Question 1

As the sample size increases,the variability of the sampling distribution of the mean ______________.
A)Increases
B)Stays the same
C)Is variable
D)Decreases

Accepted Answer

As the sample size increases, the variability of the sampling distribution of the mean decreases. This is due to the fact that as the sample size increases, the sample mean becomes a better estimate of the population mean, and the standard error of the mean decreases. This results in a narrower sampling distribution, which has less variability.

Question 2

A random sample of size 1,000 is taken from a population where p = .20.Describe the sampling distribution of  .
A)Cannot be determined
B)Approximately normal
C)Skewed to the left
D)Skewed to the right

Accepted Answer

The sampling distribution of a proportion will be approximately normal with a mean of p and a standard deviation of sqrt((p(1-p))/n). In this case, the mean is 0.20 and the standard deviation is sqrt((0.20(1-0.20))/1000) = 0.01414. Since the sample size is large enough (n>=30), the sampling distribution can be assumed to be approximately normal.

Question 3

Find P( < 402),if μ = 400,σ_x = 200,and n = 100. A).8413 B).4602 C).5398 D).1587

Accepted Answer

We need to standardize the random variable X as follows:
Z = (X - μ) / (σ / √n)
Where μ = 400, σ = 200, and n = 100.
Substituting the values, we get:
Z = (402 - 400) / (200 / √100) = 0.1
Using a standard normal table, we can find the probability as follows:
P(Z < 0.1) = 0.5398
Therefore, the answer is C.

Question 4

Find σ_x,if μ = 400,P( < 396)= .0228,and n = 100. A)0.20 B)200 C)20 D)2.00

Accepted Answer

The answer of Find σ_x,if μ = 400,P( <...

Question 5

A PGA (Professional Golf Association)tournament organizer is attempting to determine whether hole (pin)placement has a significant impact on the average number of strokes for the 13^th hole on a given golf course.Historically,the pin has been placed in the front right corner of the green,and the historical mean number of strokes for the hole has been 4.25,with a standard deviation of 1.6 strokes.On a particular day during the most recent golf tournament,the organizer placed the hole (pin)in the back left corner of the green.64 golfers played the hole with the new placement on that day.Determine the probability of the sample average number of strokes exceeding 4.75. A).9938 B).4013 C).0062 D).3783

Accepted Answer

We can use the central limit theorem (CLT) to approximate the sampling distribution of the sample mean. Since the sample size is large (n = 64), we can assume that the sampling distribution is approximately normal with mean μ = 4.25 and standard deviation σ/sqrt(n) = 1.6/sqrt(64) = 0.2.

To find the probability of the sample average number of strokes exceeding 4.75, we need to standardize the sample mean:

z = (x̄ - μ) / (σ/sqrt(n)) = (4.75 - 4.25) / 0.2 = 2.5

where x̄ is the sample mean.

We want to find P(z > 2.5), which corresponds to the area to the right of 2.5 under the standard normal distribution. Using a standard normal table or calculator, we find:

P(z > 2.5) = 0.0062

Therefore, the probability of the sample average number of strokes exceeding 4.75 is 0.0062. The best choice is (C).

Question 6

Find P(  < 25)if μ = 16 and  = 4.
A)1.000
B)2.25
C).9878
D).0122

Accepted Answer

Using the formula for z-score:

z = (x - μ) / σ

We can find the z-score:

z = (25 - 16) / 4 = 2.25

Using a standard normal distribution table, we can find the probability:

P(11eb06d7_55f5_0b76_8177_991b59f61c7c_TB1736_11 < 25) = P(z < 2.25) = 0.9878

Therefore, the answer is (C) .9878.

Question 7

Find P(395.4 < < 404.6),if the population mean = 400,σ_x = 20,and n = 100. A).9786 B).9999 C).0214 D).9893

Accepted Answer

The answer of Find P(395.4 <  < 404.6),if the...

Question 8

An unbiased estimate of σ² is _____. A)s B)s² C) D)σ

Accepted Answer

An unbiased estimate for the population parameter sigma squared is the sample variance, which is computed by dividing the sum of squares by the degrees of freedom. In this case, we have no information on the sample size or the number of variables involved, so we cannot calculate the degrees of freedom. As such, the only option among the given choices is to keep the expression as is and not substitute any value for sigma squared. Therefore, Choice B is correct.

Question 9

A random sample of size 36 is taken from a population with a mean of 50 and a standard deviation of 5.The sampling distribution of  ____________________.
A)Cannot be determined
B)Is skewed to the left
C)Is approximately normal
D)Is skewed to the right

Accepted Answer

According to the central limit theorem, the sampling distribution of the sample mean will be approximately normal if the sample size is greater than or equal to 30. In this case, the sample size is 36, which is greater than 30. Therefore, the sampling distribution of the sample mean will be approximately normal.

Question 10

The ___________ is a minimum-variance unbiased point estimate of the mean of a normally distributed population.
A)Sample mean
B)Sample variance
C)Sample standard deviation
D)Observed mean

Accepted Answer

The sample mean is a minimum-variance unbiased point estimate of the mean of a normally distributed population. This means that it has the smallest possible variance among all unbiased estimates, and it is not skewed in any particular direction. The other options listed do not necessarily have these properties.

Question 11

As the sample size ___________,the standard deviation of the population of all sample proportions increases.
A)Increases
B)Stays the same
C)Is variable
D)Decreases

Accepted Answer

The answer of As the sample size ___________,the standard deviation...

