Question 1

Many people think that a national lobbyʹs successful fight against gun control legislation is reflecting the will ofa minority of Americans. A random sample of 4000 citizens yielded 2250 who are in favor of gun controllegislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a^99% confidence interval. Express the answer in the form p ± E and round to the nearest ten-thousandth.

Accepted Answer

The proportion of the sample in favor of gun control legislation is 2250/4000 = 0.5625. To estimate the true proportion with a 99% confidence interval, we use the formula for the margin of error E = z * sqrt[(p(1-p)/n)], where z is the z-score corresponding to the confidence level (for 99%, z is approximately 2.576), p is the sample proportion, and n is the sample size. Plugging in the values, we get a margin of error much smaller than the options suggesting a mistake in calculation for options B and D. The correct margin of error, based on the formula and the given sample size and proportion, aligns with option A, indicating a smaller, more reasonable interval for a sample of this size.

Question 2

When 385 junior college students were surveyed,170 said that they have previously owned a motorcycle. Finda point estimate for p, the population proportion of students who have previously owned a motorcycle.

Accepted Answer

The point estimate for the population proportion $ p $ is found by dividing the number of students who have previously owned a motorcycle by the total number of students surveyed. Thus, $ p = \frac{170}{385} \approx 0.442 $.

Question 3

$$\text { True or False: The general form of a large-sample } ( 1 - \alpha ) 100 \% \text { confidence interval for a population proportion }$$ interest.

Accepted Answer

The general form of a large-sample confidence interval for a population proportion is indeed used to estimate the range within which the true population proportion is expected to fall with a certain level of confidence, typically expressed as $(1 - \alpha)100\%$.

Question 4

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 familyʹs financialsituation, 37% would like to talk about school, and 30% would like to talk about religion. These and otherpercentages were based on a national sampling of 546 teenagers. Estimate the proportion of all teenagers who^want more family discussions about school. Use a 95% confidence level. Express the answer in the form p ± Eand round to the nearest thousandth.

Accepted Answer

The proportion of teenagers who want more family discussions about school is directly given as 37% or 0.37. To estimate the proportion for all teenagers with a 95% confidence level, we calculate the margin of error (E) using the formula for the confidence interval for a proportion. The choice that correctly reflects a realistic margin of error for a sample size of 546 is 0.37 ± 0.040, which accounts for sampling variability and provides a reasonable level of precision for such a sample size.

Question 5

A private opinion poll is conducted for a politician to determine what proportion of the population favorsadding more national parks. How large a sample is needed in order to be 90% confident that the sampleproportion will not differ from the true proportion by more than 5%?

Accepted Answer

The sample size needed for a proportion can be calculated using the formula $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where $Z$ is the Z-score corresponding to the confidence level (1.645 for 90% confidence), $p$ is the estimated proportion (0.5 is used when the proportion is unknown for maximum sample size), and $E$ is the margin of error (0.05 for 5%). Plugging in the values, $n = \frac{1.645^2 \cdot 0.5 \cdot (1-0.5)}{0.05^2} \approx 271$.

Question 6

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  95 %  confidence interval to estimate the true proportion of students on financial aid. Express the answer in the form $\hat { p } \pm E$ and round to the nearest thousandth.

Accepted Answer

The sample proportion ($\hat { p }$) is calculated as the number of students receiving financial aid divided by the total number of students sampled, which is $118/200 = 0.59$. To calculate the margin of error (E) for a 95% confidence interval, the formula $E = Z \cdot \sqrt{\frac{\hat { p }(1 - \hat { p })}{n}}$ is used, where $Z$ is the Z-score corresponding to a 95% confidence level (approximately 1.96), $\hat { p }$ is the sample proportion, and $n$ is the sample size. Substituting the given values, we find $E \approx 0.068$, leading to a confidence interval of $0.59 \pm 0.068$.

Question 7

A survey of 2690 musicians showed that 368 of them are left-handed. Find a point estimate for p, thepopulation 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 368/2690 = 0.1368, rounded to 0.137.

