Question 1

Choose the one alternative that best completes the statement or answers the question.
-The cell phone conversations of a random sample of 50 students has a standard deviation of 9.7 minutes. Find the margin of error, E, using a 90% confidence interval.

Accepted Answer

The margin of error is E = z * (s/√n), where z is the z-score corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size. For a 90% confidence level, z = 1.645. Plugging in the given values, we get E = 1.645 * (9.7/√50) = 2.26.

Question 2

Choose the one alternative that best completes the statement or answers the question.
-As the level of confidence increases the margin of error decreases.

Accepted Answer

As the level of confidence increases, the margin of error decreases because a higher level of confidence requires a wider range of values to be included in the confidence interval, which in turn reduces the margin of error.

Question 3

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error
-As the level of confidence increases the number of item to be included in a sample will decrease when the error and the standard deviation are held constant.

Accepted Answer

As the level of confidence increases, the margin of error decreases, which means that the sample size needed to achieve a specified margin of error will also decrease.

Question 4

Choose the one alternative that best completes the statement or answers the question.
-What statistical values does the margin of error depend upon

Accepted Answer

The answer of Choose the one alternative that best completes...

Question 5

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error
-$\mathrm { E } = 4 , \sigma = 35 \text {, } ( 1 - \alpha ) = 0.90$

Accepted Answer

To determine the sample size necessary for estimating the population mean within a specified margin of error, we use the formula $n = \left( \frac{Z \sigma}{E} \right)^2$, where $n$ is the sample size, $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $E$ is the margin of error. For a 90% confidence level, the Z-score is approximately 1.645. Plugging in the values: $n = \left( \frac{1.645 \times 35}{4} \right)^2 \approx 207.4$. Since the sample size must be a whole number, we round up to the nearest whole number, which is 208.

Question 6

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error
-$$\mathrm { E } = 3 , \sigma = 7.75 , ( 1 - \alpha ) = 0.95$$

Accepted Answer

The answer of Determine the Sample Size Necessary for Estimating...

Question 7

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error
-The grade point averages for 10 randomly selected students in an algebra class with 125 students are listed below. What is the effect on the width of the confidence interval if the sample size is increased to 20 ?
$\begin{array} { l l l l l l l l l l } 2.0 & 3.2 & 1.8 & 2.9 & 0.9 & 4.0 & 3.3 & 2.9 & 3.6 & 0.8 \end{array}$

Accepted Answer

Increasing the sample size generally results in a narrower confidence interval because the standard error of the mean decreases as the sample size increases.

Question 8

State the Properties of Student s t-Distribution
-A computer package was used to generate the following printout for estimating the sale price of condominiums in a particular neighborhood.

A friend suggests that the mean sale price of homes in this neighborhood is $\$ 49,000$. Comment on your friend's suggestion.

Accepted Answer

The answer of State the Properties of Student s t-Distribution
-A...

Question 9

State the Properties of Student s t-Distribution
-Which of the following is not a characteristic of Students t distribution

Accepted Answer

The mean of Student's t-distribution is zero, not 1.

Question 10

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error
-In order to set rates, an insurance company is trying to estimate the number of sick days that full time workers at a local bank take per year. A previous study indicated that the standard deviation was 2.5 days. a) How large a sample must be selected if the company wants to be 98% confident that the true mean differs from the sample mean by no more than 1 day b) Repeat part (a) using a 99% confidence interval. Which level of confidence requires a larger sample size Explain.

Accepted Answer

The answer of Determine the Sample Size Necessary for Estimating...

Question 11

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error
-A doctor at a local hospital is interested in estimating the birth weight of infants. How large a sample must she select if she desires to be 98% confident that the true mean is within 2 ounces of the sample mean The standard deviation of the birth weights is known to be 8 ounces.

Accepted Answer

To determine the sample size necessary for estimating the population mean within a specified margin of error, we use the formula: $n = \left(\frac{Z \cdot \sigma}{E}\right)^2$, where $n$ is the sample size, $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $E$ is the margin of error. For a 98% confidence level, the Z-score is approximately 2.33. Given $\sigma = 8$ ounces and $E = 2$ ounces, substituting these values into the formula gives $n = \left(\frac{2.33 \cdot 8}{2}\right)^2 = 86.49$. Since the sample size must be a whole number, we round up to the nearest whole number, which is 87.

Question 12

State the Properties of Student s t-Distribution
-To select the correct Student's t-distribution requires knowing the degrees of freedom. How many degrees of freedom are there for a sample of size n?

