Question 1

Choose the one alternative that best completes the statement or answers the question.
-In a recent study of 42 eighth graders, the mean number of hours per week that they played video games was $19.6$ with a population standard deviation of $5.8$ hours. Compute the $98 \%$ confidence interval for $\mu$.
A) $( 17.5,21.7 )$
B) $( 14.1,23.2 )$
C) $( 18.3,20.9 )$
D) $( 19.1,20.4 )$

Accepted Answer

To compute the 98% confidence interval for the mean $\mu$, we use the formula for the confidence interval when the population standard deviation is known: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 98% confidence level, the Z-score is approximately 2.33. Plugging in the given values: $19.6 \pm 2.33 \times \frac{5.8}{\sqrt{42}}$, we get the interval $(17.5, 21.7)$.

Question 2

Choose the one alternative that best completes the statement or answers the question.
-A confidence interval with a 95% level of confidence means that the parameter will be in the interval for 95 out of 100 samples of the same size.

Accepted Answer

A 95% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect the true population parameter to fall within 95 of those intervals.

Question 3

Write the word or phrase
-Private colleges and universities rely on money contributed by individuals and corporations for their operating expenses. Much of this money is put into a fund called an endowment, and the college spends only the interest earned by the fund. A recent survey of eight private colleges in the United States revealed the following endowments (in millions of dollars): $77.1,41.6,248.6,497.2,119,173.9,108.5$, and $212.4$. What value will be used as the point estimate for the mean endowment of all private colleges in the United States?
A) $184.788$
B) $1478.3$
C) $211.186$
D) 8

Accepted Answer

The point estimate for the mean endowment of all private colleges in the United States is calculated by finding the average of the given endowments. Adding all the endowments together and dividing by the number of colleges (8) gives the mean endowment, which is $184.788$ million dollars.

Question 4

Construct and Interpret a Confidence Interval for a Pop Mean, Assuming that the Pop Std Dev is Known
-In a recent study of 86 eighth graders, the mean number of hours per week that they played video games was 21.6. Assume $\sigma = 5.7$ hours.
a) Compute the $90 \%$ confidence interval for $\mu$.
b) If the standard deviation is doubled to $11.4$, what will be the effect on the confidence interval?

Accepted Answer

The answer of Construct and Interpret a Confidence Interval for...

Question 5

Choose the one alternative that best completes the statement or answers the question.
-A point estimate is the value of a_____ that estimates the value of a_____
A) statistic; parameter
B) parameter; statistic
C) random variable; statistic
D) random variable; parameter

Accepted Answer

A point estimate is the value of a statistic (such as a sample mean) that is used to estimate the value of a parameter (such as a population mean).

Question 6

Choose the one alternative that best completes the statement or answers the question.
-A group of 49 randomly selected parrots has a mean age of $22.4$ years with a population standard deviation of 3.8. Compute the $98 \%$ confidence interval for $\mu$.
A) $( 21.1,23.7 )$
B) $( 20.3,24.5 )$
C) $( 19.8,25.1 )$
D) $( 18.8,26.3 )$

Accepted Answer

To calculate the 98% confidence interval for the mean age of the parrots, we use the formula for the confidence interval when the population standard deviation is known: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 98% confidence level, the Z-score is approximately 2.33. Plugging in the given values: $22.4 \pm 2.33 \times \frac{3.8}{\sqrt{49}}$, we get the interval $(21.1, 23.7)$.

Question 7

Choose the one alternative that best completes the statement or answers the question.
-The critical value of a distribution is the value of
A) $\mathrm { z } _ { \alpha / 2 }$
B) the parameter
C) $\mathrm { z } _ { \alpha / 2 } \frac { \sigma } { \sqrt { \mathrm { n } } }$
D) the area of the tail region of the distribution

Accepted Answer

The critical value of a distribution, often denoted as $\mathrm { z } _ { \alpha / 2 }$, is a point on the distribution beyond which lies the alpha/2 portion of the tail(s). It is used in hypothesis testing and confidence interval construction.

Question 8

Choose the one alternative that best completes the statement or answers the question.
-Determine the point estimate of the population mean for the confidence interval with a lower bound of 19 and an upper bound of 25 .
A) 22
B) 19
C) 25
D) 23

Accepted Answer

The point estimate of the population mean is the midpoint of the confidence interval, which is calculated as (19 + 25) / 2 = 22.

Question 9

Choose the one alternative that best completes the statement or answers the question.
-A random sample of 150 towns in a western state had an average annual precipitation of $2.86$ inches and a population standard deviation of $0.78$ inches. Compute the $98 \%$ confidence interval for $\mu$.
A) $( 2.71,3.01 )$
B) $( 2.51,3.53 )$
C) $( 2.43,3.79 )$
D) $( 2.31,3.88 )$

Accepted Answer

To compute the 98% confidence interval for the population mean $\mu$, we use the formula for the confidence interval when the population standard deviation is known, which is $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 98% confidence level, the Z-score is approximately 2.33. Plugging in the given values: $2.86 \pm 2.33 \times \frac{0.78}{\sqrt{150}}$, we get the interval $(2.71, 3.01)$.

