Question 1

Provide an appropriate response.
-A researcher wishes to estimate the number of households with two computers. How large a sample is needed in order to be $98 \%$ confident that the sample proportion will not differ from the true proportion by more than $2 \%$ ? A previous study indicates that the proportion of households with two computers is $23 \%$.

Accepted Answer

The sample size needed can be calculated using the formula for estimating a proportion in a population: $n = \frac{{Z^2 \cdot p \cdot (1 - p)}}{{E^2}}$, where $Z$ is the Z-score corresponding to the confidence level (2.33 for 98%), $p$ is the estimated proportion (0.23), and $E$ is the margin of error (0.02). Plugging in the values: $n = \frac{{(2.33)^2 \cdot 0.23 \cdot (1 - 0.23)}}{{(0.02)^2}} \approx 2404$.

Question 2

Provide an appropriate response.
-Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority of Americans. A random sample of 4000 citizens yielded 2140 who are in favor of gun control legislation. Find the point estimate for estimating the proportion of all Americans who are in favor of gun control legislation.

Accepted Answer

The point estimate for the proportion of all Americans who are in favor of gun control legislation is found by dividing the number of citizens in favor (2140) by the total number of citizens sampled (4000), which equals 0.5350.

Question 3

Provide an appropriate response.
-A confidence interval for $p$ can be constructed using

Accepted Answer

The correct formula for constructing a confidence interval for a proportion $p$ uses the sample proportion $\hat{p}$, the critical value $z_{\alpha/2}$, and the standard error of the sample proportion, which is $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $n$ is the sample size. This matches choice A, which correctly represents the formula for a confidence interval for a proportion.

Question 4

Provide an appropriate response.
-A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States that has sleep deprivation. How large a sample is needed in order to be $99 \%$ confident that the sample proportion will not differ from the true proportion by more than $4 \%$ ?

Accepted Answer

To estimate the sample size needed for a proportion with a specific margin of error (E) at a certain confidence level, the formula is $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (for 99% confidence, $Z \approx 2.576$), $p$ is the estimated proportion (if unknown, 0.5 is used as it maximizes the sample size), and $E$ is the margin of error (0.04 in this case). Plugging in the values: $n = \frac{(2.576)^2 \cdot 0.5 \cdot (1-0.5)}{(0.04)^2} \approx 1037$.

Question 5

Provide an appropriate response.
-A survey of 100 fatal accidents showed that in 32 cases the driver at fault was inadequately insured. Find a point estimate for p, the population proportion of accidents where the driver at fault was inadequately insured

Accepted Answer

The point estimate for p, the population proportion, is found by dividing the number of cases where the condition is met (inadequately insured drivers at fault, 32 cases) by the total number of cases (100 fatal accidents), which equals 0.32.

Question 6

Provide an appropriate response.
-A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a $99 \%$ confidence interval to estimate the true proportion of students on financial aid.

Accepted Answer

The proportion of sampled students receiving financial aid is 118/200 = 0.59. Using a 99% confidence interval for proportions, the margin of error is calculated using the formula for the standard error of a proportion and the z-score for a 99% confidence level. The given option closest to the correct calculation is $0.59 \pm 0.090$, indicating a 99% confidence interval.

Question 7

Provide an appropriate response.
-Many people think that a national lobby's successful fight against gun control legislation is reflecting the will of a minority of Americans. A random sample of 4000 citizens yielded 2250 who are in favor of gun control legislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a $95 \%$ confidence interval.

Accepted Answer

The proportion of the sample in favor of gun control legislation is $2250 / 4000 = 0.5625$. Using the formula for a $95\%$ confidence interval for a proportion, $p \pm z \sqrt{\frac{p(1-p)}{n}}$, where $p = 0.5625$, $n = 4000$, and $z$ is approximately $1.96$ for $95\%$ confidence, the margin of error is approximately $0.0154$. Therefore, the correct confidence interval is $0.5625 \pm 0.0154$.

