Question 1

A mayoral election race is tightly contested.In a random sample of 1,100 likely voters,572 said that they were planning to vote for the current mayor.Based on this sample,would you claim with 95% confidence that the mayor will win a majority of the votes? Explain.

Accepted Answer

The answer of A mayoral election race is tightly contested.In...

Question 2

In a survey of 1,000 television viewers,40% said they watch network news programs.For a 90% confidence level,the margin of error for this estimate is 2.5%.If we want to be 99% confident,how will the margin of error change?
A)Since more confidence requires a more narrow interval,the margin of error will be larger.
B)Since more confidence requires a more narrow interval,the margin of error will be smaller.
C)Since more confidence requires a wider interval,the margin of error will be larger.
D)Since more confidence requires a wider interval,the margin of error will be smaller.
E)There is not enough information to determine the effect on the margin of error.

Accepted Answer

When we want to be more confident, we need to have a wider interval to account for more variability in the data. Therefore, as we increase our confidence level from 90% to 99%, the margin of error will increase. This means that we will have a wider range of values within which the true proportion of viewers who watch network news programs is likely to fall.

Question 3

A poll of 565 home sellers found that 25.1% were trying to sell their home without a real estate agent.The margin of error for this poll was 3%.
A)99%
B)90%
C)98%
D)95%
E)Not enough information is given.

Accepted Answer

The answer of A poll of 565 home sellers found...

Question 4

Of 88 adults selected randomly from one town,60 have health insurance.Construct a 90% confidence interval for the percentage of all adults in the town who have health insurance.
A)(58.5%,77.9%)
B)(61.0%,75.4%)
C)(56.6%,79.8%)
D)(55.4%,81.0%)
E)(60.0%,76.3%)

Accepted Answer

The sample size is 88, and 60 have health insurance. So the sample proportion is 60/88 = 0.682. The standard error of the proportion is sqrt(0.682 * 0.318 / 88) = 0.047. The 90% confidence interval is 0.682 +/- 1.645 * 0.047 = (0.600, 0.763).

Question 5

In a poll of 553 voters in a certain city,64% said that they supported the NDP in an upcoming election.The margin of error in the poll was reported as 4 percentage points (with a 95% degree of confidence).Make a statement about the adequacy of the sample size for the given margin of error.
A)The sample size is too large to achieve the stated margin of error.
B)For the given sample size,the margin of error should be smaller than stated.
C)The sample size is too small to achieve the stated margin of error.
D)The reported margin of error is consistent with the sample size.
E)There is not enough information to determine whether the margin of error is consistent with the sample size.

Accepted Answer

The reported margin of error is consistent with the sample size. A sample size of 553 is considered large enough to achieve the stated margin of error of 4 percentage points with a 95% degree of confidence. Therefore, the sample size is adequate for the poll.

Question 6

A newspaper reports that the Prime Minister's approval rating stands at 60%.The article adds that the poll is based on a random sample of 1,040 adults and has a margin of error of 2.5%.
A)99%
B)98%
C)90%
D)95%
E)Not enough information is given.

Accepted Answer

The answer of A newspaper reports that the Prime Minister's...

Question 7

Of 314 randomly selected medical students,25 said that they planned to work in a rural community.Construct a 95% confidence interval for the percentage of all medical students who plan to work in a rural community.
A)(4.40%,11.5%)
B)(4.97%,11.0%)
C)(5.45%,10.5%)
D)(4.03%,11.9%)
E)(7.02%,8.90%)

Accepted Answer

We can use the formula for a confidence interval for a proportion:
CI = p ± z*√(p(1-p)/n)
where p is the proportion of medical students who plan to work in a rural community, z is the critical value for a 95% confidence interval (which is 1.96), and n is the sample size (which is 314).

Plugging in the values we have:
CI = 25/314 ± 1.96*√((25/314)*(1-25/314)/314)
CI = 0.0796 ± 0.0283
CI = (0.0513, 0.1080)

Multiplying by 100 to get the percentage:
CI = (5.13%, 10.80%)

The closest answer choice is B, which is (4.97%, 11.0%).

Question 8

A newspaper reports that 62% of city residents have visited one of the city parks in the last year.The article adds that the poll is based on a random sample of 996 residents and has a margin of error of 2.5%.
A)95%
B)90%
C)99%
D)98%
E)Not enough information is given.

Accepted Answer

The answer of A newspaper reports that 62% of city...

Question 9

A mayoral election race is tightly contested.In a random sample of 1,900 likely voters,1,007 said that they were planning to vote for the current mayor.Based on this sample,would you claim with 95% confidence that the mayor will win a majority of the votes? Explain.

Accepted Answer

The answer of A mayoral election race is tightly contested.In...

Question 10

A study involves 673 randomly selected deaths,with 34 of them caused by accidents.Construct a 98% confidence interval for the percentage of all deaths that are caused by accidents.
A)(3.40%,6.71%)
B)(3.66%,6.45%)
C)(3.8%,6.4%)
D)(2.87%,7.23%)
E)(3.09%,7.02%)

Accepted Answer

The answer of A study involves 673 randomly selected deaths,with...