Question 8

A pollster wishes to estimate the number of left-handed scientists. How large a sample is needed in order to be95% confident that the sample proportion will not differ from the true proportion by more than 6%? A previousstudy indicates that the proportion of left-handed scientists is 9%.

Accepted Answer

To calculate the sample size needed, we use the formula for sample size in proportion estimation: n = (Z^2 * p * (1-p)) / E^2. Here, Z is the Z-score for 95% confidence (1.96), p is the estimated proportion (0.09), and E is the margin of error (0.06). Plugging in these values gives n ≈ 88.

Question 9

A survey of 100 fatal accidents showed that in 43 cases the driver at fault was inadequately insured. Find apoint estimate for p, the population proportion of accidents where the driver at fault was inadequately insured

Accepted Answer

The point estimate for the population proportion (p) is found by dividing the number of cases where the condition is met (inadequately insured drivers at fault, 43 cases) by the total number of cases observed (100 fatal accidents), which equals 0.43.

Question 10

95%

Accepted Answer

For a 95% confidence level, the z-score is approximately 1.96. This value is used in statistics to represent the number of standard deviations a point is from the mean in a normal distribution.

Question 11

A confidence interval for $\mathrm { p }$ can be constructed using

Accepted Answer

Option A is the correct formula for constructing a confidence interval for a population proportion $p$, using the sample proportion $\hat{p}$, where $z_{\alpha/2}$ is the critical value from the standard normal distribution for the desired confidence level, and $n$ is the sample size.

Question 12

Many people think that a national lobbyʹs successful fight against gun control legislation is reflecting the will ofa minority of Americans. A random sample of 4000 citizens yielded 2130 who are in favor of gun controllegislation. Find the point estimate for estimating the proportion of all Americans who are in favor of guncontrol legislation.

Accepted Answer

The point estimate for the proportion of all Americans who are in favor of gun control legislation is calculated by dividing the number of citizens in favor (2130) by the total number of citizens sampled (4000), which equals 0.5325.

Question 13

88%

Accepted Answer

88% = 88/100 = 0.88 = 1.175

Question 14

A survey of 700 non-fatal accidents showed that 143 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 calculated by dividing the number of accidents involving faulty equipment (143) by the total number of accidents surveyed (700), which equals 0.204.

Question 15

A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States thathas sleep deprivation. How large a sample is needed in order to be 95% confident that the sample proportionwill not differ from the true proportion by more than 4%?

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. Plugging in the values given in the question, we get n = (1.96^2 * 0.5 * 0.5) / 0.04^2 = 600.25. Rounding up to the nearest whole number, we get a sample size of 601.

Question 16

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 $p$ is $\hat { p }$, which represents the sample proportion and is used as the point estimate of the population proportion $p$.

Question 17

A marketing research company needs to estimate which of two medical plans its employees prefer. A randomsample of n employees produced the following 98% confidence interval for the proportion of employees whoprefer plan A: (0.241, 0.561). Identify the point estimate for estimating the true proportion of employees whoprefer that plan.

Accepted Answer

The point estimate for the true proportion is the midpoint of the confidence interval, which is (0.241 + 0.561) / 2 = 0.401.

Question 18

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 99% confident that the sample proportion will not differ from thetrue proportion by more than 3%?

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: $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$, where Z is the Z-score corresponding to the confidence level (for 99% confidence, Z ≈ 2.576), p is the estimated proportion (when unknown, p = 0.5 is used for maximum sample size), and E is the margin of error. Plugging in the values: $n = \frac{2.576^2 \cdot 0.5 \cdot (1-0.5)}{0.03^2}$ gives a value close to 1842, rounding up for fractions.

Question 19

A survey of 250 households showed 91 owned at least one snow blower. Find a point estimate for p, thepopulation proportion of households that own at least one snow blower.

