Question 1

A method of selecting a sample by randomly selecting the first element and then selecting every K^th element thereafter is A)area sampling B)stratified sampling C)systematic sampling D)stratified simple random sampling

Accepted Answer

This is systematic sampling, where every K^th element is selected after randomly selecting the first element.

Question 2

Exhibit 22-2
A simple random sample of 43 elements has been selected from a population of size 800. The sample mean is 500, and the sample standard deviation is 60.
-Refer to Exhibit 22-2. An approximate 95% confidence interval for the population total is
A)482 to 517
B)385,759 to 414,241
C)400,000 to 500,000
D)350,000 to 450,000

Accepted Answer

To calculate the confidence interval for population total, we use the formula:
CI = sample mean ± Z* (sample standard deviation /sqrt(sample size)) * sqrt(population size)
Here, sample mean (x-bar) = 500, sample standard deviation (s) = 60, sample size (n) = 43, population size (N) = 800, and Z-value for 95% confidence level is 1.96 (from standard normal distribution table).
Plugging these values in the formula, we get:
CI = 500 ± 1.96 * (60/sqrt(43)) * sqrt(800)
CI = 500 ± 9,022
CI = (500 - 9,022) to (500 + 9,022)
CI = 385,978 to 414,022
Therefore, the 95% confidence interval for the population total is 385,978 to 414,022. Hence, the best choice is B.

Question 3

Exhibit 22-2
A simple random sample of 43 elements has been selected from a population of size 800. The sample mean is 500, and the sample standard deviation is 60.
-Refer to Exhibit 22-2. An approximate 95% confidence interval for the population mean is
A)482.2 to 517.8
B)440.0 to 560.0
C)500.0 to 560.0
D)440.0 to 500.0

Accepted Answer

We will use the formula: 
Confidence Interval = sample mean ± (critical value) * (standard error)
where critical value = 1.96 (for approximate 95% confidence interval)
standard error = sample standard deviation /√(sample size)
Plugging in the values, we get:
Confidence Interval = 500 ± (1.96) * (60/√43)
Confidence Interval = 500 ± 17.8
Confidence Interval = (500 - 17.8) to (500 + 17.8)
Confidence Interval = 482.2 to 517.8
Therefore, answer is (A) 482.2 to 517.8.

Question 4

Which of the following is(are) the most common type(s) of surveys?
A)only mail surveys
B)only telephone surveys
C)only personal interviews
D)All of these answers are correct.

Accepted Answer

All three types of surveys (mail, telephone, and personal interviews) are common and widely used in survey research. The choice of survey method depends on factors such as sample size, target population, budget, and research objectives. Therefore, it is best to use a combination of survey methods rather than relying on only one type.

Question 5

A version of cluster sampling in which the elements are formed into clusters on the basis of their geographic proximity is
A)stratified simple random sampling
B)random sampling
C)judgment sampling
D)area sampling

Accepted Answer

Area sampling is the best choice as it involves grouping the elements based on their physical proximity, such as geographic location, and then randomly selecting clusters. This method accounts for the inherent variability within different regions or areas, making it easier to obtain a representative sample.

Question 6

Sampling errors can
A)be avoided by increasing the sample size to at least 30
B)be avoided if the sample is increased so that it will be at least 5% of the population
C)be avoided by using probabilistic sampling
D)not be avoided

Accepted Answer

Sampling errors cannot be completely avoided because they are inherent in the process of selecting a sample from a population. Even with probabilistic sampling methods and increasing sample sizes, there is always a chance of error in estimating population parameters.

Question 7

Exhibit 22-4
Simple random sampling has been used to obtain a sample of size 50 from a population of size 500. The sample proportion was 0.7.
-Refer to Exhibit 22-4. The estimate of the standard error of proportion is
A)0.210
B)0.700
C)0.300
D)0.062

Accepted Answer

The standard error of proportion can be estimated using the formula √(p(1-p)/n), where p is the sample proportion and n is the sample size. Plugging in the values given in the question, we get √(0.7(1-0.7)/50) = 0.062. Therefore, the answer is D.

Question 8

Exhibit 22-1
A simple random sample of size 64 is taken from a population of size 800. The sample mean is determined to be 2,550 with a standard deviation of 500.
-Refer to Exhibit 22-1. The point estimate of the population total is
A)51,200
B)1,275,000
C)102,000
D)2,040,000

Accepted Answer

The answer of Exhibit 22-1
A simple random sample of size...

Question 9

Exhibit 22-3
From a population of size 600, a simple random sample of size 58 is selected. The sample mean is 400, and the sample standard deviation is 40.
-Refer to Exhibit 22-3. An approximate 95% confidence interval for the population total is
A)350,000 to 450,000
B)234,010 to 245,990
C)350 to 400
D)390 to 409

Accepted Answer

The 95% confidence interval for the population total is calculated using the sample mean and standard deviation, considering the finite population correction factor. The interval B) 234,010 to 245,990 is the correct range for the population total based on the given sample statistics.

