Question 1

Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$ . Assume that the population has a normal distribution. $n = 30 , \bar { x } = 84.6 , \mathrm {~s} = 10.5,90 \%$ confidence

Accepted Answer

C

Question 2

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. $$n = 195 , x = 162 ; 95 \% \text { confidence }$$

Accepted Answer

B

Question 3

In constructing a confidence interval for $\sigma$ or $\sigma ^ { 2 }$, a table is used to find the critical values $\chi \frac { 2 } { L }$ and $\chi \frac { 2 } { R }$ for values of $n \leq 101$. For larger values of $n , \chi _ { L } ^ { 2 }$ and $\chi _ { R } ^ { 2 }$ can be approximated by using the following formula: $\chi ^ { 2 } =$ $\frac { 1 } { 2 } [ \pm z / 2 / 2 + \sqrt { 2 k - 1 } ] 2$ where $k$ is the number of degrees of freedom and $z \alpha / 2$ is the critical $z$-score. Construct the $90 \%$ confidence interval for $\sigma$ using the following sample data: a sample of size $n = 232$ yields a mean weight of $154 \mathrm { lb }$ and a standard deviation of $25.5 \mathrm { lb }$. Round the confidence interyal limits to the nearest hundredth.

Accepted Answer

C

Question 4

Fifty people are selected randomly from a certain population and it is found that 12 people in the sample are over six feet tall. What is the best point estimate of the proportion of people in the population who are over 6
Feet tall?

Accepted Answer

To determine the best point estimate of the proportion of people in the population who are over six feet tall, we use the formula: 
Point estimate = number of successes / sample size
In this case, the number of successes (people over six feet tall) is 12, and the sample size is 50. 
Point estimate = 12/50 = 0.24 
Therefore, the best point estimate of the proportion of people in the population who are over six feet tall is 0.24. Therefore, the answer is choice C.

Question 5

Find the critical value $\chi _ { \mathrm { R } } ^ { 2 }$ corresponding to a sample size of 19 and a confidence level of 99 percent.

Accepted Answer

Using a chi-square distribution table with 19 degrees of freedom and a cumulative probability of 0.99, we find the critical value to be approximately 37.156. Therefore, the correct answer is choice D.

Question 6

Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.
A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 185 milligrams with s = 17.6 milligrams. Construct a 95% confidence interval for the true mean cholesterol content of all such eggs.

Accepted Answer

We use the formula for a confidence interval for the population mean:

$
\bar{X} \pm t_{\alpha/2}\left(\dfrac{s}{\sqrt{n}}ight)
$

where $\bar{X}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, $t_{\alpha/2}$ is the critical value from the t-distribution with $n-1$ degrees of freedom (for a 95% confidence interval and $n-1=11$, $t_{\alpha/2}=2.201$).

Plugging in the given values, we get:

$
185 \pm 2.201 \left(\dfrac{17.6}{\sqrt{12}}ight) \
185 \pm 12.099 \
\Rightarrow 	ext{CI: } (173.9, 196.1)
$

Therefore, the best choice is C.

Question 7

Find the value of ${ } ^ { z } \alpha / 2$ that corresponds to a confidence level of 89.48%.

Accepted Answer

To find the value of ${ } ^{z}\alpha / 2$ for a confidence level of 89.48%, we can use a standard normal distribution table or calculator to find the z-score that corresponds to the area to the right of the z-score being 0.0556 (since 1 - 0.8948 = 0.1052, and we want the area to the left to be 0.0556).

Using a standard normal distribution table or calculator, we can find that the z-score that corresponds to an area of 0.0556 to the right is approximately 1.62. Since we are looking for ${ }^{z}\alpha / 2$, we need to find the value of alpha that corresponds to a total area of 0.1052 (twice the area to the right of the z-score).

Using a standard normal distribution table or calculator, we can find that the z-score that corresponds to an area of 0.0526 to the right (half of 0.1052) is approximately 1.44. Thus, ${ }^{z}\alpha / 2$ is approximately 1.44, or -1.44 (since the standard normal distribution is symmetric about 0).

Since none of the answer choices are exactly 1.44 or -1.44, we can choose the best answer choice by selecting the one that is closest. In this case, B (1.62) is the closest, so the answer is B.

Question 8

If the computed minimum sample size n needed for a particular margin of error is not a whole number, round the value of n ________ (up or down) to the next ________ (smaller or larger) whole number.

Accepted Answer

When calculating sample size, it is important to round up to the next larger whole number to ensure that the sample size is large enough to provide accurate results. Therefore, if the computed minimum sample size is not a whole number, rounding up to the next larger whole number (in choice C) is the best choice.

Question 9

Find the critical value $z _ { \alpha / 2 }$ that corresponds to a 91% confidence level.

Accepted Answer

For a 91% confidence level, the critical value $z _ { \alpha / 2 }$ is approximately 1.70, as it corresponds to the area in the tails of the standard normal distribution.