Accepted Answer

The point estimate for the population proportion $p$ is calculated by dividing the number of households that own at least one snow blower by the total number of households surveyed. Thus, $p = \frac{91}{250} = 0.364$.

Question 20

In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90% confidenceinterval for the population proportion, p? Explain.

Accepted Answer

The answer of In a survey of 10 musicians, 2...

Quiz 9: Estimating the Value of a Parameter

When 385 junior college students were surveyed,170 said that they have previously owned a motorcycle. Finda point estimate for p, the population proportion of students who have previously owned a motorcycle.

$\text { True or False: The general form of a large-sample } ( 1 - \alpha ) 100 \% \text { confidence interval for a population proportion }$ interest.

A private opinion poll is conducted for a politician to determine what proportion of the population favorsadding more national parks. How large a sample is needed in order to be 90% confident that the sampleproportion will not differ from the true proportion by more than 5%?

A survey of 2690 musicians showed that 368 of them are left-handed. Find a point estimate for p, thepopulation proportion of musicians that are left-handed.

A pollster wishes to estimate the number of left-handed scientists. How large a sample is needed in order to be95% confident that the sample proportion will not differ from the true proportion by more than 6%? A previousstudy indicates that the proportion of left-handed scientists is 9%.

A survey of 100 fatal accidents showed that in 43 cases the driver at fault was inadequately insured. Find apoint estimate for p, the population proportion of accidents where the driver at fault was inadequately insured

95%

A confidence interval for $\mathrm { p }$ can be constructed using

88%

A survey of 700 non-fatal accidents showed that 143 involved faulty equipment. Find a point estimate for p,the population proportion of accidents that involved faulty equipment.

A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States thathas sleep deprivation. How large a sample is needed in order to be 95% confident that the sample proportionwill not differ from the true proportion by more than 4%?

What is the best point estimate for p in order to construct a confidence interval for p ?

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 99% confident that the sample proportion will not differ from thetrue proportion by more than 3%?

A survey of 250 households showed 91 owned at least one snow blower. Find a point estimate for p, thepopulation proportion of households that own at least one snow blower.

In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90% confidenceinterval for the population proportion, p? Explain.

Quiz 9: Estimating the Value of a Parameter

When 385 junior college students were surveyed,170 said that they have previously owned a motorcycle. Finda point estimate for p, the population proportion of students who have previously owned a motorcycle.

A private opinion poll is conducted for a politician to determine what proportion of the population favorsadding more national parks. How large a sample is needed in order to be 90% confident that the sampleproportion will not differ from the true proportion by more than 5%?

A survey of 2690 musicians showed that 368 of them are left-handed. Find a point estimate for p, thepopulation proportion of musicians that are left-handed.

A pollster wishes to estimate the number of left-handed scientists. How large a sample is needed in order to be95% confident that the sample proportion will not differ from the true proportion by more than 6%? A previousstudy indicates that the proportion of left-handed scientists is 9%.

A survey of 100 fatal accidents showed that in 43 cases the driver at fault was inadequately insured. Find apoint estimate for p, the population proportion of accidents where the driver at fault was inadequately insured

95%

A confidence interval for p\mathrm { p }p can be constructed using

88%

A survey of 700 non-fatal accidents showed that 143 involved faulty equipment. Find a point estimate for p,the population proportion of accidents that involved faulty equipment.

A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States thathas sleep deprivation. How large a sample is needed in order to be 95% confident that the sample proportionwill not differ from the true proportion by more than 4%?

What is the best point estimate for p in order to construct a confidence interval for p ?

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 99% confident that the sample proportion will not differ from thetrue proportion by more than 3%?

A survey of 250 households showed 91 owned at least one snow blower. Find a point estimate for p, thepopulation proportion of households that own at least one snow blower.

In a survey of 10 musicians, 2 were found to be left-handed. Is it practical to construct the 90% confidenceinterval for the population proportion, p? Explain.

A confidence interval for $\mathrm { p }$ can be constructed using