Question 10

Exhibit 22-2
A simple random sample of 43 elements has been selected from a population of size 800. The sample mean is 500, and the sample standard deviation is 60.
-Refer to Exhibit 22-2. The standard error of the mean is
A)9.2
B)43.0
C)1.6
D)8.9

Accepted Answer

The answer of Exhibit 22-2
A simple random sample of 43...

Question 11

Errors such as measurement error, processing error, and interviewer error are
A)sampling errors
B)nonsampling errors
C)could be either sampling or nonsampling errors
D)impossible to detect

Accepted Answer

Measurement error, processing error, and interviewer error are all types of nonsampling errors. Sampling errors, on the other hand, are errors that result from a random sample being used to estimate information about a larger population. Since these errors are not related to the act of sampling, but rather to issues that arise during data collection and analysis, they are considered nonsampling errors.

Question 12

Exhibit 22-3
From a population of size 600, a simple random sample of size 58 is selected. The sample mean is 400, and the sample standard deviation is 40.
-Refer to Exhibit 22-3. An approximate 95% confidence interval for the population mean is
A)395 to 405
B)380 to 420
C)360 to 440
D)390 to 410

Accepted Answer

We can use the formula: 
CI = sample mean ± (tCritical * (sample standard deviation /√sample size))
where tCritical for a 95% confidence level with 57 degrees of freedom (58-1) is approximately 2.002.

Plugging in the values: 
CI = 400 ± (2.002 * (40/√58))
CI = 400 ± 10.92
CI = (400-10.92 , 400+10.92)
CI = (389.08 , 410.92)

So the best choice is D: 390 to 410 as it is the only choice that contains the entire confidence interval.

Question 13

Exhibit 22-2
A simple random sample of 43 elements has been selected from a population of size 800. The sample mean is 500, and the sample standard deviation is 60.
-Refer to Exhibit 22-2. The population total is
A)34,400
B)21,500
C)400,000
D)500,000

Accepted Answer

To find the population total, we can use the formula: population total = sample mean * population size. Rearranging this equation gives us population size = population total / sample mean. Plugging in the values given, we get population size = population total / 500. To solve for population total, we need to find population size. The formula for standard error of the mean is: standard deviation / square root of sample size. Plugging in the values given, we get standard error of the mean = 60 / sqrt(43) = 9.16. We can use this to construct a 95% confidence interval for the population mean by adding and subtracting 1.96 times the standard error of the mean from the sample mean. This gives us an interval of 500 +/- 1.96*9.16, or (482.7, 517.3). To calculate population size, we can use the lower and upper bounds of this interval to get two estimates of population size: 482.7*n and 517.3*n, where n is the population size. Since we know n = 800, we can solve for the population total by plugging this into our two equations: 482.7*800 = 386,160 and 517.3*800 = 413,840. Taking the average of these two estimates gives us a population total of (386,160 + 413,840) / 2 = 400,000, which is answer choice C.

Question 14

The advantage of nonprobabilistic sampling is that it is inexpensive and
A)provides statistically valid statements about the precision of the estimate
B)can provide valid parameter estimates
C)is error free
D)None of these alternatives is correct.

Accepted Answer

Nonprobabilistic sampling does not provide statistically valid statements about precision or valid parameter estimates. It also introduces potential bias and is generally not recommended for research purposes.

Question 15

In a sample survey, it is common practice to use a t value of (when approximating a 95% confidence interval)
A)1
B)2
C)3
D)4

Accepted Answer

When approximating a 95% confidence interval, it is common practice to use a t value with 1 degree of freedom (for a small sample size) which corresponds to a value of approximately 2. Therefore, the best choice is B.

Question 16

An unbiased estimate of the population total is given by the
A)sample mean multiplied by the size of the sample
B)sample mean multiplied by the size of the population
C)sample size multiplied by the standard error
D)standard deviation divided by the square root of the sample size

Accepted Answer

The unbiased estimate of the population total is obtained by multiplying the sample mean by the size of the population. This method assumes that the sample mean is a good representation of the population mean, thus scaling it up to the population size gives an estimate of the total value in the population.

Question 17

Exhibit 22-1
A simple random sample of size 64 is taken from a population of size 800. The sample mean is determined to be 2,550 with a standard deviation of 500.
-Refer to Exhibit 22-1. An estimate of the standard error of the mean (for the total) is
A)62,500.50
B)47,958.32
C)7.8125
D)50,000.00

Accepted Answer

The answer of Exhibit 22-1
A simple random sample of size...

Question 18

With nonprobabilistic sampling
A)it is possible to make estimates about the precision of the population parameters
B)it is not possible to make statements about the precision of estimates made concerning the population parameters
C)the precision can be estimated if the sample is larger than 30
D)None of these alternatives is correct.

