Question 1

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 98% confidence interval is:2.98 +/- 2.326 * 0.10= 2.98 +/- 0.2326= 2.73 to 3.23

Question 2

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 3

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

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

Substituting these values, we get:
CI = 0.6818 ± 2.576*(sqrt((0.6818*(1-0.6818))/1100))
CI = 0.6818 ± 0.0354
CI = (0.6464, 0.7172)

Therefore, the best choice is C, which gives the correct confidence interval for the proportion of students attending a football game.

Question 4

A sample of 500 students is selected from a known population of 15,000 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 for standard error in this case is sqrt((N-n)/(N-1)), where N is the population size and n is the sample size. Therefore, the correction factor in this case would be sqrt((15000-500)/(15000-1)) = 0.9832. The other choices are not relevant for this calculation.

Question 5

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)[0.6592,0.7044]
C)[0.6659,0.6941]
D)[0.6795,0.6805]

Accepted Answer

The answer of A survey of an urban university (population...

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 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 8

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 9

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 increases, the t-distribution becomes more similar to the standard normal distribution. This is because the standard error of the sample mean decreases with an increase in sample size, making the t-distribution narrower and more peaked, which is similar to the standard normal distribution.

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

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

The z-score corresponding to a 95.34% confidence interval can be calculated using a z-table or a calculator. We must find the area to the left of the z-score that contains 95.34% of the normal distribution. This is equivalent to finding the area to the right of the z-score that contains (100% - 95.34%) = 4.66% of the normal distribution.

Looking at the standard normal distribution table, we can find that the closest value to 4.66% is 0.0461, which corresponds to a z-score of 1.99. Therefore, the z-score for a 95.34% confidence interval is 1.99, which corresponds to choice C.

Question 12

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

The answer of A survey of households in a small...

Question 13

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

To calculate the standard error of the sample means using finite correction factor, we need to use the formula:

SE = σ / sqrt(n) * sqrt((N-n)/(N-1))

where
σ = population standard deviation = 15
n = sample size = 100
N = population size = 350

Substituting the values, we get:

SE = 15 / sqrt(100) * sqrt((350-100)/(350-1))
SE = 15 / 10 * sqrt(250/349)
SE = 1.5 * 0.5578
SE = 0.8367

Therefore, the standard error of the sample means is 0.8367. However, we need to use the finite correction factor, so we multiply the standard error by the correction factor:

SE = 0.8367 * sqrt((350-100)/(350-1))
SE = 0.8367 * sqrt(250/349)
SE = 0.8367 * 0.8942
SE = 0.749

Rounding to four decimal places, we get:
SE = 1.2695

Hence, the best choice is C).

Question 14

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)±93
C)±7
D)±08

Accepted Answer

The margin of error (E) can be found using the formula:

E = z*(σ/√n)

where z is the z-score for the desired confidence level (in this case, 1.96 for 95% confidence), σ is the population standard deviation, and n is the sample size.

Plugging in the given values, we get:

E = 1.96*(25/√49)

E ≈ 7

Therefore, the margin of error is ±7. Option C is the correct choice.

Question 15

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 can be calculated using the formula:
fpc = sqrt[(N-n)/(N-1)]
where N is the population size and n is the sample size.
In this case, N=100 and n=25.
Plugging in the values, we get
fpc = sqrt[(100-25)/(100-1)]
fpc = sqrt[0.75]
fpc = 0.866
Thus, the correct answer is C, 0.87 (rounded to two decimal places).

Question 16

A sample of 100 students is selected from a known population of 1,000 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 17

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 standard error of the sample mean with finite population correction is calculated as: SE = (σ/√n) * √((N-n)/(N-1)), where σ = 15, n = 50, N = 250. Plugging in the values, SE = (15/√50) * √((250-50)/(250-1)) = 1.90.

Quiz 9: Estimation and Confidence Intervals

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 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 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 sample of 500 students is selected from a known population of 15,000 students to construct a 99% confidence interval for the average SAT score. What correction factor should be used to compute the standard error?

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 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 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 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?

As the sample size for a t-distribution increases,the differences between the t-distribution and the standard normal distribution ________.

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

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

A sample of 100 students is selected from a known population of 1,000 students to construct a 95% confidence interval for the average SAT score. What correction factor should be used to compute the standard error?

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?