Question 1

A random sample of 25 roundtrip flights between Philadelphia and Dallas has an average airfare of $393.50 with a sample standard deviation of $50.30. The critical value for an 80% confidence interval around this sample mean is ------.

Accepted Answer

The critical value for an 80% confidence interval can be found using a t-distribution table or calculator because the population standard deviation is unknown and the sample size is relatively small (n=25). For an 80% confidence level and a sample size of 25, the degrees of freedom (df) would be 24 (n-1). Looking up the critical value for 80% confidence and 24 df in a t-distribution table gives us a value close to 1.318.

Question 2

A random sample of 13 hotels in Boston had an average nightly room rate of $172.10 with a sample standard deviation of $23.90. The approximate standard error of the mean for this sample is --------.

Accepted Answer

The standard error of the mean (SEM) is calculated using the formula SEM = s / √n, where s is the sample standard deviation and n is the sample size. Plugging in the values given: SEM = 23.90 / √13 ≈ 6.62.

Question 3

Steve'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 ------.

Accepted Answer

The 90% confidence interval for a population mean, when the population standard deviation is known, is 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 desired confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 90% confidence level, the Z-score is approximately 1.645. Plugging in the given values: $93.6 \pm 1.645 \frac{4.2}{\sqrt{34}}$, we get the interval (92.4, 94.8).

Question 4

John Charleston owns a heating oil delivery company that services 1,540 residences. John would like to estimate the number of customers that have oil storage tanks that are less than 100 gallons in capacity. A random sample of 200 customers found that 48 had an oil tank that was less than 100 gallons. The 99% confidence interval to estimate the proportion of customers with oil tanks less than 100 gallons is ------.

Accepted Answer

The 99% confidence interval for the proportion is calculated using the formula for a proportion confidence interval, which includes the sample proportion, the z-value for the confidence level (which is approximately 2.576 for 99%), and the standard error of the proportion. The sample proportion is 48/200 = 0.24, and the standard error is sqrt((0.24*(1-0.24))/200). Plugging these values into the formula gives an interval that matches choice B, which is wider than the others, reflecting the higher confidence level.

Question 5

A random sample of 150 mortgages in the state of Mississippi was randomly selected. From this sample, 17 were found to be delinquent on their current payment. The approximate standard error of the proportion for this sample is -------.

Accepted Answer

The standard error of the proportion is calculated using the formula $\sqrt{\frac{p(1-p)}{n}}$, where $p$ is the sample proportion and $n$ is the sample size. Here, $p = \frac{17}{150}$ and $n = 150$. Plugging these values into the formula gives $\sqrt{\frac{(17/150)(1-17/150)}{150}}$, which approximates to 0.0258.

Question 6

A local church has a congregation consisting of 480 adult members. The church's administrators would like to estimate the average weekly donation per member per week. A random sample of 90 members was found to have an average donation of $56.70 and a standard deviation of $10.50. The 90% confidence interval to estimate the average weekly donation is ______ .

Accepted Answer

The 90% confidence interval for the average weekly donation can be calculated using the formula for the confidence interval of the mean, which is $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $t_{\alpha/2}$ is the t-score corresponding to the desired confidence level (90% in this case), $s$ is the sample standard deviation, and $n$ is the sample size. Given $\bar{x} = 56.70$, $s = 10.50$, and $n = 90$, and knowing that the t-score for 90% confidence with 89 degrees of freedom (since $n-1 = 89$) is approximately 1.645 (though this value might slightly vary, it's close to the value for a large sample size or using a z-score as an approximation for large n), the calculation yields a confidence interval close to the values in option B, indicating that the average weekly donation is estimated with 90% confidence to be between $55.04 and $58.36.

Question 7

When calculating a confidence interval for the mean using a sample size of less than 30 with an unknown population standard deviation, the ------- is used.

Accepted Answer

When the sample size is less than 30 and the population standard deviation is unknown, the Student's t-distribution is used to calculate a confidence interval for the mean. This is because the t-distribution accounts for the additional uncertainty introduced by estimating the population standard deviation from the sample.

Question 8

Verizon Wireless would like to estimate the proportion of households that use cell phones for their phone service without a land line. A random sample of 150 households was selected and 48 relied strictly on cell phones for their service. The margin of error for a 90% confidence interval for the proportion based on this sample is ------.

Accepted Answer

The margin of error (ME) for a proportion is calculated using the formula: $ME = z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z$ is the z-score corresponding to the confidence level. Here, $\hat{p} = \frac{48}{150}$, $n = 150$, and for a 90% confidence interval, the z-score is approximately 1.645. Plugging these values into the formula gives $ME = 1.645 \times \sqrt{\frac{(48/150)(1-48/150)}{150}} \approx 0.0627$.

Question 9

A random sample of 13 hotels in Boston had an average nightly room rate of $172.10 with a sample standard deviation of $23.90. The 99% confidence interval around this sample mean is -------- .

Accepted Answer

The answer of A random sample of 13 hotels in...

Question 10

A random sample of 25 roundtrip flights between Philadelphia and Dallas has an average airfare of $393.50 with a sample standard deviation of $50.30. The margin of error for a 90% confidence interval around this sample mean is -----.

