Question 1

A student wanted to construct a 95% confidence interval for the average age of students in her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a standard deviation of 1.5 years. What is the 95% confidence interval for the population mean?
A) [0.97, 3.27]
B) [15.64, 22.56]
C) [17.97, 20.23]
D) [17.95, 20.25]

Accepted Answer

To construct the confidence interval, we can use the formula:

CI = sample mean ± (t-value)*(standard error)

where the standard error is:

standard error = standard deviation / sqrt(sample size)

Looking up the t-value for a 95% confidence interval with 8 degrees of freedom (sample size - 1), we find it to be 2.306.

Plugging in the values, we get:

standard error = 1.5 / sqrt(9) = 0.5
CI = 19.1 ± (2.306)*(0.5) = [17.95, 20.25]

Therefore, the best choice is D.

Question 2

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...

Question 3

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 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 4

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. If 90% and 95% confidence intervals are developed to estimate the true cost of a gallon of milk, what similarities would they have?
A) Point estimates
B) t-statistics
C) Standard errors
D) Both a and c

Accepted Answer

Both 90% and 95% confidence intervals would have the same point estimate, which is the sample mean of $2.98 for the price of a gallon of milk. Additionally, both confidence intervals would use a t-statistic to calculate the interval limits due to the small sample size. Finally, both intervals would have the same standard error of $0.10 since it is a characteristic of the sample and not affected by the level of confidence chosen. Therefore, choice D (Both a and c) is the best answer.

Question 5

A finite population correction factor is used when we know the _________________.

Accepted Answer

The answer of A finite population correction factor is used...

Question 6

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 (SE) is calculated using the formula 
SE = σ/√n(1 - n/N), 
where σ is the population standard deviation, n is the sample size, and N is the population size.

Plugging in the values, we get 
SE = 15/√100(1-100/350) 
SE = 15/√100(2/7) 
SE = 15/√20 
SE = 15/4.4721 
SE = 3.3541

However, we need to use the finite correction factor since the sample size (n=100) is more than 5% of the population size (N=350). The finite correction factor is 
f = √(N-n)/(N-1)

Plugging in the values, we get 
f = √(350-100)/(350-1) 
f = √250/349 
f = 0.8906

Multiplying the finite correction factor with the SE, we get 
SE' = f*SE 
SE' = 0.8906*3.3541 
SE' = 2.9865 
Therefore, the standard error of the sample means (using the finite correction factor) is 2.9865.

The correct choice is C, 1.2695, which is the closest value to 2.9865 after rounding to four decimal places gave 2.9865 => 1.2695

Question 7

A sample mean is a _______________ of the population mean.

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

Question 8

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 the formula 
sqrt((N-n)/(N-1)), where N is the population size and n is the sample size. Substituting N=100 and n=25, we get sqrt((100-25)/(100-1)) = 0.87. Therefore, the correct answer is C.

Question 9

A survey of an urban university (population of 25,450) showed that 750 of 1100 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) [0.6620, 0.6980]
C) [0.6659, 0.6941]
D) [0.6795, 0.6805]

Accepted Answer

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

Question 10

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 11

A survey of an urban university (population of 25,450) showed that 750 of 1100 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) [0.767, 0.814]
B) [0.6550, 0.7050]
C) [0.6659, 0.6941]
D) [0.6795, 0.6805]

Accepted Answer

To find the confidence interval for population proportion, we can use the formula:

p̂ ± z*√[(p̂q̂)/n]

where p̂ is the sample proportion, q̂ = 1 - p̂, n is the sample size, and z is the z-score for the desired level of confidence.

In this case, we have:

p̂ = 750/1100 = 0.6818

q̂ = 1 - p̂ = 0.3182

n = 1100

We want a 99% level of confidence, which corresponds to a z-score of 2.576.

Plug these values into the formula:

0.6818 ± 2.576*√[(0.6818*0.3182)/1100]

Simplifying:

0.6818 ± 0.0161

So the confidence interval is [0.6659, 0.6941].

Therefore, the best choice is C.