Accepted Answer

The degrees of freedom for a sample of size n in the context of Student's t-distribution is $n - 1$. This is because one degree of freedom is lost when calculating the sample mean.

Question 13

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error
-$$\mathrm { E } = 20 , \sigma = 175 , ( 1 - \alpha ) = 0.99$$

Accepted Answer

To determine the sample size necessary for estimating the population mean within a specified margin of error, we use the formula: $n = \left(\frac{Z_{\alpha/2} \cdot \sigma}{E}\right)^2$. For a confidence level of 99% ($1 - \alpha = 0.99$), the $Z_{\alpha/2}$ value is approximately 2.576 (from Z-tables for a two-tailed test). Given $\sigma = 175$ and $E = 20$, we substitute these values into the formula: $n = \left(\frac{2.576 \cdot 175}{20}\right)^2$, which calculates to approximately 508.

Question 14

State the Properties of Student s t-Distribution
-Every Student s t-distribution with n < N, n the number in the sample and N the number in the population, will be less peaked and have thinner tails.

Accepted Answer

The Student's t-distribution is less peaked and has thinner tails than the normal distribution, especially when the sample size is small.

Question 15

State the Properties of Student s t-Distribution
-A computer package was used to generate the following printout for estimating the sale price of condominiums in a particular neighborhood.

What assumptions are necessary for any inferences derived from this printout to be valid?

Accepted Answer

The answer of State the Properties of Student s t-Distribution
-A...

Question 16

Explain the Role of Margin of Error in Constructing a Confidence Interval
-A certain confidence in interval is 8.15 < < 9.45. Find the sample mean x and the margin of error E.

Accepted Answer

The answer of Explain the Role of Margin of Error...

Question 17

State the Properties of Student s t-Distribution
-Suppose a $99 \%$ confidence interval for $\mu$ turns out to be $( 1000,2100 )$. If this interval was based on a sample of size $n = 26$, explain what assumptions are necessary for this interval to be valid.

Accepted Answer

The assumption necessary for the interval to be valid is that the population from which the sample is drawn must have an approximately normal distribution, especially since the sample size is relatively small (n = 26).

Question 18

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error
-The principal at Riverside High School would like to estimate the mean length of time each day that it takes all the buses to arrive and unload the students. How large a sample is needed if the principal would like to assert with 90% confidence that the sample mean is off by, at most, 7 minutes. Assume = 14 minutes.

Accepted Answer

The sample size needed is given by the formula:n = (Z^2 * s^2) / E^2where Z is the z-score for the desired confidence level, s is the sample standard deviation, and E is the margin of error.In this case, the confidence level is 90%, so the z-score is 1.645. The sample standard deviation is 14 minutes, and the margin of error is 7 minutes. Plugging these values into the formula, we get:n = (1.645^2 * 14^2) / 7^2= 11.06Since we cannot have a fraction of a sample, we round up to the nearest whole number, which is 11. Therefore, the principal needs a sample size of 11 to estimate the mean length of time each day that it takes all the buses to arrive and unload the students with 90% confidence that the sample mean is off by, at most, 7 minutes.

Question 19

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error
-A standardized test has a mean of 104 and a standard deviation of 16. We want to be 98% certain that we are within 4 points of the true mean. Determine the required sample size.

Accepted Answer

The answer of Determine the Sample Size Necessary for Estimating...

Question 20

Explain the Role of Margin of Error in Constructing a Confidence Interval
-Suppose a 90% confidence interval for turns out to be (120, 280). Based on the interval, do you believe the average is equal to 300

Accepted Answer

The confidence interval of (120, 280) does not include 300, indicating that based on the data, we are 90% confident the true average is not 300.

Quiz 9: Estimating the Value of a Parameter Using Confidence Intervals

Choose the one alternative that best completes the statement or answers the question. -The cell phone conversations of a random sample of 50 students has a standard deviation of 9.7 minutes. Find the margin of error, E, using a 90% confidence interval.

Choose the one alternative that best completes the statement or answers the question. -As the level of confidence increases the margin of error decreases.

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error -As the level of confidence increases the number of item to be included in a sample will decrease when the error and the standard deviation are held constant.

Choose the one alternative that best completes the statement or answers the question. -What statistical values does the margin of error depend upon

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error - $\mathrm { E } = 4 , \sigma = 35 \text {, } ( 1 - \alpha ) = 0.90$

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error - $\mathrm { E } = 3 , \sigma = 7.75 , ( 1 - \alpha ) = 0.95$

State the Properties of Student s t-Distribution -Which of the following is not a characteristic of Students t distribution

State the Properties of Student s t-Distribution -To select the correct Student's t-distribution requires knowing the degrees of freedom. How many degrees of freedom are there for a sample of size n?

