Question 1

A medical devices company wants to know the number of hours its MRI machines are used per day. A previous study found a standard deviation of four hours. How many MRI machines must the company find data for in order to have a margin of error of at most 0.5 hour when calculating a 98% confidence interval?
A) 346.26
B) 347
C) 424.69
D) 425

Accepted Answer

The answer of A medical devices company wants to know...

Question 2

Suppose we wish to find the required sample size to find a 90% confidence interval for the population proportion with the desired margin of error. If there is no rough estimate  of the population proportion, what value should be assumed for  ?
A) 0.05
B) 0.10
C) 0.50
D) 0.90

Accepted Answer

The answer of Suppose we wish to find the required...

Question 3

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) [$292.48, $331.32]
C) [$278.09. $345.71]
D) [$266.82, $356.98]

Accepted Answer

We need to use the t-distribution since the sample size is small (<30) and the population standard deviation is unknown. The formula for the confidence interval is:
(sample mean) ± t(α/2, n-1) * (sample standard deviation / sqrt(n))
Where t(α/2, n-1) is the t-value obtained from the t-distribution table with a degree of freedom of n-1 and a confidence level of 90%. From the table, t(0.05, 20) = 1.725.
Plugging in the values, we get:
311.90 ± 1.725 * (51.60 / sqrt(21))
= 311.90 ± 14.42
Therefore, the 90% confidence interval is [$292.48, $331.32].

Question 4

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 5

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) [484.63, 674.37]
B) [489.90, 670.10]
C) [508.46, 651.54]
D) [476.30, 683.70]

Accepted Answer

To calculate a 95% confidence interval for the mean sale price, we first find the mean ($\bar{x}$) and the standard deviation (s) of the sample. The mean sale price is $\frac{549 + 449 + 705 + 529 + 639 + 609}{6} = 580$. The standard deviation can be calculated using the formula for sample standard deviation, which gives us approximately 92.87. For a 95% confidence interval and a sample size of 6, we use the t-distribution because the population standard deviation is unknown. The degrees of freedom (df) is 5 (n-1), and the critical t-value for a 95% confidence interval with df = 5 from the t-distribution table is approximately 2.571. The margin of error (E) is calculated as $E = t \times \frac{s}{\sqrt{n}} = 2.571 \times \frac{92.87}{\sqrt{6}} \approx 97.37$. Therefore, the 95% confidence interval is $\bar{x} \pm E = 580 \pm 97.37$, which gives us [482.63, 677.37]. However, rounding to the nearest whole number as provided in the options, the closest match is [484.63, 674.37].

Question 6

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 a proportion, we first need to calculate the sample proportion, which is the relative frequency of people who plan a one-week vacation this summer. From the table, the relative frequency for one week vacation is 0.36.

Next, we need to calculate the standard error of the proportion:

SE = sqrt(p*(1-p)/n)
SE = sqrt(0.36 * 0.64 / 3057)
SE = 0.013

To find the 95% confidence interval, we use the formula:

CI = p ± z*SE

where z is the z-score for a 95% confidence interval, which is 1.96.

CI = 0.36 ± 1.96*0.013
CI = [0.343, 0.377]

Therefore, the 95% confidence interval for the proportion of people who plan a one-week vacation this summer is [0.343, 0.377]. Option C is the closest choice.

Question 7

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 8

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) [0.044, 0.171]
B) [0.036, 0.180]
C) [0.029, 0.188]
D) [0.015, 0.201]

Accepted Answer

To calculate the 98% confidence interval for the proportion, we use the formula for the confidence interval for a proportion: $\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level, and $n$ is the sample size. Here, $\hat{p} = \frac{14}{130}$, $n = 130$, and for a 98% confidence level, $Z_{\alpha/2} \approx 2.33$. Plugging these values into the formula gives a confidence interval of approximately [0.044, 0.171].

Question 9

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

The answer of The minimum sample size n required to...