Question 10

Choose the one alternative that best completes the statement or answers the question.
-The margin of error of a confidence interval is
A) $\pm z _ { \alpha / 2 } \frac { \sigma } { \sqrt { n } }$
B) $\mathrm { z } _ { \alpha / 2 }$
C) $\bar { x }$
D) $\frac { \sigma } { \sqrt { n } }$

Accepted Answer

The margin of error of a confidence interval is given by $\pm z _ { \alpha / 2 } \frac { \sigma } { \sqrt { n } }$, where $z _ { \alpha / 2 }$ is the critical value from the standard normal distribution, $\sigma$ is the population standard deviation, and $n$ is the sample size. This formula adjusts the critical value by the standard error of the mean to calculate the margin of error.

Question 11

Choose the one alternative that best completes the statement or answers the question.
-A random sample of 40 part-time workers has a mean annual earnings of $\$ 3120$ and a population standard deviation of $\$ 677$. Compute the $95 \%$ confidence interval for $\mu$.
A) $( \$ 2910 , \$ 3330 )$
B) $( \$ 210 , \$ 110 )$
C) $( \$ 4812 , \$ 5342 )$
D) $( \$ 1987 , \$ 2346 )$

Accepted Answer

To compute the 95% confidence interval for the mean ($\mu$) when the population standard deviation ($\sigma$) is known, we use the formula for the confidence interval based on the Z-distribution since the sample size is large enough (n=40). The formula is $\bar{x} \pm Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z_{\alpha/2}$ is the Z-score corresponding to the desired confidence level (for 95% confidence, $Z_{\alpha/2} = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values, we find that the interval $(\$2910, \$3330)$ is the correct answer, which corresponds to choice A.

Question 12

Construct and Interpret a Confidence Interval for a Pop Mean, Assuming that the Pop Std Dev is Known
-Compute the critical value $\mathrm { z } \alpha / 2$ that corresponds to a $99 \%$ level of confidence.
A) $2.575$
B) $1.96$
C) $1.645$
D) $2.33$

Accepted Answer

For a 99% confidence level, the critical value $z_{\alpha/2}$ is found in the z-table corresponding to the tails of the distribution (1 - 0.99 = 0.01, so 0.005 in each tail since it's two-tailed), which is approximately 2.575.

Question 13

Choose the one alternative that best completes the statement or answers the question.
-A random sample of 56 lithium batteries has a mean life of 645 hours with a population standard deviation of 31 hours. Compute the $95 \%$ confidence interval for $\mu$.
A) $( 636.9,653.1 )$
B) $( 539.6,551.2 )$
C) $( 112.0,118.9 )$
D) $( 712.0,768.0 )$

Accepted Answer

The $95\%$ confidence interval for the mean $\mu$ can be calculated using the formula for the confidence interval when the population standard deviation is known: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, 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: $645 \pm 1.96 \frac{31}{\sqrt{56}}$, we get the interval $(636.9, 653.1)$.

Question 14

Choose the one alternative that best completes the statement or answers the question.
-In a random sample of 60 laptop computers, the mean repair cost was $\$ 150$ with a population standard deviation of $\$ 36$. Compute the $99 \%$ confidence interval for $\mu$.
A) $( \$ 138 , \$ 162 )$
B) $( \$ 18 , \$ 54 )$
C) $( \$ 238 , \$ 274 )$
D) $( \$ 537 , \$ 654 )$

Accepted Answer

To compute the 99% confidence interval for the mean repair cost, we use the formula for the confidence interval when the population standard deviation is known: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-value corresponding to the desired confidence level (for 99%, $Z \approx 2.576$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $150 \pm 2.576 \frac{36}{\sqrt{60}}$, we get the interval $(\$138, \$162)$.

Question 15

Explain the Role of Margin of Error in Constructing a Confidence Interval
-A 90% confidence interval for the mean percentage of airline reservations being canceled on the day of the flight is (1.2%, 4.5%). What is the point estimator of the mean percentage of reservations that are canceled on the day of the flight
A) 2.85%
B) 1.65%
C) 2.25%
D) 3.3%

Accepted Answer

The point estimator of the mean percentage of reservations that are canceled on the day of the flight is the midpoint of the confidence interval, which is calculated as (1.2% + 4.5%) / 2 = 2.85%.

Question 16

Write the word or phrase that best completes each statement or answers the question.
-The grade point averages for 10 randomly selected students in a statistics class with 125 students are listed below. What can you say about the mean score $\mu$ of all 125 students?
$$\begin{array} { l l l l l l l l l l } 
3.0 & 2.7 & 3.2 & 2.8 & 2.2 & 3.8 & 3.7 & 2.1 & 3.6 & 3.4
\end{array}$$

Accepted Answer

The answer of Write the word or phrase that best...

Question 17

Choose the one alternative that best completes the statement or answers the question.
-A confidence interval for a parameter is
A) An interval of numbers combined with the likelihood the interval contains the unknown parameter.
B) A point estimate plus a margin of error.
C) An interval of probabilities concerning a parameter.
D) A statement of believability of a statistical result.