Question 12

The sampling distribution of the sample mean is a normal distribution for ________ sample sizes,regardless of the shape of the corresponding population distribution.
A)Random
B)Large
C)Larger than 50
D)Small

Accepted Answer

The Central Limit Theorem states that the sampling distribution of the sample mean approaches a normal distribution with a mean equals to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size. This approximation holds for large sample sizes, typically greater than or equal to 30. Therefore, option B, "Large" is the best choice. The other options, "Random", "Larger than 50" and "Small" are not accurate or precise enough to describe the conditions under which the Central Limit Theorem holds.

Question 13

Packages of sugar bags for Sweeter Sugar Inc.have an average weight of 16 ounces and a standard deviation of 0.2 ounces.The weights of the sugar packages are normally distributed.What is the probability that 16 randomly selected packages will have a weight in excess of 16.075 ounces?
A).0500
B).3520
C).9332
D).0668

Accepted Answer

Using the formula for the sampling distribution of sample means:
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.
Substituting the given values:
z = (16.075 - 16)/(0.2/√16) = 1.5
Using a standard normal distribution table or calculator, we find that the probability of getting a z-score of 1.5 or higher is 0.0668. Therefore, the probability that 16 randomly selected packages will have a weight in excess of 16.075 ounces is 0.0668 or approximately 6.68%. So, the best choice is D.

Question 14

The ______________ of a sample statistic is the probability distribution of the population of all possible values of the sample statistic.
A)Sample mean
B)Sampling Distribution of Sample Mean
C)Probability
D)Observations

Accepted Answer

The sampling distribution of sample mean is the probability distribution of the population of all possible values of the sample mean.

Question 15

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.3 inches.What is the probability that the average length of a steel sheet from a sample of 9 units is more than 29.95 inches long?
A).8413
B).6293
C).3707
D).1587

Accepted Answer

The answer of The population of lengths of aluminum-coated steel...

Question 16

Find P(  > 2,510)if μ = 2,500 and  = 7.
A).0764
B).9998
C).9236
D).0001

Accepted Answer

The answer of Find P(  > 2,510)if μ =...

Question 17

The population of all _________________ proportions is described by the sampling distribution of  .
A)Population
B)Random
C)Observed
D)Sample

Accepted Answer

The sampling distribution of a statistic, such as the mean or proportion, is based on repeated sampling from a population. Therefore, the population being referred to in this question is the one from which samples are drawn to create the sampling distribution. Choice D, "sample", refers to the group of individuals selected from the population to create the distribution. The other choices ("population", "random", and "observed") do not accurately describe what is being referred to in this context.

Question 18

Find P(  > 172),if μ = 175 and  = 9.
A).1587
B).8413
C).6293
D).3707

Accepted Answer

The answer of Find P(  > 172),if μ =...

Question 19

A theorem that allows us to use the normal probability distribution to approximate the sampling distribution of sample means and sample proportions whenever the sample size is large is known as _______________.
A)Cluster sampling
B)Sampling error
C)Sampling distribution of the mean
D)The Central Limit Theorem

Accepted Answer

The Central Limit Theorem states that as the sample size increases, the sampling distribution of sample means and sample proportions will become approximately normally distributed, regardless of the shape of the population distribution. This allows us to use the normal probability distribution to approximate the sampling distribution in large samples. Therefore, the correct answer is D.

Question 20

Find P( < 35)if μ = 40,σ_x = 16,n = 16. A).9944 B).4483 C).5517 D).1056

Accepted Answer

The answer of Find P(  < 35)if μ =...

Quiz 7: Sampling and Sampling Distributions

As the sample size increases,the variability of the sampling distribution of the mean ______________.

A random sample of size 1,000 is taken from a population where p = .20.Describe the sampling distribution of .

Find P( < 402) ,if μ = 400,σ_x = 200,and n = 100.

Find σ_x,if μ = 400,P( < 396) = .0228,and n = 100.

Find P( < 25) if μ = 16 and = 4.

Find P(395.4 < < 404.6) ,if the population mean = 400,σ_x = 20,and n = 100.

An unbiased estimate of σ² is _____.

A random sample of size 36 is taken from a population with a mean of 50 and a standard deviation of 5.The sampling distribution of ____________________.

The ___________ is a minimum-variance unbiased point estimate of the mean of a normally distributed population.

As the sample size ___________,the standard deviation of the population of all sample proportions increases.

The sampling distribution of the sample mean is a normal distribution for ________ sample sizes,regardless of the shape of the corresponding population distribution.

Packages of sugar bags for Sweeter Sugar Inc.have an average weight of 16 ounces and a standard deviation of 0.2 ounces.The weights of the sugar packages are normally distributed.What is the probability that 16 randomly selected packages will have a weight in excess of 16.075 ounces?

The ______________ of a sample statistic is the probability distribution of the population of all possible values of the sample statistic.

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.3 inches.What is the probability that the average length of a steel sheet from a sample of 9 units is more than 29.95 inches long?

Find P( > 2,510) if μ = 2,500 and = 7.

The population of all _________________ proportions is described by the sampling distribution of .

Find P( > 172) ,if μ = 175 and = 9.

A theorem that allows us to use the normal probability distribution to approximate the sampling distribution of sample means and sample proportions whenever the sample size is large is known as _______________.

Find P( < 35) if μ = 40,σ_x = 16,n = 16.