Question 10

Of 380 randomly selected medical students, 21 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

Accepted Answer

The sample proportion is 21/380 = 0.0553. Using the formula for a confidence interval for a proportion, the 95% confidence interval is calculated as 0.0553 ± 1.96 * sqrt((0.0553*(1-0.0553))/380), which results in the interval 0.0323 < p < 0.0782.

Question 11

Do one of the following, as appropriate: (a) Find the critical value $z _ { \alpha / 2 }$ , (b) find the critical value $t _ { \alpha / 2 }$ , (c) state that neither the normal nor the t distribution applies. $90 \%$ confidence interval for $\mu ; n = 9 ; \sigma = 4.2$; population appears to be very skewed.

Accepted Answer

Since the population appears to be very skewed, we cannot assume that it follows a normal distribution. Additionally, the sample size is small (n=9), so we cannot rely on the central limit theorem to justify the use of the normal distribution. Therefore, neither the normal nor the t distribution applies, and we cannot find the critical value for either. Instead, we would need to use alternative methods, such as bootstrapping or a non-parametric test.

Question 12

A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the true population proportion of all New York State union
Members who favor the Republican candidate.

Accepted Answer

The answer of A survey of 300 union members in...

Question 13

Identify the distribution that applies to the following situation: In constructing a confidence interval ofσ you have 50 sample values and they appear to be from a population with
A skewed distribution.

Accepted Answer

Since the sample values appear to be from a population with a skewed distribution, the use of the Student t distribution is appropriate as it is designed to handle situations where the population distribution is unknown or non-normal.

Question 14

Use the given data to find the minimum sample size required to estimate the population proportion. Margin of error: 0.028; confidence level: 99%; $p \text { and } q \text { unknown }$

Accepted Answer

The answer of Use the given data to find the...

Question 15

Use the given degree of confidence and sample data to construct a confidence interval for the population mean µ. Assume that the population has a normal distribution.
Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6,
Construct a 99% confidence interval for the mean score of all students.

Accepted Answer

The correct answer is A.The confidence interval is:95 ± 2.576 * (6.6 / sqrt(30))= 95 ± 3.32= (91.68, 98.32)

Question 16

Express the confidence interval $( 0.432,0.52 )$ in the form of $\hat { p } \pm \mathrm { E }$.

Accepted Answer

The midpoint of the interval $(0.432, 0.52)$ is $\hat{p} = \frac{0.432 + 0.52}{2} = 0.476$. The margin of error $E$ is half the width of the interval: $E = \frac{0.52 - 0.432}{2} = 0.044$. Thus, the confidence interval can be expressed as $0.476 \pm 0.044$.

Question 17

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. $$95 \% \text { confidence; } n = 2388 , x = 1672$$

Accepted Answer

First, we need to find the sample proportion: 
$$\hat{p}=\frac{x}{n}=\frac{1672}{2388}=0.7003$$
Next, we need to find the standard error: 
$$SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.7003(1-0.7003)}{2388}}=0.0132$$
To find the margin of error, we need to multiply the standard error by the appropriate z-score for the given confidence level. For 95% confidence, the z-score is 1.96. 
$$E=1.96	imes SE=1.96	imes 0.0132=0.0259\approx 0.026$$
Therefore, the margin of error is approximately 0.026. The closest answer choice is B, which is 0.0184.

Question 18

Which critical value is appropriate for a 99% confidence level where n = 17; $\sigma$ is unknown and the population appears to be normally distributed?

Accepted Answer

Since $\sigma$ is unknown and the sample size is small (n = 17), the t-distribution should be used. For a 99% confidence level with 16 degrees of freedom (n-1), the critical value is approximately 2.921.

Question 19

The margin of error ________ (increases or decreases) with an increased sample size and ________ (increases or decreases) with an increase in confidence level.

Accepted Answer

The margin of error decreases with an increased sample size, as a larger sample size provides a more accurate representation of the population. However, the margin of error increases with an increase in confidence level, as a higher confidence level requires a wider interval estimation. Therefore, choice C is correct as it correctly identifies how the margin of error changes with both sample size and confidence level.

Question 20

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places.
98% confidence; the sample size is 800, of which 40% are successes

Accepted Answer

The margin of error E is calculated using the formula E = Z * sqrt[(p(1-p))/n], where Z is the Z-score for the confidence level (2.33 for 98%), p is the sample proportion (0.40), and n is the sample size (800). Plugging in these values gives E ≈ 0.0404.

Quiz 7: Estimating Parameters and Determining Sample Sizes

Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$ . Assume that the population has a normal distribution. $n = 30 , \bar { x } = 84.6 , \mathrm {~s} = 10.5,90 \%$ confidence

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. $n = 195 , x = 162 ; 95 \% \text { confidence }$

Fifty people are selected randomly from a certain population and it is found that 12 people in the sample are over six feet tall. What is the best point estimate of the proportion of people in the population who are over 6 Feet tall?

Find the critical value $\chi _ { \mathrm { R } } ^ { 2 }$ corresponding to a sample size of 19 and a confidence level of 99 percent.