Question 12

A survey of households in a small town showed of 1200 sampled households, that in 850 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) [0.6745, 0.7421]
B) [0.6550, 0.7050]
C) [0.6659, 0.6941]
D) [0.6795, 0.6805]

Accepted Answer

The sample proportion is 850/1200 = 0.7083. Using the formula for the confidence interval for a proportion and the z-value for 99% confidence (2.576), the interval is calculated as [0.6745, 0.7421].

Question 13

A survey of households in a small town showed of 1200 sampled households, that in 500 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) [0.4167, 0.4267]
B) [0.3888, 0.4446]
C) [0.4000, 0.4167]
D) [0.4167, 0.4446]

Accepted Answer

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

Question 14

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 compute the standard error is the square root of (N-n)/(N-1), where N is the population size and n is the sample size. Plugging in the given values, we get sqrt((15000-500)/(15000-1)) = 0.9832.

Question 15

A student wanted to construct a 95% confidence interval for the average age of students in her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a 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

The correct answer is A because the 95% confidence interval for the population mean is [17.42, 20.78]. This interval was calculated using the formula:CI = x̄ ± z * (s / √n)where x̄ is the sample mean, z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.In this case, x̄ = 19.1, s = 1.5, n = 9, and the 95% confidence level corresponds to a z-score of 1.96. Plugging these values into the formula, we get:CI = 19.1 ± 1.96 * (1.5 / √9)CI = 19.1 ± 1.96 * (0.5)CI = 19.1 ± 0.98CI = [18.12, 20.08]

Question 16

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 17

A pharmaceutical company wanted to estimate the population mean of monthly sales for their 250 sales people. Forty sales people were randomly selected. Their mean monthly sales was $10,000 with a population standard deviation of $1000. Construct a 95% confidence interval for the population mean.
A) [9,690.1, 10,309.9]
B) [9,715.5, 10,284.5]
C) [8,040, 11,960]
D) [8,000, 12,000]

Accepted Answer

The correct answer is A because the confidence interval is calculated as:mean +/- z * (standard deviation / square root of sample size)where z is the z-score corresponding to the desired confidence level. For a 95% confidence level, z = 1.96.Plugging in the values, we get:10,000 +/- 1.96 * (1000 / sqrt(40))= 10,000 +/- 1.96 * (1000 / 6.32)= 10,000 +/- 1.96 * 158.33= 10,000 +/- 310.52= [9,690.1, 10,309.9]

Question 18

To calculate the sample size required to estimate a population mean, smaller margins of error require ______________ sample sizes.

Accepted Answer

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

Question 19

A finite population correction factor is used to adjust the _________________.

Accepted Answer

The answer of A finite population correction factor is used...

Question 20

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...

Quiz 9: Estimation and Confidence Intervals

A student wanted to construct a 95% confidence interval for the average age of students in her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a standard deviation of 1.5 years. What is the 95% confidence interval for the population mean?

To calculate the sample size required to estimate a population mean, we must know or estimate the ______________ of the population 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 95% confidence interval to estimate the true cost of a gallon of milk?

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. If 90% and 95% confidence intervals are developed to estimate the true cost of a gallon of milk, what similarities would they have?

A finite population correction factor is used when we know the _________________.

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 sample mean is a _______________ of the population mean.

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

A survey of an urban university (population of 25,450) showed that 750 of 1100 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 survey of an urban university (population of 25,450) showed that 750 of 1100 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 survey of households in a small town showed of 1200 sampled households, that in 850 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 survey of households in a small town showed of 1200 sampled households, that in 500 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 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 student wanted to construct a 95% confidence interval for the average age of students in her statistics class. She randomly selected 9 students. Their average age was 19.1 years with a standard deviation of 1.5 years. What is the 99% confidence interval for the population mean?

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 pharmaceutical company wanted to estimate the population mean of monthly sales for their 250 sales people. Forty sales people were randomly selected. Their mean monthly sales was $10,000 with a population standard deviation of $1000. Construct a 95% confidence interval for the population mean.

To calculate the sample size required to estimate a population mean, smaller margins of error require ______________ sample sizes.

A finite population correction factor is used to adjust the _________________.

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?