Question 10

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 use a one-sample t-test to test the hypothesis that the population mean filling weight is equal to 1.20 pounds.

Ho: μ = 1.20 
Ha: μ ≠ 1.20

The sample mean, x̄ = 1.22 pounds, and the population standard deviation, σ = 0.06 pounds, are given.

We can calculate the t-value as follows:

t = (x̄ - μ) / (σ / sqrt(n)) 
t = (1.22 - 1.20) / (0.06 / sqrt(36)) 
t = 2

Using a t-table with 35 degrees of freedom (n - 1), we find that the critical values for a two-tailed test at a 5% significance level are ±2.03.

Since our calculated t-value (2) falls within this range, we fail to reject the null hypothesis. This means we do not have enough evidence to conclude that the machine is operating improperly because the interval contains the target (1.20 pounds). Therefore, the correct conclusion is choice B, "The machine is operating properly because the interval contains the target."

Question 11

A marketing firm wants to estimate how much root beer the average teenager drinks per year. A previous study found a standard deviation of 1.12 liters. How many teenagers must the firm interview in order to have a margin of error of at most 0.1 liter when constructing a 99% confidence interval?
A) 831.39
B) 832
C) 832.39
D) 833

Accepted Answer

To find the sample size, use the formula for margin of error in a confidence interval: ME = Z * (σ/√n). For a 99% confidence interval, Z is approximately 2.576. Solving for n gives n = (Z * σ / ME)^2 = (2.576 * 1.12 / 0.1)^2 ≈ 832.39. Since the sample size must be a whole number, round up to 833.

Question 12

According to a report in USA Today, more and more parents are helping their young adult children buy homes. You would like to construct a 90% confidence interval of the proportion of all young adults in Louisville, Kentucky, who received help from their parents in buying a home. How large a sample should you take so that the margin of error is no more than 0.06?
A) n = 114
B) n = 188
C) n = 267
D) Cannot be determined.

Accepted Answer

To determine the sample size needed for a confidence interval with a specific margin of error, you can use the formula for the sample size of a proportion, which is $n = \left(\frac{Z^2 \cdot p \cdot (1-p)}{E^2}\right)$, where $Z$ is the Z-score corresponding to the desired confidence level, $p$ is the estimated proportion (if unknown, 0.5 is used for maximum sample size), and $E$ is the desired margin of error. For a 90% confidence interval, the Z-score is approximately 1.645. Assuming $p = 0.5$ for maximum variability, and wanting $E = 0.06$, the calculation is $n = \left(\frac{1.645^2 \cdot 0.5 \cdot (1-0.5)}{0.06^2}\right)$, which simplifies to approximately 188. Therefore, a sample size of 188 is needed.

Question 13

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 minimum sample size can be written as:
n = [Z^2 * p *(1-p)] / E^2 
Where:
Z = the Z-score associated with the desired level of confidence (in this case, 1.96 for 95% confidence)
p = the estimated proportion of the population that likes the slogan (since no prior estimate is available, we use 0.5 as a conservative estimate)
E = the desired margin of error (0.08 in this case)

Substituting these values into the formula, we get:
n = [1.96^2 * 0.5 * (1-0.5)] / 0.08^2
n ≈ 151

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

Question 14

Students who graduated from college last year 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

The answer of Students who graduated from college last year...

Question 15

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 16

The minimum sample size n required to estimate a population mean with 95% confidence and the desired margin of error 1.5 was found to be 198. Which of the following is the approximate value of the assumed estimate of the population standard deviation?
A) 10.7690
B) 12.8309
C) 115.9671
D) 164.6326

Accepted Answer

We know that the formula for calculating the minimum sample size n is given by:

n = [(z-value)² * σ²] / E²

Where,
z-value = 1.96 (for 95% confidence level)
E = 1.5 (desired margin of error)

We can rearrange this formula to solve for σ (population standard deviation) as follows:

σ = sqrt[(n * E²) / (z-value)²]

Substituting the given values, we get:

σ = sqrt[(198 * 1.5²) / 1.96²] = 10.7690 (approx.)