Accepted Answer

A confidence interval is an interval of numbers, usually derived from sample statistics, that is likely to contain the true parameter of a population with a certain level of confidence. This definition aligns with option A, as it combines an interval of numbers (the confidence interval itself) with the likelihood (confidence level) that this interval actually contains the unknown parameter.

Question 18

Explain the Role of Margin of Error in Constructing a Confidence Interval
-Suppose a 95% confidence interval for turns out to be (130, 300). To make more useful inferences from the data, it is desired to reduce the width of the confidence interval. Which of the following will result in a reduced interval width
A) All of these.
B) Increase the sample size.
C) Decrease the confidence level.
D) Increase the sample size and decrease the confidence level.

Accepted Answer

Increasing the sample size and decreasing the confidence level both reduce the width of the confidence interval, so all options listed contribute to a reduced interval width.

Question 19

Choose the one alternative that best completes the statement or answers the question.
-In a sample of 10 randomly selected employees, it was found that their mean height was $63.4$ inches. From previous studies, it is assumed that the standard deviation, $\sigma$, is $2.4$. Compute the $95 \%$ confidence interval for $\mu .$
A) $( 61.9,64.9 )$
B) $( 58.1,67.3 )$
C) $( 59.7,66.5 )$
D) $( 60.8,65.4 )$

Accepted Answer

To compute the 95% confidence interval for the population mean $\mu$, we use 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 (for 95%, $Z = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $63.4 \pm 1.96 \frac{2.4}{\sqrt{10}}$ gives the interval $(61.9, 64.9)$.

Question 20

Choose the one alternative that best completes the statement or answers the question.
-A group of 40 car owners calculated that their average repair bill was $\$ 192$ with a population standard deviation of $\$ 8$. Compute the $95 \%$ confidence interval for the mean repair bill of all the owners.
A) $( \$ 189.50 , \$ 194.50 )$
B) $( \$ 186.50 , \$ 197.50 )$
C) $( \$ 188.50 , \$ 195.60 )$
D) $( \$ 187.30 , \$ 196.10 )$

Accepted Answer

The $95\%$ confidence interval for the mean 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 confidence level (for $95\%$, $z = 1.96$), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values: $192 \pm 1.96 \times \frac{8}{\sqrt{40}}$ results in the interval $(\$189.50, \$194.50)$.

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

Choose the one alternative that best completes the statement or answers the question. -A confidence interval with a 95% level of confidence means that the parameter will be in the interval for 95 out of 100 samples of the same size.

Choose the one alternative that best completes the statement or answers the question. -A point estimate is the value of a_ that estimates the value of a_

Choose the one alternative that best completes the statement or answers the question. -A group of 49 randomly selected parrots has a mean age of $22.4$ years with a population standard deviation of 3.8. Compute the $98 \%$ confidence interval for $\mu$ .

Choose the one alternative that best completes the statement or answers the question. -The critical value of a distribution is the value of

Choose the one alternative that best completes the statement or answers the question. -Determine the point estimate of the population mean for the confidence interval with a lower bound of 19 and an upper bound of 25 .

Choose the one alternative that best completes the statement or answers the question. -A random sample of 150 towns in a western state had an average annual precipitation of $2.86$ inches and a population standard deviation of $0.78$ inches. Compute the $98 \%$ confidence interval for $\mu$ .

Choose the one alternative that best completes the statement or answers the question. -The margin of error of a confidence interval is

Choose the one alternative that best completes the statement or answers the question. -A random sample of 40 part-time workers has a mean annual earnings of $\$ 3120$ and a population standard deviation of $\$ 677$ . Compute the $95 \%$ confidence interval for $\mu$ .

Construct and Interpret a Confidence Interval for a Pop Mean, Assuming that the Pop Std Dev is Known -Compute the critical value $\mathrm { z } \alpha / 2$ that corresponds to a $99 \%$ level of confidence.

Choose the one alternative that best completes the statement or answers the question. -A random sample of 56 lithium batteries has a mean life of 645 hours with a population standard deviation of 31 hours. Compute the $95 \%$ confidence interval for $\mu$ .

Choose the one alternative that best completes the statement or answers the question. -In a random sample of 60 laptop computers, the mean repair cost was $\$ 150$ with a population standard deviation of $\$ 36$ . Compute the $99 \%$ confidence interval for $\mu$ .

Choose the one alternative that best completes the statement or answers the question. -A confidence interval for a parameter is

Choose the one alternative that best completes the statement or answers the question. -A group of 40 car owners calculated that their average repair bill was $\$ 192$ with a population standard deviation of $\$ 8$ . Compute the $95 \%$ confidence interval for the mean repair bill of all the owners.

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

Choose the one alternative that best completes the statement or answers the question. -A confidence interval with a 95% level of confidence means that the parameter will be in the interval for 95 out of 100 samples of the same size.

Choose the one alternative that best completes the statement or answers the question. -A point estimate is the value of a_____ that estimates the value of a_____

Choose the one alternative that best completes the statement or answers the question. -A group of 49 randomly selected parrots has a mean age of 22.422.422.4 years with a population standard deviation of 3.8. Compute the 98%98 \%98% confidence interval for μ\muμ .