Question 8

Provide an appropriate response.
-A private opinion poll is conducted for a politician to determine what proportion of the population favors adding more national parks. How large a sample is needed in order to be 99% confident that the sample proportion will not differ from the true proportion by more than 5%

Accepted Answer

The sample size needed is given by the formula n = (Z^2 * p * q) / e^2, where Z is the z-score for the desired confidence level, p is the estimated proportion, q is 1 - p, and e is the margin of error. In this case, we have Z = 2.576 (for a 99% confidence level), p = 0.5 (since we have no prior information to suggest otherwise), q = 0.5, and e = 0.05. Plugging these values into the formula, we get n = (2.576^2 * 0.5 * 0.5) / 0.05^2 = 663.55. Rounding up to the nearest integer, we get a sample size of 664.

Question 9

Provide an appropriate response.
-A survey of 700 non-fatal accidents showed that 272 involved uninsured drivers. Construct a $99 \%$ confidence interval for the proportion of fatal accidents that involved uninsured drivers.

Accepted Answer

The answer of Provide an appropriate response.
-A survey of 700...

Question 10

Provide an appropriate response.
-What is the best point estimate for $p$ in order to construct a confidence interval for $p$ ?

Accepted Answer

The best point estimate for $p$ when constructing a confidence interval for a population proportion $p$ is $\hat { p }$, which represents the sample proportion. This is because $\hat { p }$ is calculated directly from sample data and serves as the most direct estimate of the population proportion $p$ that we're trying to infer about.

Question 11

Provide an appropriate response.
-A survey of 2060 musicians showed that 301 of them are left-handed. Find a point estimate for p, the population proportion of musicians that are left-handed.

Accepted Answer

The point estimate for the population proportion $p$ is found by dividing the number of left-handed musicians by the total number of musicians surveyed, which is $301/2060 = 0.146$.

Question 12

Provide an appropriate response.
-A senator wishes to estimate the proportion of United States voters who favor abolishing the Electoral College. How large a sample is needed in order to be $98 \%$ confident that the sample proportion will not differ from the true proportion by more than $5 \%$ ?

Accepted Answer

The sample size needed for estimating a proportion with a specific margin of error (E) at a certain confidence level can be calculated using the formula: $n = \left(\frac{Z^2 \cdot p \cdot (1-p)}{E^2}\right)$, where $Z$ is the Z-score corresponding to the confidence level, $p$ is the estimated proportion (when unknown, $p = 0.5$ is used for maximum sample size), and $E$ is the margin of error. For a $98\%$ confidence level, the Z-score is approximately $2.33$ (from Z-tables). Assuming $p = 0.5$ for maximum variability, and $E = 0.05$ (5%), the calculation is $n = \left(\frac{(2.33)^2 \cdot 0.5 \cdot (1-0.5)}{(0.05)^2}\right)$, which simplifies to approximately $543$.

Question 13

Provide an appropriate response.
-True or False: The general form of a large-sample $( 1 - \alpha ) 100 \%$ confidence interval for a population proportion interest.

Accepted Answer

The general form of a large-sample $( 1 - \alpha ) 100 \%$ confidence interval for a population proportion interest is:$$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$where $\hat{p}$ is the sample proportion, $z_{\alpha/2}$ is the critical value from the standard normal distribution with area $\alpha/2$ in the upper tail, and $n$ is the sample size.

Question 14

Provide an appropriate response.
-When 425 junior college students were surveyed, 190 said they have a passport. Construct a $95 \%$ confidence interval for the proportion of junior college students that have a passport.

Accepted Answer

The answer of Provide an appropriate response.
-When 425 junior college...

Question 15

Provide an appropriate response.
-A marketing research company needs to estimate which of two medical plans its employees prefer. A random sample of $n$ employees produced the following $90 \%$ confidence interval for the proportion of employees who prefer plan A: $( 0.332,0.612 )$. Identify the point estimate for estimating the true proportion of employees who prefer that plan.

Accepted Answer

The point estimate for the proportion is the midpoint of the confidence interval, which is calculated as $(0.332 + 0.612) / 2 = 0.472$.

Question 16

Provide an appropriate response.
-In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the $90 \%$ confidence interval for the population proportion, p? Explain.

Accepted Answer

The answer of Provide an appropriate response.
-In a survey of...

Question 17

Provide an appropriate response.
-A pollster wishes to estimate the number of left-handed scientists. How large a sample is needed in order to be 99% confident that the sample proportion will not differ from the true proportion by more than 6% A previous study indicates that the proportion of left-handed scientists is 9%.

