Question 1

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-square distribution table or calculator, given the degrees of freedom (df) and the confidence level. For a sample size of 19, the degrees of freedom is $df = n - 1 = 18$. At a 99% confidence level, the critical value for 18 degrees of freedom is approximately 37.156.

Question 2

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 $\mu ; \quad n = 9 ; \sigma = 4.2$ population appears to be very skewed.
A) $z _ { a / 2 } = 2.896$
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 for constructing a confidence interval. The central limit theorem, which allows for the use of the normal distribution in constructing confidence intervals, typically requires a larger sample size and/or a population that is not significantly skewed.

Question 3

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: $\bar{x} \pm t_{\alpha/2} \times \frac{s}{\sqrt{n}}$. With $\bar{x} = 185$, $s = 17.6$, $n = 12$, and using the t-distribution for 11 degrees of freedom, the interval is $185 \pm 11.2$, which gives $173.8 \mathrm{mg} < \mu < 196.2 \mathrm{mg}$.

Question 4

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 is 9% (100% - 91%), so each tail has 4.5% (9% / 2). Looking up 4.5% in the z-table or using a calculator to find the z-value that corresponds to the cumulative area of 0.955 (0.5 + 0.455), we find $z _ { \alpha / 2 }$ is approximately 1.70.

Question 5

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, $E = \frac{0.52 - 0.432}{2} = 0.044$. Therefore, the interval can be expressed as $0.476 \pm 0.044$.

Question 6

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 $\hat{p} = \frac{x}{n}$ is the sample proportion, $n$ is the sample size, and $z$ is the z-score corresponding to the desired level of confidence. For a 95% confidence level, $z \approx 1.96$.Given $n = 195$ and $x = 162$, we find $\hat{p} = \frac{162}{195} = 0.8308$.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, $0.778 < p < 0.883$.

Question 7

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

The answer of The margin of error ________ ________ (increases...

Question 8

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

The confidence level of 89.48% means that the area in the two tails (outside the confidence interval) is 100% - 89.48% = 10.52%. Since this area is split between the two tails, each tail has 10.52%/2 = 5.26% of the area. Looking up 5.26% in the z-table (or using a calculator), we find that the z-value that corresponds to an area of 5.26% in one tail is approximately 1.62.

Question 9

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 10

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) $z _ { a / 2 } = 2.583$
C) $t _ { \alpha / 2 } = 2.921$
D) $z _ { \alpha / 2 } = 2.567$

Accepted Answer

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

Question 11

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 12

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 13

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 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). Calculating this gives E ≈ 0.0404.

Question 14

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.
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 for the confidence interval: $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, 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. For a 99% confidence level, the z-score is approximately 2.576. Plugging in the given values ($\bar{x} = 95$, $\sigma = 6.6$, $n = 30$), we get the interval $95 \pm 2.576\frac{6.6}{\sqrt{30}}$, which calculates to the interval $91.68 < \mu < 98.32$.

Question 15

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 observed in the sample, which is 12 out of 50, or 0.24.

Question 16

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 17

Identify the distribution that applies to the following situation: In constructing a confidence interval of $\sigma$ 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 Student t distribution is used when the population standard deviation is unknown and the sample size is small (less than 30). In this case, the sample size is 50, which is greater than 30, so the Student t distribution is not appropriate. The normal distribution is used when the population standard deviation is known and the sample size is large (greater than 30). In this case, the population standard deviation is unknown, so the normal distribution is not appropriate. The chi-square distribution is used to test the goodness of fit of a distribution to data. In this case, we are not testing the goodness of fit of a distribution, so the chi-square distribution is not appropriate. Therefore, the only distribution that is appropriate is the Student t distribution.

Question 18

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 the 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 19

In constructing a confidence interval for $\sigma \text { or } \sigma ^ { 2 }$ 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 , $\chi _ { \mathrm { L } } ^ { 2 } \text { and } \chi _ { \mathrm { R } } ^ { 2 }$
approximated by using the following formula: $\chi ^ { 2 } = \frac { 1 } { 2 } \pm z _ { \alpha / 2 } + \sqrt { 2 k - 1 } ^ { 2 }$
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 20

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 rounded up to the next larger whole number to ensure the sample size is sufficient to achieve the desired margin of error.

Quiz 7: Estimating Parameters and Determining Sample Sizes

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

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

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 proportion p. $n = 195 , x = 162 ; 95 \%$ confidence

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

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

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

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

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

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

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

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?

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.

Identify the distribution that applies to the following situation: In constructing a confidence interval of $\sigma$ you have 50 sample values and they appear to be From a population with a skewed distribution.

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.

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.

Quiz 7: Estimating Parameters and Determining Sample Sizes

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

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 proportion p. n=195,x=162;95%n = 195 , x = 162 ; 95 \%n=195,x=162;95% confidence

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

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