Question 1

The margin of error ________ ________ (increases or decreases)with an increase in confidence level.
A) increases; increases
B) increases; decreases
C) decreases; increases
D) decreases; decreases

Accepted Answer

C

Question 2

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?
A) 0.76
B) 0.24
C) 0.50
D) 0.18

Accepted Answer

The best point estimate of the proportion of people in the population who are over 6 feet tall is the proportion found in the sample, which is 12 out of 50, or 0.24.

Question 3

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
A) $0.778 < p < 0.883$
B) $0.788 < p < 0.873$
C) $0.789 < p < 0.873$
D) $0.777 < p < 0.884$

Accepted Answer

To construct a 95% confidence interval for the population proportion $p$, we use the formula: $\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p} = \frac{x}{n}$, $z$ is the z-score corresponding to the confidence level (for 95% confidence, $z \approx 1.96$), $n$ is the sample size, and $x$ is the number of successes.Given $n = 195$ and $x = 162$, we find $\hat{p} = \frac{162}{195} \approx 0.8308$.The margin of error (ME) is calculated as: $ME = 1.96\sqrt{\frac{0.8308(1-0.8308)}{195}} \approx 0.052$.Thus, the confidence interval is $0.8308 \pm 0.052$, which simplifies to $0.7788 < p < 0.8828$, or rounded to three decimal places, $0.778 < p < 0.883$.

Question 4

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) $0.0323 < p < 0.0782$
B) $0.0360 < p < 0.0745$
C) $0.0251 < p < 0.0854$
D) $0.0280 < p < 0.0826$

Accepted Answer

To find a 95% confidence interval for the proportion, we use the formula: $\hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p} = \frac{21}{380}$, $n = 380$, and $z$ is the z-score for a 95% confidence level, which is approximately 1.96. Plugging in the values, we get the interval $0.0323 < p < 0.0782$.

Question 5

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.
A) $z _ { \alpha / 2 } = 2.896$
B) $z _ { \alpha / 2 } = 2.365$
C) $z _ { \alpha / 2 } = 2.306$
D) Neither the normal nor the t distribution applies.

Accepted Answer

Given the small sample size (n=9) and the population being very skewed, neither the normal (z) nor the t distribution would be appropriate without assuming the population distribution is approximately normal, which is contradicted by the statement of skewness.

Question 6

Express the confidence interval  (0.432,0.52)  in the form of $\hat { p } \pm \mathrm { E }$
A) $0.476 \pm 0.044$
B) $0.476 \pm 0.088$
C) $0.432 \pm 0.044$
D) $0.432 \pm 0.088$

Accepted Answer

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

Question 7

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 , $\bar { x } = 84.6$  s=10.5,90 % confidence
A) $79.32 < \mu < 89.88$
B) $81.36 < \mu < 87.84$
C) $81.34 < \mu < 87.86$
D) $80.68 < \mu < 88.52$

Accepted Answer

The answer of Use the given degree of confidence and...

Question 8

Find the critical value $\chi _ { \mathrm { R } } ^ { 2 }$ corresponding to a sample size of 19 and a confidence level of 99 percent.
A)  37.156 
B)  6.265 
C)  34.805 
D)  7.015

Accepted Answer

The critical value $\chi^2$ can be found using a $\chi^2$ distribution table or calculator, corresponding to a confidence level of 99% and degrees of freedom (df). The degrees of freedom for a sample size of 19 is 18 (df = n - 1). Looking up the critical value for 99% confidence and 18 degrees of freedom gives us 37.156.

Question 9

Which critical value is appropriate for a  99% confidence level where  n=17 ;$\sigma$ is unknown and the population appears to be normally distributed?
A) $t _ { \alpha / 2 } = 2.898$
B) $z _ { \alpha / 2 } = 2.583$
C) $t _ { \alpha / 2 } = 2.921$
D) $z _ { \alpha / 2 } = 2.567$

Accepted Answer

Since the sample size is small (n=17), $\sigma$ is unknown, and the population is normally distributed, the t-distribution should be used. For a 99% confidence level with 16 degrees of freedom (n-1), the critical value is $t _ { \alpha / 2 } = 2.921$.

Question 10

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
A) 0.0355
B) 0.0339
C) 0.0404
D) 0.0446

Accepted Answer

The answer of Assume that a sample is used to...

Question 11

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.
A) up; smaller
B) up; larger
C) down; smaller
D) down; larger

Accepted Answer

When calculating the minimum sample size needed for a particular margin of error, if the result is not a whole number, it is standard practice to round up to the next larger whole number. This ensures that the sample size is not smaller than needed for the desired margin of error.

Question 12

Find the value of $z _ { \alpha / 2 }$ that corresponds to a confidence level of  89.48%.
A)  1.25 
B)  1.62 
C)  0.0526 
D)  -1.62

Accepted Answer

For a confidence level of 89.48%, the corresponding $z_{\alpha/2}$ value is approximately 1.62, as it represents the critical value for the two-tailed test.

