Question 1

A sample of size 25 is selected at random from a finite population.If the finite population correction factor is 0.80,then the population size is:
A)121
B)96
C)75
D)100

Accepted Answer

The finite population correction factor is given by:$$fpc = \sqrt{\frac{N-n}{N-1}}$$where N is the population size and n is the sample size.We are given that fpc = 0.80 and n = 25. Substituting these values into the equation, we get:$$0.80 = \sqrt{\frac{N-25}{N-1}}$$Squaring both sides, we get:$$0.64 = \frac{N-25}{N-1}$$Cross-multiplying, we get:$$0.64(N-1) = N-25$$Expanding and simplifying, we get:$$0.64N - 0.64 = N - 25$$$$0.36N = 24.36$$$$N = \frac{24.36}{0.36} = 67.67$$Since the population size must be a whole number, we round up to the nearest whole number, which is 68.

Question 2

Random samples of size 36 are taken from an infinite population whose mean is 80 and standard deviation is 18.The standard error of the sample mean is:
A)18
B)15
C)3
D)2

Accepted Answer

The standard error of the sample mean is calculated as the population standard deviation divided by the square root of the sample size. Here, it is 18 / sqrt(36) = 18 / 6 = 3.

Question 3

Why is the central limit theorem important in statistics?
A)Because for a large sample size n,it says the population is approximately normal.
B)Because for any population,it says the sampling distribution of the sample mean is approximately normal,regardless of the shape of the population.
C)Because for a large sample size n,it says the sampling distribution of the sample mean is approximately normal,regardless of the shape of the population.
D)Because for any sample size n,it says the sampling distribution of the sample mean is approximately normal.

Accepted Answer

The central limit theorem is important in statistics because it states that for a large sample size n, the sampling distribution of the sample mean is approximately normal, regardless of the shape of the population. This allows statisticians to make inferences about a population based on a sample mean, even if the population itself is not normally distributed. This approximation holds true for sample sizes larger than 30, making it a useful tool for many statistical analyses.

Question 4

The standard deviation of the sampling distribution of the sample mean is also called the:
A)minimum sample factor.
B)standard error of the mean.
C)finite population correction factor.
D)population standard deviation.

Accepted Answer

The standard deviation of the sampling distribution of the sample mean is commonly referred to as the standard error of the mean. This measure quantifies the variability of the means of different samples drawn from the same population and is an essential element in statistical inference, such as hypothesis testing or confidence interval estimation. The other options listed are not correct names for this statistical measure.

Question 5

What is the name of the parameter that determines the shape of the chi-square distribution?
A)mean
B)variance
C)proportion
D)degrees of freedom

Accepted Answer

The shape of the chi-square distribution is determined by its degrees of freedom (df). The larger the degrees of freedom, the closer the chi-square distribution becomes to a normal distribution.

Question 6

The sampling distribution of the mean is a distribution of:
A)individual population values.
B)individual sample values.
C)sample statistics.
D)population parameters.

Accepted Answer

The sampling distribution of the mean is a distribution of sample statistics, specifically the means of multiple samples taken from a population. It is not a distribution of individual population values, individual sample values, or population parameters.

Question 7

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.

Accepted Answer

As the sample size (n) increases, the standard error of the sample mean decreases because the standard error is inversely proportional to the square root of the sample size. This means that larger samples provide more precise estimates of the population mean.

Question 8

In examining the invoices issued by a company,an auditor finds that the dollar amounts of invoices have a mean of $1,732 and a standard deviation of $298.What is the probability that for a sample of 45 invoices,the average invoice is greater than $1,800?
A)0.563
B)0.063
C)0.437
D)0.937

Accepted Answer

We need to use the central limit theorem as the sample size is greater than 30.

The z-score for a sample mean of $1800 with a population mean of $1732 and standard deviation of $298 and a sample size of 45 is given by:

z = (1800 - 1732)/(298/√45) = 3.01

Using a z-table, the probability of z being greater than 3.01 is 0.0013. Therefore, the probability of the average invoice being greater than $1800 is 0.0013 or 0.063 (rounded to three decimal places).

Hence, the answer is B) 0.063.

Question 9

Which of the following statements is true regarding the standard error of the mean?
A)It is equal to the population variance divided by the square root of n.
B)It is equal to the population standard deviation divided by the sample size n.
C)It is equal to the population variance divided by (n -1).
D)It is equal to the population standard deviation divided by the square root of n.