Accepted Answer

The sample size needed is 151. This is calculated using the formula n = (Z^2 * p * q) / e^2, where Z is the z-score for the desired confidence level (99% = 2.576), p is the estimated proportion of left-handed scientists (0.09), q is the complement of p (1 - p = 0.91), and e is the margin of error (0.06).

Question 18

Provide an appropriate response.
-A survey of 500 non-fatal accidents showed that 141 involved faulty equipment. Find a point estimate for p. the population proportion of accidents that involved faulty equipment.

Accepted Answer

The point estimate for the population proportion $p$ is found by dividing the number of accidents involving faulty equipment (141) by the total number of accidents surveyed (500), which gives $0.282$.

Question 19

Provide an appropriate response.
-A survey of 250 households showed 41 owned at least one snow blower. Find a point estimate for p, the population proportion of households that own at least one snow blower.

Accepted Answer

The point estimate for p, the population proportion, is found by dividing the number of households that own at least one snow blower (41) by the total number of households surveyed (250), which equals 0.164.

Question 20

Provide an appropriate response.
-An article a Florida newspaper reported on the topics that teenagers most want to discuss with their parents. The findings, the results of a poll, showed that $46 \%$ would like more discussion about the family's financial situation, $37 \%$ would like to talk about school, and $30 \%$ would like to talk about religion. These and other percentages were based on a national sampling of 502 teenagers. Estimate the proportion of all teenagers who want more family discussions about school. Use a $95 \%$ confidence level.

Accepted Answer

The proportion of teenagers who want more family discussions about school is $37\%$, or $0.37$. To estimate the proportion with a $95\%$ confidence level, we use the formula for the confidence interval for a proportion: $\hat{p} \pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p} = 0.37$, $n = 502$, and $z$ is the z-score corresponding to a $95\%$ confidence level (approximately 1.96). Calculating the margin of error gives a result that matches choice A, indicating a confidence interval of $0.37 \pm 0.042$.

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

Provide an appropriate response. -A confidence interval for ppp can be constructed using

Provide an appropriate response. -A survey of 100 fatal accidents showed that in 32 cases the driver at fault was inadequately insured. Find a point estimate for p, the population proportion of accidents where the driver at fault was inadequately insured

Provide an appropriate response. -A survey of 700 non-fatal accidents showed that 272 involved uninsured drivers. Construct a 99%99 \%99% confidence interval for the proportion of fatal accidents that involved uninsured drivers.

Provide an appropriate response. -What is the best point estimate for ppp in order to construct a confidence interval for ppp ?

Provide an appropriate response. -A survey of 2060 musicians showed that 301 of them are left-handed. Find a point estimate for p, the population proportion of musicians that are left-handed.

Provide an appropriate response. -True or False: The general form of a large-sample (1−α) 100%( 1 - \alpha ) 100 \%(1−α) 100% confidence interval for a population proportion interest.

Provide an appropriate response. -When 425 junior college students were surveyed, 190 said they have a passport. Construct a 95%95 \%95% confidence interval for the proportion of junior college students that have a passport.

Provide an appropriate response. -In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90%90 \%90% confidence interval for the population proportion, p? Explain.

Provide an appropriate response. -A survey of 500 non-fatal accidents showed that 141 involved faulty equipment. Find a point estimate for p. the population proportion of accidents that involved faulty equipment.

Provide an appropriate response. -A survey of 250 households showed 41 owned at least one snow blower. Find a point estimate for p, the population proportion of households that own at least one snow blower.

Provide an appropriate response. -A confidence interval for $p$ can be constructed using

Provide an appropriate response. -A survey of 700 non-fatal accidents showed that 272 involved uninsured drivers. Construct a $99 \%$ confidence interval for the proportion of fatal accidents that involved uninsured drivers.

Provide an appropriate response. -What is the best point estimate for $p$ in order to construct a confidence interval for $p$ ?

Provide an appropriate response. -True or False: The general form of a large-sample $( 1 - \alpha ) 100 \%$ confidence interval for a population proportion interest.

Provide an appropriate response. -When 425 junior college students were surveyed, 190 said they have a passport. Construct a $95 \%$ confidence interval for the proportion of junior college students that have a passport.

Provide an appropriate response. -In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the $90 \%$ confidence interval for the population proportion, p? Explain.