Question 13

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.
A) $0.316 < p < 0.430$
B) $0.301 < p < 0.445$
C) $0.308 < p < 0.438$
D) $0.304 < p < 0.442$

Accepted Answer

The sample proportion is 112/300 = 0.373. Using the formula for a confidence interval for a proportion and a Z-score for 98% confidence (approximately 2.33), the interval is calculated as 0.373 ± 2.33 * sqrt((0.373*(1-0.373))/300), which results in the interval 0.308 to 0.438.

Question 14

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=1672 
A)  0.0248 
B)  0.0184 
C)  0.0156 
D)  0.0206

Accepted Answer

The answer of Assume that a sample is used to...

Question 15

Find the critical value $z _ { \alpha / 2 }$ that corresponds to a  91% confidence level.
A)  1.34 
B)  1.645 
C)  1.70 
D)  1.75

Accepted Answer

For a 91% confidence level, the area in the tails (outside the confidence interval) is 9%, or 0.09. This is split between the two tails, so each tail has an area of 0.045. Looking up 0.045 in the z-table (or using a calculator or software that can invert the normal distribution), we find the z-value that corresponds to the cumulative area of 0.955 (1 - 0.045) is approximately 1.70.

Question 16

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.
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.
A) $173.7 \mathrm { mg } < \mu < 196.3 \mathrm { mg }$
B) $175.9 \mathrm { mg } < \mu < 194.1 \mathrm { mg }$
C) $173.8 \mathrm { mg } < \mu < 196.2 \mathrm { mg }$
D) $173.9 \mathrm { mg } < \mu < 196.1 \mathrm { mg }$

Accepted Answer

The confidence interval is calculated using the formula: $\bar{x} \pm t^* \frac{s}{\sqrt{n}}$, where $\bar{x} = 185$, $s = 17.6$, $n = 12$, and $t^*$ is the t-value for 95% confidence with 11 degrees of freedom. The correct interval is $173.8 \mathrm { mg } < \mu < 196.2 \mathrm { mg }$.

Question 17

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.
A) $91.68<\mu<98.32$
B) $91.69<\mu<98.31$
C) $92.95 < \mu < 97.05$
D) $92.03 < \mu < 97.97$

Accepted Answer

To construct a 99% confidence interval for the mean score of all students, 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 confidence level, $\sigma$ is the standard deviation, and $n$ is the sample size. Given $\bar{x} = 95$, $\sigma = 6.6$, and $n = 30$, and knowing that the z-score for a 99% confidence level is approximately 2.576, we calculate the margin of error as $2.576 \times \frac{6.6}{\sqrt{30}} \approx 3.16$. Thus, the confidence interval is $95 \pm 3.16$, which gives $91.84$ and $98.16$, rounding to two decimal places as required by the options leads to $91.68 < \mu < 98.32$.

Question 18

In constructing a confidence interval for $\sigma$ or $\sigma ^ { 2 }$, a table is used to find the critical values $\chi _ { \mathrm { L } } ^ { 2 } \text { and } \chi _ { \mathrm { R } } ^ { 2 }$ for values of $n \leq 101$ For larger values of  n, $\chi _ { \mathrm { L } } ^ { 2 } \text { and } \chi _ { \mathrm { R } } ^ { 2 }$  can be
approximated by using the following formula: $\chi ^ { 2 } = \frac { 1 } { 2 } \pm z _ { \alpha / 2 } + \sqrt { 2 k - 1 } ^ { 2 }$ 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 lb  and a standard deviation of  25.5 lb. Round the confidence interval limits to the nearest hundredth.
A) $23.66 \mathrm { lb } < \sigma < 27.58 \mathrm { lb }$
B) $23.39 \mathrm { lb } < \sigma < 28.09 \mathrm { lb }$
C) $24.09 \mathrm { lb } < \sigma < 27.15 \mathrm { lb }$
D) $23.71 \mathrm { lb } < \sigma < 27.65 \mathrm { lb }$

Accepted Answer

The answer of In constructing a confidence interval for $\sigma$...

Question 19

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
A) 1116
B) 2223
C) 1939
D) 2115

Accepted Answer

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

Question 20

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.
A) Student t distribution
B) Normal distribution
C) Chi-square distribution
D) None of the three distributions is appropriate.

Accepted Answer

The answer of Identify the distribution that applies to the...

Quiz 7: Estimating Parameters and Determining Sample Sizes

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

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?

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

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.

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.

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

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 , $\bar { x } = 84.6$ s=10.5,90 % confidence

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

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

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

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

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.

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=1672

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 and q unknown

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.

Quiz 7: Estimating Parameters and Determining Sample Sizes

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

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?

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

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.

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

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\bar { x } = 84.6xˉ=84.6 s=10.5,90 % confidence

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

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?