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error - $\mathrm { E } = 20 , \sigma = 175 , ( 1 - \alpha ) = 0.99$

State the Properties of Student s t-Distribution -Every Student s t-distribution with n < N, n the number in the sample and N the number in the population, will be less peaked and have thinner tails.

State the Properties of Student s t-Distribution -A computer package was used to generate the following printout for estimating the sale price of condominiums in a particular neighborhood. What assumptions are necessary for any inferences derived from this printout to be valid?

Explain the Role of Margin of Error in Constructing a Confidence Interval -A certain confidence in interval is 8.15 < < 9.45. Find the sample mean x and the margin of error E.

State the Properties of Student s t-Distribution -Suppose a $99 \%$ confidence interval for $\mu$ turns out to be $( 1000,2100 )$ . If this interval was based on a sample of size $n = 26$ , explain what assumptions are necessary for this interval to be valid.

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error -A standardized test has a mean of 104 and a standard deviation of 16. We want to be 98% certain that we are within 4 points of the true mean. Determine the required sample size.

Explain the Role of Margin of Error in Constructing a Confidence Interval -Suppose a 90% confidence interval for turns out to be (120, 280) . Based on the interval, do you believe the average is equal to 300

Quiz 9: Estimating the Value of a Parameter Using Confidence Intervals

Choose the one alternative that best completes the statement or answers the question. -The cell phone conversations of a random sample of 50 students has a standard deviation of 9.7 minutes. Find the margin of error, E, using a 90% confidence interval.

Choose the one alternative that best completes the statement or answers the question. -As the level of confidence increases the margin of error decreases.

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error -As the level of confidence increases the number of item to be included in a sample will decrease when the error and the standard deviation are held constant.

Choose the one alternative that best completes the statement or answers the question. -What statistical values does the margin of error depend upon

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error - E=4,σ=35, (1−α) =0.90\mathrm { E } = 4 , \sigma = 35 \text {, } ( 1 - \alpha ) = 0.90E=4,σ=35, (1−α) =0.90

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error - E=3,σ=7.75,(1−α) =0.95\mathrm { E } = 3 , \sigma = 7.75 , ( 1 - \alpha ) = 0.95E=3,σ=7.75,(1−α) =0.95

State the Properties of Student s t-Distribution -Which of the following is not a characteristic of Students t distribution

State the Properties of Student s t-Distribution -To select the correct Student's t-distribution requires knowing the degrees of freedom. How many degrees of freedom are there for a sample of size n?

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error - E=20,σ=175,(1−α) =0.99\mathrm { E } = 20 , \sigma = 175 , ( 1 - \alpha ) = 0.99E=20,σ=175,(1−α) =0.99

State the Properties of Student s t-Distribution -Every Student s t-distribution with n < N, n the number in the sample and N the number in the population, will be less peaked and have thinner tails.

State the Properties of Student s t-Distribution -A computer package was used to generate the following printout for estimating the sale price of condominiums in a particular neighborhood. What assumptions are necessary for any inferences derived from this printout to be valid?

Explain the Role of Margin of Error in Constructing a Confidence Interval -A certain confidence in interval is 8.15 < < 9.45. Find the sample mean x and the margin of error E.

State the Properties of Student s t-Distribution -Suppose a 99%99 \%99% confidence interval for μ\muμ turns out to be (1000,2100) ( 1000,2100 ) (1000,2100) . If this interval was based on a sample of size n=26n = 26n=26 , explain what assumptions are necessary for this interval to be valid.

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error -A standardized test has a mean of 104 and a standard deviation of 16. We want to be 98% certain that we are within 4 points of the true mean. Determine the required sample size.

Explain the Role of Margin of Error in Constructing a Confidence Interval -Suppose a 90% confidence interval for turns out to be (120, 280) . Based on the interval, do you believe the average is equal to 300

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error - $\mathrm { E } = 4 , \sigma = 35 \text {, } ( 1 - \alpha ) = 0.90$

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error - $\mathrm { E } = 3 , \sigma = 7.75 , ( 1 - \alpha ) = 0.95$

Determine the Sample Size Necessary for Estimating the Population Mean within a Specified Margin of Error - $\mathrm { E } = 20 , \sigma = 175 , ( 1 - \alpha ) = 0.99$

State the Properties of Student s t-Distribution -Suppose a $99 \%$ confidence interval for $\mu$ turns out to be $( 1000,2100 )$ . If this interval was based on a sample of size $n = 26$ , explain what assumptions are necessary for this interval to be valid.