Therefore, the assumed estimate of the population standard deviation is approximately 10.7690, which is closest to option A).

Question 17

Find the minimum sample size when we want to construct a 95% confidence interval on the population proportion for the support of candidate A in the following mayoral election. Candidate A is facing two opposing candidates. In a preselected poll of 100 residents, 22 supported candidate B and 14 supported candidate C. The desired margin of error is 0.06.
A) n = 173
B) n = 174
C) n = 245
D) n = 246

Accepted Answer

We need to find the minimum sample size, which means we want to be as efficient as possible without sacrificing the desired level of confidence.

First, we need to find the proportion of people who support candidate A in the preselected poll:

Proportion of support for A = 1 - Proportion of support for B or C 
= 1 - (22/100 + 14/100) 
= 1 - 0.36 
= 0.64

Next, we need to find the critical value for a 95% confidence interval. Using a standard normal distribution table or calculator, we find the critical value to be 1.96.

Then, we use the formula for the margin of error:

Margin of Error = Z* * sqrt(pq/n)

where Z* is the critical value, p is the proportion of support for candidate A, q is 1-p, and n is the sample size.

Re-arranging the formula gives us:

n = (Z*/ ME)^2 * pq

Plugging in the values, we get:

n = (1.96/0.06)^2 * 0.64 * (1-0.64) 
n = 246.13

Rounding up to the nearest whole number gives us a minimum sample size of 246, which is answer choice D.

Question 18

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

The answer of A Monster.com pool of 3,057 individuals asked:...

Question 19

The minimum sample size n required to estimate a population mean with 95% confidence and the assumed estimate of the population standard deviation 6.5 was found to be 124. Which of the following is the approximate value of the assumed desired margin of error?
A) D = 0.9220
B) D = 0.9602
C) D = 1.1441
D) D = 1.3090

Accepted Answer

The answer of The minimum sample size n required to...

Question 20

Find the minimum sample size when we want to construct a 90% confidence interval on the population proportion for the support of candidate A in the following mayoral election. Candidate A is facing two opposing candidates. In a preselected poll of 100 residents, 22 supported her. The desired margin of error is 0.08.
A) n = 44
B) n = 72
C) n = 73
D) n = 103

Accepted Answer

We can use the formula for the margin of error, E, of a confidence interval for a population proportion:

E = z*sqrt(p̂(1-p̂)/n)

where p̂ is the sample proportion, n is the sample size, and z is the z-score corresponding to the desired confidence level.

Since we want a 90% confidence interval, we use a z-score of 1.645 (from a standard normal distribution table). We also know that p̂ = 22/100 = 0.22 (the proportion of support for candidate A in the sample).

Plugging in these values and solving for n, we get:

0.08 = 1.645*sqrt(0.22(1-0.22)/n)

Squaring both sides and multiplying by n gives:

n = (1.645^2)*0.22(1-0.22)/(0.08^2)

n ≈ 72.8708

Rounding up to the next integer, we get a minimum sample size of 73, which corresponds to choice C.

Quiz 8: Interval Estimation

Suppose we wish to find the required sample size to find a 90% confidence interval for the population proportion with the desired margin of error. If there is no rough estimate of the population proportion, what value should be assumed for ?

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 ________.

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 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 ________.

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 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 ________.

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 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 marketing firm wants to estimate how much root beer the average teenager drinks per year. A previous study found a standard deviation of 1.12 liters. How many teenagers must the firm interview in order to have a margin of error of at most 0.1 liter when constructing a 99% confidence interval?

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 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 mean with 95% confidence and the desired margin of error 1.5 was found to be 198. Which of the following is the approximate value of the assumed estimate of the population standard deviation?

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 ________.

The minimum sample size n required to estimate a population mean with 95% confidence and the assumed estimate of the population standard deviation 6.5 was found to be 124. Which of the following is the approximate value of the assumed desired margin of error?