Find the value of ${ } ^ { z } \alpha / 2$ that corresponds to a confidence level of 89.48%.

If the computed minimum sample size n needed for a particular margin of error is not a whole number, round the value of n (up or down) to the next (smaller or larger) whole number.

Find the critical value $z _ { \alpha / 2 }$ that corresponds to a 91% confidence level.

Of 380 randomly selected medical students, 21 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the true population proportion of all New York State union Members who favor the Republican candidate.

Identify the distribution that applies to the following situation: In constructing a confidence interval ofσ you have 50 sample values and they appear to be from a population with A skewed distribution.

Use the given data to find the minimum sample size required to estimate the population proportion. Margin of error: 0.028; confidence level: 99%; $p \text { and } q \text { unknown }$

Express the confidence interval $( 0.432,0.52 )$ in the form of $\hat { p } \pm \mathrm { E }$ .

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. $95 \% \text { confidence; } n = 2388 , x = 1672$

Which critical value is appropriate for a 99% confidence level where n = 17; $\sigma$ is unknown and the population appears to be normally distributed?

The margin of error (increases or decreases) with an increased sample size and (increases or decreases) with an increase in confidence level.

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. 98% confidence; the sample size is 800, of which 40% are successes

Quiz 7: Estimating Parameters and Determining Sample Sizes

Use the given degree of confidence and sample data to construct a confidence interval for the population mean μ\muμ . Assume that the population has a normal distribution. n=30,xˉ=84.6, s=10.5,90%n = 30 , \bar { x } = 84.6 , \mathrm {~s} = 10.5,90 \%n=30,xˉ=84.6, s=10.5,90% confidence

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n=195,x=162;95% confidence n = 195 , x = 162 ; 95 \% \text { confidence }n=195,x=162;95% confidence

Fifty people are selected randomly from a certain population and it is found that 12 people in the sample are over six feet tall. What is the best point estimate of the proportion of people in the population who are over 6 Feet tall?

Find the critical value χR2\chi _ { \mathrm { R } } ^ { 2 }χR2​ corresponding to a sample size of 19 and a confidence level of 99 percent.

Find the value of zα/2{ } ^ { z } \alpha / 2zα/2 that corresponds to a confidence level of 89.48%.

If the computed minimum sample size n needed for a particular margin of error is not a whole number, round the value of n ________ (up or down) to the next ________ (smaller or larger) whole number.

Find the critical value zα/2z _ { \alpha / 2 }zα/2​ that corresponds to a 91% confidence level.

Of 380 randomly selected medical students, 21 said that they planned to work in a rural community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct the 98% confidence interval for the true population proportion of all New York State union Members who favor the Republican candidate.

Identify the distribution that applies to the following situation: In constructing a confidence interval ofσ you have 50 sample values and they appear to be from a population with A skewed distribution.

Use the given data to find the minimum sample size required to estimate the population proportion. Margin of error: 0.028; confidence level: 99%; p and q unknown p \text { and } q \text { unknown }p and q unknown

Express the confidence interval (0.432,0.52) ( 0.432,0.52 ) (0.432,0.52) in the form of p^±E\hat { p } \pm \mathrm { E }p^​±E .

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. 95% confidence; n=2388,x=167295 \% \text { confidence; } n = 2388 , x = 167295% confidence; n=2388,x=1672

Which critical value is appropriate for a 99% confidence level where n = 17; σ\sigmaσ is unknown and the population appears to be normally distributed?

The margin of error ________ (increases or decreases) with an increased sample size and ________ (increases or decreases) with an increase in confidence level.

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. 98% confidence; the sample size is 800, of which 40% are successes

Use the given degree of confidence and sample data to construct a confidence interval for the population mean $\mu$ . Assume that the population has a normal distribution. $n = 30 , \bar { x } = 84.6 , \mathrm {~s} = 10.5,90 \%$ confidence

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. $n = 195 , x = 162 ; 95 \% \text { confidence }$

Find the critical value $\chi _ { \mathrm { R } } ^ { 2 }$ corresponding to a sample size of 19 and a confidence level of 99 percent.

Find the value of ${ } ^ { z } \alpha / 2$ that corresponds to a confidence level of 89.48%.

If the computed minimum sample size n needed for a particular margin of error is not a whole number, round the value of n (up or down) to the next (smaller or larger) whole number.

Find the critical value $z _ { \alpha / 2 }$ that corresponds to a 91% confidence level.

Use the given data to find the minimum sample size required to estimate the population proportion. Margin of error: 0.028; confidence level: 99%; $p \text { and } q \text { unknown }$

Express the confidence interval $( 0.432,0.52 )$ in the form of $\hat { p } \pm \mathrm { E }$ .

Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. $95 \% \text { confidence; } n = 2388 , x = 1672$

Which critical value is appropriate for a 99% confidence level where n = 17; $\sigma$ is unknown and the population appears to be normally distributed?

The margin of error (increases or decreases) with an increased sample size and (increases or decreases) with an increase in confidence level.