Accepted Answer

The margin of error (ME) for a confidence interval can be calculated using the formula: ME = z * (σ/√n), where z is the z-score corresponding to the desired confidence level (for 90% confidence, z ≈ 1.645), σ is the sample standard deviation, and n is the sample size. Plugging in the given values: ME = 1.645 * (50.3/√25) = 1.645 * 10.06 ≈ 17.2.

Question 11

Steve's average golf score at his local course is 92.9 from a random sample of 60 rounds of golf. Assume that the population standard deviation for his golf score is 4.2. The standard error of the mean for this sample is ---------- .

Accepted Answer

The standard error of the mean (SEM) is calculated by dividing the population standard deviation by the square root of the sample size: SEM = σ / √n. Here, σ = 4.2 and n = 60. Thus, SEM = 4.2 / √60 ≈ 0.54.

Question 12

A random sample of 16 roundtrip flights between Philadelphia and Dallas has an average airfare of $422.10 with a sample standard deviation of $56.30. The approximate standard error of the mean for this sample is --------.

Accepted Answer

The standard error of the mean (SEM) is calculated by dividing the sample standard deviation by the square root of the sample size: SEM = $56.30 / sqrt(16) = $56.30 / 4 = $14.075. Rounded to two decimal places, this gives $14.08.

Question 13

A random sample of 16 roundtrip flights between Philadelphia and Dallas has an average airfare of $422.10 with a sample standard deviation of $56.30. The 98% confidence interval around this sample mean is --------.

Accepted Answer

To calculate the 98% confidence interval for the mean, we use the formula: mean ± (critical value) * (standard deviation / sqrt(n)). For a 98% confidence level and a sample size of 16, we use the t-distribution because the population standard deviation is unknown. The degrees of freedom (df) is n-1 = 15. Looking up the t-table for df = 15 and a 98% confidence level (or 1% on each tail since it's two-tailed), the critical value is approximately 2.602. Plugging in the numbers: $422.10 ± 2.602 * ($56.30 / sqrt(16)) gives us the interval ($385.48, $458.72).

Question 14

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 -------.

Accepted Answer

The sample size needed for estimating a proportion with a specified margin of error (E) at a certain confidence level can be calculated using the formula: $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (for 95% confidence, $Z = 1.96$), $p$ is the estimated proportion (if unknown, 0.5 is used for maximum sample size), and $E$ is the margin of error. Plugging the values into the formula: $n = \frac{1.96^2 \cdot 0.5 \cdot (1-0.5)}{0.04^2} = 600.25$, which rounds up to 601.

Question 15

A random sample of 25 roundtrip flights between Philadelphia and Dallas has an average airfare of $393.50 with a sample standard deviation of $50.30. The 95% confidence interval around this sample mean is ------ .

Accepted Answer

The answer of A random sample of 25 roundtrip flights...

Question 16

The Girl Scouts would like to estimate the proportion of people who consider Thin Mints to be their favorite Girl Scout cookie. A pilot sample of 60 cookie customers found that 15 favored Thin Mint cookies. The total sample size needed to construct a 90% confidence interval for the proportion of customers who favor Thin Mint cookies with a margin of error equal to 6% is ----.

Accepted Answer

The sample size needed to construct a 90% confidence interval for the proportion of customers who favor Thin Mint cookies with a margin of error equal to 6% is 141. This can be calculated using the formula: n = (Z^2 * p * q) / e^2, where Z is the z-score for the desired confidence level (90% in this case), p is the sample proportion (15/60 in this case), q is 1-p, and e is the margin of error (0.06 in this case).

Question 17

A random sample of 17 hotels in Boston had an average nightly room rate of $165.40 with a sample standard deviation of $21.70. The critical value for a 98% confidence interval around this sample mean is -------.

Accepted Answer

The critical value for a 98% confidence interval with 16 degrees of freedom is 2.583.

Question 18

The U.S. Census Bureau would like to estimate the average square footage of a new single-family home with a 95% confidence interval and a margin of error within plus or minus 100 square feet. Assuming the standard deviation for the square footage of new single-family homes is 850 square feet, the required sample size is ----------.

Accepted Answer

To calculate the required sample size for estimating a population mean with a specific confidence level and margin of error, the formula is: $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, 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 95% confidence interval, the Z-score is approximately 1.96. Given $\sigma = 850$ square feet and $E = 100$ square feet, substituting these values into the formula gives: $n = \left(\frac{1.96 \cdot 850}{100}\right)^2 \approx 278$. Therefore, a sample size of approximately 278 is required.

Question 19

A ________ for the mean is an interval estimate around a sample mean that provides us with a range of where the true population mean lies.

Accepted Answer

A confidence interval for the mean is an interval estimate that is used to indicate the reliability of an estimate. It provides a range of values within which the true population mean is expected to lie with a certain level of confidence.

Question 20

The U.S. Census Bureau would like to estimate the average square footage of a new single-family home with a 95% confidence interval and a margin of error within plus or minus 200 square feet. Assuming the standard deviation for the square footage of new single-family homes is 850 square feet, the required sample size is ---------.

Accepted Answer

The formula for the sample size is: n = (Z^2 * s^2) / e^2, where Z is the z-score for the desired confidence level, s is the standard deviation, and e is the margin of error. Plugging in the given values, we get: n = (1.96^2 * 850^2) / 200^2 = 70.45. Rounding up to the nearest whole number, the required sample size is 70.

Quiz 8: Nutrition