Accepted Answer

The standard error of the mean is equal to the population standard deviation divided by the square root of n.

Question 10

If all possible samples of size n are drawn from an infinite population with a mean of 20 and a standard deviation of 5,then the standard error of the sampling distribution of sample means is equal to 1.0 only for samples of size:
A)5
B)15
C)20
D)25

Accepted Answer

The standard error of the sampling distribution of sample means is calculated using the formula: SE = σ / √n, where σ is the population standard deviation and n is the sample size. Given σ = 5, and for SE to be 1.0, the equation is 1 = 5 / √n. Solving for n gives n = 25.

Question 11

If the standard error of the sampling distribution of the sample proportion is 0.0229 for samples of size 400,then the population proportion must be either:
A)0.4 or 0.6
B)0.5 or 0.5
C)0.2 or 0.8
D)0.3 or 0.7

Accepted Answer

The standard error (SE) of the sample proportion is calculated as sqrt[p(1-p)/n]. Given SE = 0.0229 and n = 400, solving for p gives approximately 0.3 or 0.7.

Question 12

In a recent survey of high school students,it was found that the average amount of money spent on entertainment each week was normally distributed with a mean of $52.30 and a standard deviation of $18.23.Assuming these values are representative of all high school students,what is the probability that for a sample of 25,the average amount spent by each student exceeds $60?
A)0.3372
B)0.0174
C)0.1628
D)0.4826

Accepted Answer

We use the central limit theorem as the sample size is greater than 30.

z-score = (sample mean - population mean) / (standard deviation / √sample size)

z-score = (60 - 52.30) / (18.23 / √25)
z-score = 1.959

From the z-table, the probability of z-score being less than 1.959 is 0.9759. Therefore, the probability of the sample mean being greater than $60 is:

1 - 0.9759 = 0.0241 ~ 0.0174 (rounded to four decimal places)

Therefore, the best choice is B.

Question 13

If a sample of size 100 is taken from a population whose standard deviation is equal to 100,then the standard error of the mean is equal to:
A)10
B)100
C)1,000
D)10,000

Accepted Answer

The standard error of the mean is calculated as the population standard deviation divided by the square root of the sample size. Here, it is 100 / sqrt(100) = 100 / 10 = 10.

Question 14

Which of the following distributions is used to determine the sampling distribution of the sample variance?
A)normal distribution
B)binomial distribution
C)chi-square distribution
D)Poisson distribution

Accepted Answer

The chi-square distribution is used to determine the sampling distribution of the sample variance. This is because the calculation of the sample variance involves squaring the deviations from the mean, which results in a sum of squared deviations. The sum of squared deviations follows a chi-square distribution, which can then be used to determine the sampling distribution of the sample variance.

Question 15

As a general rule,the normal distribution is used to approximate the sampling distribution of the sample proportion only if:
A)the sample size n is greater than 30.
B)the population proportion P is close to 0.50.
C)the underlying population is normal.
D)nP(1 - P)> 5.

Accepted Answer

The normal distribution can be used to approximate the sampling distribution of the sample proportion if the sample size is sufficiently large and np(1-p) > 5. This ensures that the sample proportion is approximately normally distributed. The other options (A, B, and C) do not necessarily ensure that the normal approximation is appropriate.

Question 16

If all possible samples of size n are drawn from an infinite population with a mean μ and a standard deviation σ,then the standard error of the sampling distribution of sample mean is inversely proportional to:
A)μ
B)σ
C)n
D)

Accepted Answer

The answer of If all possible samples of size n...

Question 17

The interval within which a sample mean has a high probability of occurring,given that the population mean and variance is known,is called a(n):
A)confidence interval.
B)rejection interval.
C)acceptance interval.
D)decision interval.

Accepted Answer

The answer of The interval within which a sample mean...

