is not extreme. Thus, the answer is B (False).

Question 7

A sample of n = 16 scores is selected from a population with μ = 70 and σ = 8.It is very unlikely that the sample mean will be greater than 78.

Accepted Answer

The answer of A sample of n = 16 scores...

Question 8

A researcher obtained M = 27 for a sample of n = 36 scores selected from a population with µ = 30 and σ = 18.This sample mean corresponds to a z-score of z = -1.00.

Accepted Answer

This is true as the formula for calculating the z-score is (M-µ)/(σ/√n), which when substituted with the given values becomes (-3)/(18/√36) = -1. Therefore, a z-score of -1.00 corresponds to a sample mean of M = 27.

Question 9

The mean for a sample of n = 9 scores has a standard error of 2 points.This sample was selected from a population with a standard deviation of σ = 18.

Accepted Answer

The standard error of the mean is calculated as the population standard deviation (σ) divided by the square root of the sample size (n), which in this case would be 18 / sqrt(9) = 18 / 3 = 6, not 2.

Question 10

A sample of n = 9 scores is selected from a population with μ = 50 and σ = 12.The probability of obtaining a sample mean greater than 46 is p = 0.8413.

Accepted Answer

This statement is true. The probability of obtaining a sample mean greater than 46 (denoted as X̄) can be found by standardizing X̄ using the formula z = (X̄ - μ) / (σ / √n), where μ = 50, σ = 12, and n = 9. Then, we can use a z-table or a calculator to find that the corresponding probability is 0.8413, which is given in the problem statement. Therefore, the statement is true.

Question 11

As the population standard deviation increases,the standard error will also increase.

Accepted Answer

The standard error is the standard deviation of the sampling distribution of a statistic. As the population standard deviation increases, the variability of the data increases, which in turn increases the standard deviation of the sampling distribution and thus the standard error.

Question 12

A sample of n = 9 scores is selected from a population with µ = 40 and σ = 18.If the sample mean corresponds to z = -1.00,then the sample mean is M = 31.

Accepted Answer

The correct sample mean (M) is calculated using the formula for the z-score: $z = \frac{M - \mu}{\sigma / \sqrt{n}}$. Plugging in the given values ($z = -1$, $\mu = 40$, $\sigma = 18$, $n = 9$), we solve for M, which gives us $M = 34$, not 31.

Question 13

A sample of n = 25 scores is selected from a population with μ = 50 and σ = 10.The probability of obtaining a sample mean greater than 55 is p = 0.3085.

Accepted Answer

The probability of obtaining a sample mean greater than 55 would typically be calculated using the standard error of the mean and the Z-score formula. Given the population standard deviation (σ = 10) and the sample size (n = 25), the standard error (σM) would be σ/√n = 10/√25 = 2. The Z-score for a sample mean of 55 would be (55 - μ) / σM = (55 - 50) / 2 = 2.5. Looking up this Z-score in a standard normal distribution table or using a calculator would likely yield a probability much lower than 0.3085 for the area to the right of this Z-score, indicating the probability of obtaining a sample mean greater than 55 is less than 0.3085.

Question 14

A sample is obtained from a population with σ = 20.If the sample mean has a standard error of 5 points,then the sample size is n = 4.

Accepted Answer

The formula for standard error is SE = σ/√n, therefore n = (σ/SE)² = (20/5)² = 16.

Question 15

A sample of n = 25 scores is selected from a population with μ = 70 and σ = 20.It is very unlikely that the sample mean will be greater than 72.

Accepted Answer

We can use the central limit theorem to determine that the distribution of sample means for samples of size 25 will have a mean of μ = 70 and a standard deviation of σ/√n = 4. Therefore, the probability of obtaining a sample mean greater than 72 can be calculated as follows:

z = (72 - 70) / 4 = 0.5
P(x > 72) = P(z > 0.5) = 0.3085

This probability is not very low, indicating that it is actually quite possible to obtain a sample mean greater than 72 by chance alone. Therefore, the statement is false and the answer is B.

Question 16

If the sample size is increased,then the standard error for sample means will also increase.

Accepted Answer

Increasing the sample size will actually decrease the standard error for sample means. This is because as the sample size increases, the sample mean becomes a more accurate estimate of the population mean, resulting in less variation in sample means.

Question 17

A sample of n = 4 scores is selected from a normal population with μ = 30 and σ = 8.The probability of obtaining a sample mean greater than 34 is equal to the probability of obtaining a z-score greater than z = 2.00.

Accepted Answer

To find the probability of obtaining a sample mean greater than 34, we use the formula for the standard error of the mean (σM = σ / √n) and then calculate the z-score for the sample mean of 34. With σ = 8 and n = 4, the standard error σM = 8 / √4 = 4. The z-score for a sample mean of 34 is (34 - 30) / 4 = 1. A z-score of 1 does not equal a z-score of 2.00, thus the probabilities are not equal.

Question 18

If the sample size is doubled,the standard error will be cut in half.

Accepted Answer

Doubling the sample size decreases the standard error, but it does so by a factor of the square root of 2 (approximately 1.41), not by cutting it in half.

Question 19

On average,a sample of n = 16 scores from a population with σ = 10 will provide a better estimate of the population mean than you would get with a sample of n = 16 scores from a population with σ = 5.

Accepted Answer

A smaller population standard deviation (σ) indicates that the scores in the population are more tightly clustered around the mean, meaning that a sample of a given size is more representative of the population. Therefore, a sample of n = 16 scores from a population with σ = 5 will provide a better estimate of the population mean than a sample of n = 16 scores from a population with σ = 10.

Question 20

A sample of n = 9 scores is randomly selected from a population with µ = 80 and ? = 9.If the sample mean is M = 83,then the corresponding z-score is z = +3.00.

Accepted Answer

The z-score is calculated using the formula $z = \frac{M - \mu}{\sigma / \sqrt{n}}$, where $M$ is the sample mean, $\mu$ is the population mean, $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $z = \frac{83 - 80}{9 / \sqrt{9}} = \frac{3}{3} = 1$. Thus, the correct z-score is $z = +1.00$, not $z = +3.00$.

Quiz 7: Probability and Samples: The Distribution of Sample Means.