Question 1

A sample of 500 students is selected from a known population of 15000 students to construct a 99% confidence interval for the average SAT score. What correction factor should be used to compute the standard error?
A) 0.9499
B) 0.9832
C) 2.5760
D) Cannot be determined

Accepted Answer

The correction factor to be used when constructing confidence intervals for means when the population size is known is 1- n/N, where n is the sample size and N is the population size. 
In this case, n = 500 and N = 15000. Therefore, the correction factor is 1 - 500/15000 = 0.9667.
For a 99% confidence interval, we use a z-score of 2.576. 
So the standard error = (population standard deviation)/sqrt(sample size)
= (unknown)/sqrt(500) = (unknown)/22.36
We know that 99% confidence interval is given by Mean ± (z-score)(standard error)
Therefore, in order to calculate the confidence interval, we need to use the correction factor of 0.9667 to compute the standard error. 
Thus, the answer is B) 0.9832.

Question 2

A sample of 25 is selected from a known population of 100 elements. What is the finite population correction factor?
A) 8.66
B) 75
C) 0.87
D) Cannot be determined

Accepted Answer

The finite population correction factor is given by sqrt((N-n)/(N-1)), where N is the population size and n is the sample size. Plugging in the values, we get sqrt((100-25)/(100-1)) = sqrt(0.75/0.99) = 0.866 or 0.87 (rounded to two decimal places). Therefore, the answer is C.

Question 3

A population has a known standard deviation of 25. A simple random sample of 49 items is taken from the selected population. The sample mean (x-bar) is 300. What is the margin of error at the 95% confidence level?
A) 8
B) 293
C) 7
D) 308

Accepted Answer

The formula for calculating margin of error is:
Margin of Error = z* (σ/√n)
where z* is the z-score for the desired confidence level, σ is the population standard deviation, and n is the sample size.

For a 95% confidence level, the z-score is 1.96. 
Plugging in the values, we get:
Margin of Error = 1.96*(25/√49) = 7.14

Rounding the answer to the nearest whole number, we get a margin of error of 7. Therefore, the correct answer is C.

Question 4

A sample of 100 is selected from a known population of 350 elements. The population standard deviation is 15. Using the finite correction factor, what is the standard error of the sample means?
A) 12.6773
B) 0.8452
C) 1.2695
D) Cannot be determined

Accepted Answer

The standard error of the sample means can be calculated using the formula:
Standard error = population standard deviation / square root of sample size * finite correction factor
Finite correction factor = square root of (N - n) / (N - 1)
where N is the population size and n is the sample size.

In this case, N = 350, n = 100, and population standard deviation = 15. 
Substituting the values, we get:
Finite correction factor = square root of (350 - 100) / (350 - 1) = 0.9691
Standard error = 15 / sqrt(100) * 0.9691 = 1.2695

Therefore, the correct answer is C.

Question 5

A student wanted to construct a 99% confidence interval for the mean age of students in her statistics class. She randomly selected nine students. Their mean age was 19.1 years with a sample standard deviation of 1.5 years. What is the 99% confidence interval for the population mean?
A) [17.42, 20.78]
B) [17.48, 20.72]
C) [14.23, 23.98]
D) [0.44, 3.80]

Accepted Answer

Given the sample size (n=9), sample mean ($\bar{x}=19.1$), and sample standard deviation (s=1.5), along with the confidence level (99%), we can calculate the confidence interval for the population mean using the t-distribution (since the population standard deviation is unknown and the sample size is small). The degrees of freedom (df) is $n-1 = 8$. For a 99% confidence level, the t-value from the t-distribution table corresponding to df=8 is approximately 3.355 (this value might slightly vary depending on the source, but it's around this value for a two-tailed test). The formula for the confidence interval is:$$\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$$Plugging in the values:$$19.1 \pm 3.355 \times \frac{1.5}{\sqrt{9}}$$$$19.1 \pm 3.355 \times 0.5$$$$19.1 \pm 1.6775$$This calculation results in the interval $[17.4225, 20.7775]$, which rounds to $[17.42, 20.78]$, matching option A.

Question 6

A survey of 25 grocery stores revealed that the mean price of a gallon of milk was $2.98, with a standard error of $0.10. What is the 95% confidence interval to estimate the true cost of a gallon of milk?
A) $2.81 to $3.15
B) $2.94 to $3.02
C) $2.77 to $3.19
D) $2.95 to $3.01

Accepted Answer

The answer of A survey of 25 grocery stores revealed...

