Question 1

A Monster.com pool of 3,057 individuals asked: "What's the longest vacation you plan to take this summer?" The following relative frequency distribution summarizes the results.  The 99% confidence interval for the proportion of people who plan a one-week or two-week vacation this summer is ______.
A) [0.563,0.603]
B) [0.557,0.597]
C) [0.563,0.597]
D) [0.557,0.603]

Accepted Answer

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Question 2

The Retail Advertising and Marketing Association would like to estimate the average amount of money that a person spends for Mother's Day with 99% confidence interval and a margin of error within plus or minus $6.Assuming the standard deviation for spending on Mother's Day is $36,the required sample size is ______.
A) 210
B) 239
C) 284
D) 313

Accepted Answer

To calculate the required sample size for estimating the average amount of money spent on Mother's Day with a specific margin of error and confidence level, we use the formula for sample size in estimating a mean: $n = \left(\frac{Z\sigma}{E}\right)^2$, where $n$ is the sample size, $Z$ is the Z-score corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $E$ is the margin of error. For a 99% confidence interval, the Z-score is approximately 2.576. Given $\sigma = $36$ and $E = $6$, substituting these values into the formula gives $n = \left(\frac{2.576 \times 36}{6}\right)^2$, which calculates to approximately 239. Therefore, the required sample size is 239.

Question 3

The snowfall (in inches)during the month of January in a particular geographic region is normally distributed with a standard deviation of 16.75.In the last 12 years,the sample average of snowfall is computed as 122.50.a.Construct a 90% confidence interval of the average snowfall in this region.b.Construct a 95% confidence interval of the average snowfall in this region.c.Comment on the width of the preceding confidence intervals.

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Question 4

A machine that is programmed to package 1.20 pounds of cereal is being tested for its accuracy in a sample of 36 cereal boxes,the sample mean filling weight is calculated as 1.22 pounds.The population standard deviation is known to be 0,06 pounds.Which of the following conclusions is correct?
A) The machine is operating improperly because the target is below the lower limit.
B) The machine is operating properly because the interval contains the target.
C) The machine is operating improperly because the target is above the upper limit.
D) There is not enough information to make a conclusion.

Accepted Answer

We can calculate the margin of error using the formula:

margin of error = z*(standard deviation/square root of sample size)

where z is the z-score for a certain level of confidence, which in this case we will use 95% confidence level and thus z = 1.96.

Substituting the known values we get:

margin of error = 1.96*(0.06/square root of 36) = 0.012

Therefore, the confidence interval for the true mean filling weight of the cereal boxes is:

1.22 ± 0.012

which means that we are 95% confident that the population mean filling weight is between 1.208 and 1.232 pounds.

Since the target weight of 1.20 pounds falls within this interval, we can conclude that the machine is operating properly within the acceptable level of variation. Therefore, the correct answer is B.

Question 5

A machine that is programmed to package 1.20 pounds of cereal is being tested for its accuracy in a sample of 36 cereal boxes,the sample mean filling weight is calculated as 1.22 pounds.The population standard deviation is known to be 0,06 pounds.Find the 95% confidence interval for the mean.
A) [1.20,1.24]
B) [1.18,1.26]
C) [1.16,1.28]
D) [1.14,1.30]

Accepted Answer

The 95% confidence interval for the mean can be calculated using the formula: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the confidence level (for 95%, $Z = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $1.22 \pm 1.96 \frac{0.06}{\sqrt{36}} = 1.22 \pm 0.02$, which gives the interval [1.20, 1.24].

Question 6

The minimum sample size n required to estimate a population proportion,with 95% confidence and the assumed most conservative  = 0.5,was found to be 122.Which of the following is the approximate value of the assumed desired margin of error?
A) D = 0.0055
B) D = 0.0078
C) D = 0.0745
D) D = 0.0887

Accepted Answer

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Question 7

A random sample of 130 mortgages in the state of Maryland was randomly selected.From this sample,14 were found to be delinquent on their current payment.The 98% confidence interval for the proportion based on this sample is _______.
A) [0.063,0.153]
B) [0.036,0.180]
C) [0.029,0.188]
D) [0.015,0.201]

Accepted Answer

The sample proportion is 14/130 = 0.1077. Using the formula for a confidence interval for a proportion and a Z-score for 98% confidence (approximately 2.33), the interval is calculated as [0.063, 0.153].

Question 8

Students who graduated from college in 2010 owed an average of $25,250 in student loans.An economist wants to determine if average debt has changed.She takes a sample of 40 recent graduates and finds that their average debt is $27,500 with a standard deviation of $9,120.Use 90% confidence interval.Which of the following conclusion is correct?
A) The average debt decreased.
B) The average debt increased.
C) The average debt has not changed.
D) There is not enough information.

Accepted Answer

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Question 9

Recently,six single-family homes in San Luis Obispo County in California sold at the following prices (in $1,000s): 549,449,705,529,639,609.Find a 95% confidence interval for the mean sale price in San Luis Obispo County.
A) [485.34,674.66]
B) [489.90,670.10]
C) [508.46,651.54]
D) [476.30,683.70]

Accepted Answer

The sample mean is (549+449+705+529+639+609)/6 = 574. The sample standard deviation is 72.9. The 95% confidence interval for the mean sale price is 574 +/- 1.96 * 72.9 = [485.34,674.66].