Accepted Answer

Nonprobabilistic sampling methods do not provide a random and representative sample of the population, which means it is not possible to calculate the precision of estimates made concerning population parameters. Therefore, option B is the correct answer. Option A is incorrect because nonprobabilistic sampling does not provide a basis for making generalizations about the population. Option C is also incorrect because sample size alone cannot guarantee precision in nonprobabilistic sampling. Option D is incorrect because option B is the correct answer.

Question 19

Exhibit 22-3
From a population of size 600, a simple random sample of size 58 is selected. The sample mean is 400, and the sample standard deviation is 40.
-Refer to Exhibit 22-3. The population total is
A)3,944
B)16,000
C)240,000
D)24,000

Accepted Answer

The formula to calculate population total is:
Population Total = Sample Mean x Population Size
Here, we have the sample mean (400) and we know that the sample size is 58, but we do not know the population size. However, we are given that the population size is 600. Therefore, we can set up a proportion:
Sample Size / Population Size = Sample Mean / Population Mean
58 / 600 = 400 / Population Mean
Population Mean = (400 x 600) / 58 = 4,129.31
Population Total = Sample Mean x Population Size = 400 x 4,129.31 = 1,651,724
Rounding this to the nearest thousand gives us a population total of 1,652,000, which is closest to choice C (240,000).

Question 20

Exhibit 22-3
From a population of size 600, a simple random sample of size 58 is selected. The sample mean is 400, and the sample standard deviation is 40.
-Refer to Exhibit 22-3. An estimate of the standard error of the mean is
A)4.0
B)5.0
C)6.0
D)7.0

Accepted Answer

The standard error of the mean is calculated as the sample standard deviation divided by the square root of the sample size: 40 / √58 ≈ 5.24, which rounds to 5.0.

Quiz 22: Sample Survey on Website

A method of selecting a sample by randomly selecting the first element and then selecting every Kth element thereafter is

Exhibit 22-2 A simple random sample of 43 elements has been selected from a population of size 800. The sample mean is 500, and the sample standard deviation is 60. -Refer to Exhibit 22-2. An approximate 95% confidence interval for the population total is

Exhibit 22-2 A simple random sample of 43 elements has been selected from a population of size 800. The sample mean is 500, and the sample standard deviation is 60. -Refer to Exhibit 22-2. An approximate 95% confidence interval for the population mean is

Which of the following is(are) the most common type(s) of surveys?

A version of cluster sampling in which the elements are formed into clusters on the basis of their geographic proximity is

Sampling errors can

Exhibit 22-4 Simple random sampling has been used to obtain a sample of size 50 from a population of size 500. The sample proportion was 0.7. -Refer to Exhibit 22-4. The estimate of the standard error of proportion is

Exhibit 22-1 A simple random sample of size 64 is taken from a population of size 800. The sample mean is determined to be 2,550 with a standard deviation of 500. -Refer to Exhibit 22-1. The point estimate of the population total is

Exhibit 22-3 From a population of size 600, a simple random sample of size 58 is selected. The sample mean is 400, and the sample standard deviation is 40. -Refer to Exhibit 22-3. An approximate 95% confidence interval for the population total is

Exhibit 22-2 A simple random sample of 43 elements has been selected from a population of size 800. The sample mean is 500, and the sample standard deviation is 60. -Refer to Exhibit 22-2. The standard error of the mean is

Errors such as measurement error, processing error, and interviewer error are

Exhibit 22-3 From a population of size 600, a simple random sample of size 58 is selected. The sample mean is 400, and the sample standard deviation is 40. -Refer to Exhibit 22-3. An approximate 95% confidence interval for the population mean is

Exhibit 22-2 A simple random sample of 43 elements has been selected from a population of size 800. The sample mean is 500, and the sample standard deviation is 60. -Refer to Exhibit 22-2. The population total is

The advantage of nonprobabilistic sampling is that it is inexpensive and

In a sample survey, it is common practice to use a t value of (when approximating a 95% confidence interval)

An unbiased estimate of the population total is given by the

Exhibit 22-1 A simple random sample of size 64 is taken from a population of size 800. The sample mean is determined to be 2,550 with a standard deviation of 500. -Refer to Exhibit 22-1. An estimate of the standard error of the mean (for the total) is

With nonprobabilistic sampling

Exhibit 22-3 From a population of size 600, a simple random sample of size 58 is selected. The sample mean is 400, and the sample standard deviation is 40. -Refer to Exhibit 22-3. The population total is

Exhibit 22-3 From a population of size 600, a simple random sample of size 58 is selected. The sample mean is 400, and the sample standard deviation is 40. -Refer to Exhibit 22-3. An estimate of the standard error of the mean is

A method of selecting a sample by randomly selecting the first element and then selecting every K^th element thereafter is