Question 7

A survey of an urban university (population of 25,450) showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game?
A) [0.7510, 0.8290]
B) [0.6591, 0.7045]
C) [0.6659, 0.6941]
D) [0.6795, 0.6805]

Accepted Answer

The sample proportion is 750/1100 = 0.6818. Using the formula for the confidence interval for a proportion and a 90% confidence level, the interval is approximately [0.6591, 0.7045].

Question 8

A sample of 50 is selected from a known population of 250 elements. The population standard deviation is 15. Using the finite correction factor, what is the standard error of the sample means?
A) 2.89
B) 1.90
C) 2.12
D) Cannot be determined

Accepted Answer

The answer of A sample of 50 is selected from...

Question 9

A sample of 100 students is selected from a known population of 1000 students to construct a 95% confidence interval for the average SAT score. What correction factor should be used to compute the standard error?
A) 0.949
B) 0.901
C) 1.96
D) Cannot be determined

Accepted Answer

The correction factor, also known as the finite population correction (FPC), is used when sampling without replacement from a finite population. It is calculated as sqrt((N-n)/(N-1)), where N is the population size and n is the sample size. Here, N = 1000 and n = 100, so the FPC = sqrt((1000-100)/(1000-1)) = sqrt(900/999) ≈ 0.949.

Question 10

Local government officials are interested in knowing if taxpayers are willing to support a school bond initiative that will require an increase in property taxes. A random sample of 750 likely voters was taken. Four hundred fifty of those sampled favored the school bond initiative. The 95% confidence interval for the true proportion of voters favoring the initiative is ________.
A) [0.541, 0.639]
B) [0.400, 0.600]
C) [0.500, 0.700]
D) [0.565, 0.635]

Accepted Answer

The answer of Local government officials are interested in knowing...

Question 11

A survey of 25 grocery stores revealed that the mean price of a gallon of milk was $2.98, with a standard error of $0.10. If 90% and 95% confidence intervals were developed to estimate the true cost of a gallon of milk, what similarities would they have?
A) Both have the same confidence level
B) Both use the same t statistic
C) Both use the same z statistic
D) Both use the same point estimate of the population mean

Accepted Answer

Both confidence intervals would use the same point estimate of the population mean, which is $2.98, as the center of their intervals.

Question 12

A sample mean is a _______________ estimate of the population mean.

Accepted Answer

The answer of A sample mean is a _______________ estimate...

Question 13

As the sample size for a t distribution increases, the differences between the t distribution and the standard normal distribution:
A) are unchanged and remain the same.
B) become smaller, as the t distribution approaches the standard normal distribution.
C) become greater.
D) are evident because the tails of the t distribution become thicker.

Accepted Answer

As the sample size for a t distribution increases, the t distribution becomes more normal and the differences between the t distribution and the standard normal distribution become smaller. This is because the t distribution approaches the standard normal distribution as the sample size becomes larger.

Question 14

A survey of 25 grocery stores revealed that the average price of a gallon of milk was $2.98, with a standard error of $0.10. What is the 98% confidence interval to estimate the true cost of a gallon of milk?
A) $2.73 to $3.23
B) $2.85 to $3.11
C) $2.94 to $3.02
D) $2.95 to $3.01

Accepted Answer

The answer of A survey of 25 grocery stores revealed...

Question 15

A survey of households in a small town showed that in 500 of 1,200 sampled households, at least one member attended a town meeting during the year. Using the 95% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?
A) [0.417, 0.427]
B) [0.389, 0.445]
C) [0.400, 0.417]
D) [0.417, 0.445]

Accepted Answer

To find the confidence interval for the proportion of households represented at a town meeting, we need to use the formula:

CI = p ± z * √(p * (1-p)/n)

where:
p = proportion of households represented at a town meeting (500/1200 = 0.417)
z = z-score for 95% confidence level (1.96)
n = sample size (1200)

Plugging in the values, we get:

CI = 0.417 ± 1.96 * √(0.417 * (1-0.417)/1200)
CI = [0.389, 0.445]

So the confidence interval for the proportion of households represented at a town meeting is [0.389, 0.445]. The best choice is B.

Question 16

A survey of households in a small town showed that in 850 of 1,200 sampled households, at least one member attended a town meeting during the year. Using the 99% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?
A) [0.674, 0.742]
B) [0.655, 0.705]
C) [0.665, 0.694]
D) [0.679, 0.680]

Accepted Answer

The proportion of households represented at a town meeting is 850/1200 = 0.708. At the 99% confidence level, the Z-score is approximately 2.576. The standard error (SE) is calculated using the formula SE = sqrt[p(1-p)/n], where p is the sample proportion and n is the sample size. So, SE = sqrt[0.708(1-0.708)/1200] ≈ 0.013. The margin of error (ME) is Z*SE, so ME = 2.576 * 0.013 ≈ 0.034. Therefore, the confidence interval is 0.708 ± 0.034, which gives us [0.674, 0.742].