Question 10

The average natural gas bill for a random sample of 21 homes in 19810 zip code during the month of March was $311.90 with a sample standard deviation of $51.60.The 90% confidence interval around this sample mean is _______.
A) [$285.98,$337.82]
B) [$292.46,$331.34]
C) [$278.09.$345.71]
D) [$266.82,$356.98]

Accepted Answer

To find the 90% confidence interval, we use the formula:

Sample Mean ± (t-value)*(standard error of the mean)

where t-value comes from the t-distribution with n-1 degrees of freedom and a 90% confidence level. For a sample size of 21, the degrees of freedom is 20 and the t-value is 1.725.

The standard error of the mean is found by dividing the sample standard deviation by the square root of the sample size:

standard error of the mean = 51.60 / sqrt(21) = 11.22

Plugging in the values, we get:

311.90 ± 1.725*11.22

which simplifies to:

311.90 ± 19.34

Therefore, the 90% confidence interval is [$292.46, $331.34], which is closest to choice B.

Question 11

The height of high school basketball players is known to be normally distributed with a standard deviation of 1.75 inches.In a random sample of eight high school basketball players,the heights (in inches)are recorded as 75,82,68,74,78,70,77,and 76.Construct a 95% confidence interval on the average height of all high school basketball players.

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Question 12

Bob's average golf score at his local course is 93.6 from a random sample of 34 rounds of golf.Assume that the population standard deviation for his golf score is 4.2.The 90% confidence interval around this sample mean is ______.
A) [89.0,98.3]
B) [90.5,96.7]
C) [92.4,94.8]
D) [93.2,94.0]

Accepted Answer

We use the formula for a confidence interval for a population mean:
CI = x̄ ± z*(σ/√n)
where x̄ is the sample mean, z is the z-score corresponding to the confidence level (1.645 for 90%), σ is the population standard deviation, and n is the sample size.

Plugging in the given values, we get:
CI = 93.6 ± 1.645*(4.2/√34)
CI = 93.6 ± 1.096
CI = [92.504, 94.696]

Rounding to one decimal place gives us the answer of [92.4, 94.8], which corresponds to choice C.

Question 13

An employee of the Bureau of Transportation Statistics has been given the task of estimating the proportion of on-time arrivals of a budget airline.A prior study has estimated this on-time arrival rate as 78.5%.What is the minimum number of arrivals this employee must include in the sample to ensure that the margin of error for a 95% confidence interval is no more than 0.05?
A) 270
B) 259
C) 260
D) 269

Accepted Answer

The margin of error formula for a proportion is given by:

Margin of Error = z*(sqrt(p*(1-p)/n))

where z is the z-score for the desired confidence level, p is the estimated proportion, and n is the sample size.

Solving for n, we get:

n = (z / Margin of Error)^2 * p*(1-p)

Plugging in z = 1.96 for a 95% confidence level, p = 0.785, and Margin of Error = 0.05, we get:

n = (1.96 / 0.05)^2 * 0.785*(1-0.785) = 259.97

Rounding up to the nearest whole number, we get n = 260, which is option C. Therefore, the employee must include at least 260 arrivals in the sample to ensure a margin of error no more than 0.05 for a 95% confidence interval.

Question 14

A sample of holiday shoppers is taken randomly from a local mall.Forty-nine shoppers were selected and asked what their average spending on gifts would be during the entire holiday season.The point estimate of the population mean was calculated as $550 and the sample standard deviation was calculated as $92.a.Construct a 95% confidence interval of the population mean spending.b.Explain how the central limit theorem is used in constructing this interval.

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Question 15

A researcher in campaign finance law wants to estimate the proportion of elementary,middle,and high school teachers who contributed to a candidate during a recent election cycle.Given that no prior estimate of the population proportion is available,what is the minimum sample size such that the margin of error is no more than 0.03 for a 95% confidence interval?
A) 751.67
B) 752
C) 1067.11
D) 1068

Accepted Answer

The formula for minimum sample size needed to estimate a population proportion with a given margin of error is: 
n = (z^2 * p * q) / (E^2) 
Where: 
z = the critical value from the standard normal distribution corresponding to the desired level of confidence (1.96 for 95% confidence) 
p = the estimate of the population proportion (since there is no prior estimate, we use 0.5 for maximum variability) 
q = 1 - p 
E = the desired margin of error 
Plugging in the values, we get: 
n = (1.96^2 * 0.5 * 0.5) / (0.03^2) 
n = 1067.11
Therefore, the minimum sample size needed is 1068. Answer choice D is the closest option to this value.