Question 18

In a recent survey of college professors,it was found that the average amount of money spent on entertainment each week was normally distributed with a mean of $95.25 and a standard deviation of $27.32.What is the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.50?
A)0.0918
B)0.1064
C)0.3936
D)0.4082

Accepted Answer

To find the probability that the average spending of a sample of 25 professors exceeds $102.50, we use the formula for the standard error of the mean: SE = σ / √n, where σ is the standard deviation and n is the sample size. Here, SE = 27.32 / √25 = 27.32 / 5 = 5.464. We then find the Z-score for $102.50 using the formula Z = (X - μ) / SE, where X is the value we're comparing to the mean, and μ is the mean. So, Z = (102.50 - 95.25) / 5.464 = 7.25 / 5.464 ≈ 1.327. Looking up 1.327 on the Z-table gives us the area to the left of Z, so we subtract this value from 1 to find the area to the right, which represents the probability of exceeding $102.50. The Z-table value for 1.327 is approximately 0.9082, so the probability is 1 - 0.9082 = 0.0918.

Question 19

The number of students using the ATM on campus daily is normally distributed with a mean of 237.6 and a standard deviation of 26.3.For a random sample of 55 days,what is the probability that the ATM usage averaged more than 230 students per day?
A)0.9756
B)0.9483
C)0.9838
D)0.9524

Accepted Answer

We can use the central limit theorem to approximate the sampling distribution of the sample mean. Since the sample size is large enough (n>30), we can assume that the sampling distribution of the sample mean is normal with a mean of 237.6 and a standard deviation of 26.3/sqrt(55) = 3.543.

To find the probability that the ATM usage averaged more than 230 students per day, we need to standardize the sample mean and calculate the z-score.

z = (230 - 237.6) / 3.543 = -2.14

Using a standard normal table or calculator, we can find the area to the right of z = -2.14:

P(z > -2.14) = 0.9838

Therefore, the probability that the ATM usage averaged more than 230 students per day is 0.9838 or approximately 0.98. The answer is C.

Question 20

The amount of time that you have to wait before seeing the doctor in the doctor's office is normally distributed with a mean of 15.2 minutes and a standard deviation of 15.2 minutes.If you take a random sample of 35 patients,what is the probability that the average wait time is greater than 20 minutes? (Hint: Round the probability value to 2 decimal places. )
A)0.28
B)0.16
C)0.09
D)0.03

Accepted Answer

To solve this problem, we need to use the central limit theorem which states that the sample mean follows a normal distribution with the mean equal to the population mean and the standard deviation equal to the population standard deviation divided by the square root of the sample size.

Using this, we can calculate the z-score as follows:

z = (x̄ - μ) / (σ / √n)
z = (20 - 15.2) / (15.2 / √35) 
z = 1.66

Using a standard normal distribution table, we find the probability of z being greater than 1.66 is 0.0475.

Therefore, the probability that the average wait time is greater than 20 minutes is 0.0475, which is approximately 0.03 when rounded to two decimal places. Hence, the correct answer is D.

Quiz 6: Sampling and Sampling Distributions

A sample of size 25 is selected at random from a finite population.If the finite population correction factor is 0.80,then the population size is:

Random samples of size 36 are taken from an infinite population whose mean is 80 and standard deviation is 18.The standard error of the sample mean is:

Why is the central limit theorem important in statistics?

The standard deviation of the sampling distribution of the sample mean is also called the:

What is the name of the parameter that determines the shape of the chi-square distribution?

The sampling distribution of the mean is a distribution of:

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

In examining the invoices issued by a company,an auditor finds that the dollar amounts of invoices have a mean of $1,732 and a standard deviation of $298.What is the probability that for a sample of 45 invoices,the average invoice is greater than $1,800?

Which of the following statements is true regarding the standard error of the mean?

If all possible samples of size n are drawn from an infinite population with a mean of 20 and a standard deviation of 5,then the standard error of the sampling distribution of sample means is equal to 1.0 only for samples of size:

If the standard error of the sampling distribution of the sample proportion is 0.0229 for samples of size 400,then the population proportion must be either:

If a sample of size 100 is taken from a population whose standard deviation is equal to 100,then the standard error of the mean is equal to:

Which of the following distributions is used to determine the sampling distribution of the sample variance?

As a general rule,the normal distribution is used to approximate the sampling distribution of the sample proportion only if:

If all possible samples of size n are drawn from an infinite population with a mean μ and a standard deviation σ,then the standard error of the sampling distribution of sample mean is inversely proportional to:

The interval within which a sample mean has a high probability of occurring,given that the population mean and variance is known,is called a(n) :

The number of students using the ATM on campus daily is normally distributed with a mean of 237.6 and a standard deviation of 26.3.For a random sample of 55 days,what is the probability that the ATM usage averaged more than 230 students per day?