Question 17

When estimating a population mean with a confidence interval, a smaller margin of error requires a _________ sample size.

Accepted Answer

The answer of When estimating a population mean with a...

Question 18

The z-score or z-value corresponding to a 95.34% confidence interval is ________.
A) 1.96
B) 1.65
C) 1.99
D) 1.68

Accepted Answer

A 95.34% confidence interval requires the z-score that cuts off 2.83% in the upper-tail of the standard normal distribution, which is (100-95.34)/2. Therefore, we need to find the z-score that corresponds to the area of 0.01415 in the upper tail of the normal distribution. Using a standard normal distribution table or a calculator, the z-score that corresponds to the area of 0.01415 is approximately 1.99. Therefore, the z-score or z-value corresponding to a 95.34% confidence interval is 1.99.

Question 19

A survey of an urban university (population of 25,450) showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 99% level of confidence, what is the confidence interval for the proportion of students attending a football game?
A) [0.7671, 0.8143]
B) [0.6550, 0.7050]
C) [0.6464, 0.7172]
D) [0.6805, 0.6815]

Accepted Answer

We can use the formula for confidence interval for a proportion:
CI = p ± z*sqrt((p(1-p))/n)
where:
p = proportion of students in the sample attending the football game = 750/1100 = 0.6818
z = z-value for 99% confidence level = 2.576 (from standard normal distribution table)
n = sample size = 1100

Plugging in the values, we get:
CI = 0.6818 ± 2.576*sqrt((0.6818*(1-0.6818))/1100)
CI = [0.6464, 0.7172]

Therefore, the best choice is C, which gives the correct confidence interval with 99% level of confidence.

Question 20

To calculate the sample size required to estimate a population mean, we must know or estimate the ______________ of the population distribution.

Accepted Answer

The answer of To calculate the sample size required to...

Quiz 9: Estimation and Confidence Intervals

A sample of 500 students is selected from a known population of 15000 students to construct a 99% confidence interval for the average SAT score. What correction factor should be used to compute the standard error?

A sample of 25 is selected from a known population of 100 elements. What is the finite population correction factor?

A population has a known standard deviation of 25. A simple random sample of 49 items is taken from the selected population. The sample mean (x-bar) is 300. What is the margin of error at the 95% confidence level?

A sample of 100 is selected from a known population of 350 elements. The population standard deviation is 15. Using the finite correction factor, what is the standard error of the sample means?

A student wanted to construct a 99% confidence interval for the mean age of students in her statistics class. She randomly selected nine students. Their mean age was 19.1 years with a sample standard deviation of 1.5 years. What is the 99% confidence interval for the population mean?

A survey of 25 grocery stores revealed that the mean price of a gallon of milk was $2.98, with a standard error of $0.10. What is the 95% confidence interval to estimate the true cost of a gallon of milk?

A survey of an urban university (population of 25,450) showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game?

A sample of 50 is selected from a known population of 250 elements. The population standard deviation is 15. Using the finite correction factor, what is the standard error of the sample means?

A sample of 100 students is selected from a known population of 1000 students to construct a 95% confidence interval for the average SAT score. What correction factor should be used to compute the standard error?

A survey of 25 grocery stores revealed that the mean price of a gallon of milk was $2.98, with a standard error of $0.10. If 90% and 95% confidence intervals were developed to estimate the true cost of a gallon of milk, what similarities would they have?

A sample mean is a _______________ estimate of the population mean.

As the sample size for a t distribution increases, the differences between the t distribution and the standard normal distribution:

A survey of 25 grocery stores revealed that the average price of a gallon of milk was $2.98, with a standard error of $0.10. What is the 98% confidence interval to estimate the true cost of a gallon of milk?

A survey of households in a small town showed that in 500 of 1,200 sampled households, at least one member attended a town meeting during the year. Using the 95% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?

A survey of households in a small town showed that in 850 of 1,200 sampled households, at least one member attended a town meeting during the year. Using the 99% level of confidence, what is the confidence interval for the proportion of households represented at a town meeting?

When estimating a population mean with a confidence interval, a smaller margin of error requires a _________ sample size.

The z-score or z-value corresponding to a 95.34% confidence interval is ________.

A survey of an urban university (population of 25,450) showed that 750 of 1,100 students sampled attended a home football game during the season. Using the 99% level of confidence, what is the confidence interval for the proportion of students attending a football game?

To calculate the sample size required to estimate a population mean, we must know or estimate the ______________ of the population distribution.