Question 16

The National Center for Education would like to estimate the proportion of students who defaulted on their student loans for the state of Arizona.The total sample size needed to construct a 95% confidence interval for the proportion of student loans in default with a margin of error equal to 4% is ______.
A) 336
B) 455
C) 416
D) 601

Accepted Answer

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Question 17

A Monster.com pool of 3,057 individuals asked,"What's the longest vacation you plan to take this summer?" The following relative frequency distribution summarizes the results.  The 95% confidence interval for the proportion of people who plan a one-week vacation this summer is ______.
A) [0.345,0.375]
B) [0.345,0.382]
C) [0.343,0.377]
D) [0.338,0.382]

Accepted Answer

To find the confidence interval for the proportion of people who plan a one-week vacation this summer, we need to use the formula:
CI = p ± z*√((p(1-p))/n)
Where:
p = proportion of people who plan a one-week vacation
n = sample size
z = z-score corresponding to the level of confidence (95% confidence = 1.96)
First, we need to find the values for p and n from the given data:
p = 960/3057 = 0.314
n = 3057
Now, we can calculate the confidence interval:
CI = 0.314 ± 1.96√((0.314(1-0.314))/3057)
CI = 0.314 ± 0.017
CI = [0.297, 0.331]
This means we can be 95% confident that the true proportion of people who plan a one-week vacation this summer is between 0.297 and 0.331. However, none of the given answer choices match exactly with this interval. The closest answer choice is C, which has an interval of [0.343, 0.377]. This interval does not overlap with the calculated interval, but it is the closest option. Therefore, the best choice is C.

Question 18

A politician wants to estimate the percentage of people who like his new slogan.Given that no prior estimate of the population proportion is available,what is the minimum sample size such that the margin of error is no more than 0.08 for a 95% confidence interval?
A) 105.70
B) 106
C) 150.06
D) 151

Accepted Answer

The formula for the margin of error is:

Margin of error = z* (standard deviation / square root of sample size)

Where z* is the z-score corresponding to the level of confidence, in this case, 95% confidence level corresponds to a z-score of 1.96.

Since no prior estimate of the population proportion is available, we can use a conservative estimate of 0.5 for the population proportion, which gives the maximum margin of error.

We want the margin of error to be no more than 0.08, so we can solve for the required sample size:

0.08 = 1.96 * sqrt[(0.5 * 0.5) / n]

n = (1.96 / 0.08)^2 * (0.5 * 0.5)

n ≈ 151

Therefore, the minimum sample size required to achieve a margin of error no more than 0.08 with 95% confidence level is 151.

Question 19

To estimate the mean earnings forecast for a large company,an investor looked up earnings forecasts from six financial analysts.The six forecasts he found were 200,220,300,185,210,and 210 (in millions).Suppose the investor knows the population standard deviation is 25 million.Calculate a 95% confidence interval of the population mean earnings forecast.

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Question 20

Each portion of the SAT exam is designed to be normally distributed such that it has a population standard deviation of 100 and a mean of 500.However,the mean has changed over the years as less selective schools began requiring the SAT,and because students later began to prepare more specifically for the exam.Construct a 90% confidence interval for the population mean from the following eight scores from the math portion,using the population standard deviation of 100: 450,660,760,540,420,430,640,and 580.

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Quiz 8: Interval Estimation

A Monster.com pool of 3,057 individuals asked: "What's the longest vacation you plan to take this summer?" The following relative frequency distribution summarizes the results. The 99% confidence interval for the proportion of people who plan a one-week or two-week vacation this summer is ______.

A machine that is programmed to package 1.20 pounds of cereal is being tested for its accuracy in a sample of 36 cereal boxes,the sample mean filling weight is calculated as 1.22 pounds.The population standard deviation is known to be 0,06 pounds.Which of the following conclusions is correct?

A machine that is programmed to package 1.20 pounds of cereal is being tested for its accuracy in a sample of 36 cereal boxes,the sample mean filling weight is calculated as 1.22 pounds.The population standard deviation is known to be 0,06 pounds.Find the 95% confidence interval for the mean.

The minimum sample size n required to estimate a population proportion,with 95% confidence and the assumed most conservative = 0.5,was found to be 122.Which of the following is the approximate value of the assumed desired margin of error?

A random sample of 130 mortgages in the state of Maryland was randomly selected.From this sample,14 were found to be delinquent on their current payment.The 98% confidence interval for the proportion based on this sample is _______.

Recently,six single-family homes in San Luis Obispo County in California sold at the following prices (in $1,000s) : 549,449,705,529,639,609.Find a 95% confidence interval for the mean sale price in San Luis Obispo County.

The average natural gas bill for a random sample of 21 homes in 19810 zip code during the month of March was $311.90 with a sample standard deviation of $51.60.The 90% confidence interval around this sample mean is _______.

Bob's average golf score at his local course is 93.6 from a random sample of 34 rounds of golf.Assume that the population standard deviation for his golf score is 4.2.The 90% confidence interval around this sample mean is ______.

A Monster.com pool of 3,057 individuals asked,"What's the longest vacation you plan to take this summer?" The following relative frequency distribution summarizes the results. The 95% confidence interval for the proportion of people who plan a one-week vacation this summer is ______.

A politician wants to estimate the percentage of people who like his new slogan.Given that no prior estimate of the population proportion is available,what is the minimum sample size such that the margin of error is no more than 0.08 for a 95% confidence interval?