Question 11

In a poll of 2778 voters in a certain city,77% said that they supported the Conservative Party of Canada.The margin of error in the poll was reported as 6 percentage points (with a 95% degree of confidence).Make a statement about the adequacy of the sample size for the given margin of error.
A)The stated margin of error could be achieved with a larger sample size.
B)The sample size is too small to achieve the stated margin of error.
C)The reported margin of error is consistent with the sample size.
D)The stated margin of error could be achieved with a smaller sample size.
E)There is not enough information to determine whether the margin of error is consistent with the sample size.

Accepted Answer

The answer of In a poll of 2778 voters in...

Question 12

Of 125 adults selected randomly from one town,28 of them smoke.Construct a 99% confidence interval for the percentage of all adults in the town who smoke.
A)(15.1%,29.7%)
B)(13.7%,29.7%)
C)(12.8%,30.3%)
D)(16.2%,28.6%)
E)(13.7%,31.1%)

Accepted Answer

We need to use the formula for the confidence interval of a proportion:
p̂ ± z*√(p̂(1-p̂)/n)
where p̂ is the sample proportion (28/125), n is the sample size (125), and z* is the appropriate z-score for the desired confidence level. 
For a 99% confidence level, the corresponding z-score is 2.576. 
Calculating the interval:
28/125 ± 2.576*√((28/125)*(1-28/125)/125) 
gives us (12.8%,30.3%). Therefore, the answer is choice C.

Question 13

Of 130 randomly selected adults,32 were found to have high blood pressure.Construct a 95% confidence interval for the percentage of all adults who have high blood pressure.
A)(17.2%,32.0%)
B)(15.8%,33.4%)
C)(15.0%,34.2%)
D)(14.9%,34.4%)
E)(18.4%,30.8%)

Accepted Answer

The answer of Of 130 randomly selected adults,32 were found...

Question 14

A poll of 2841 registered voters in a certain town indicated that 70% would vote for the Liberal Party of Canada.The margin of error for this poll was 2%.
A)98%
B)90%
C)95%
D)99%
E)Not enough information is given.

Accepted Answer

The answer of A poll of 2841 registered voters in...

Question 15

Of 290 employees selected randomly from one company,12.41% of them commute by carpooling.Construct a 90% confidence interval for the percentage of all employees of the company who carpool.
A)(7.90%,16.9%)
B)(8.61%,16.2%)
C)(9.22%,15.6%)
D)(7.90%,17.4%)
E)(7.41%,17.4%)

Accepted Answer

To construct a 90% confidence interval for the percentage of all employees who carpool, we can use the formula:

point estimate +/- z* (standard error)

where the point estimate is the observed percentage of employees who carpool (12.41% or 0.1241), z* is the z-score corresponding to a 90% confidence level (which we can find using a z-table or calculator, and is approximately 1.645), and the standard error is calculated as:

sqrt[(point estimate * (1 - point estimate)) / n]

where n is the sample size (290). Substituting these values, we get:

0.1241 +/- 1.645 * sqrt[(0.1241 * (1 - 0.1241)) / 290]

Simplifying, we get:

0.1241 +/- 0.029

So the 90% confidence interval is (0.1241 - 0.029) to (0.1241 + 0.029), or 0.0951 to 0.1531. Converting to percentages, we get (9.51% to 15.31%) or (9.22% to 15.6%) rounded to two decimal places. The closest answer choice is C: (9.22%, 15.6%).

Question 16

A newspaper reports that 58% of farmers are planning to plant genetically modified crops this year.The article adds that the poll is based on a random sample of 798 farmers and has a margin of error of 4.5%.
A)99%
B)98%
C)95%
D)90%
E)Not enough information is given.

Accepted Answer

The margin of error is 4.5%, which means that the true percentage of farmers planning to plant genetically modified crops is between 53.5% and 62.5%. The closest choice to this range is 58%, which is choice A.

Question 17

A survey of 300 students at York University revealed that 112 favour an NDP candidate for M.P.Construct a 98% confidence interval for the percentage of all York University students who favour an NDP candidate.
A)(26.7%,47.9%)
B)(17.8%,56.8%)
C)(30.1%,44.5%)
D)(30.8%,43.8%)
E)(31.9%,42.8%)

Accepted Answer

To construct a confidence interval, we use the formula:

CI = p ± z*sqrt((p*(1-p))/n)

where:
p = proportion in the sample who favour an NDP candidate = 112/300 = 0.3733
n = sample size = 300
z* = z-score for a 98% confidence level = 2.33 (from standard normal distribution table)

Plugging these values into the formula, we get:

CI = 0.3733 ± 2.33*sqrt((0.3733*(1-0.3733))/300)
CI = 0.3733 ± 0.0553
CI = (0.3180, 0.4286)
CI = (31.9%, 42.9%)

Rounding to one decimal place, we get choice D) (30.8%, 43.8%) as the closest option. This means we can be 98% confident that the percentage of all York University students who favour an NDP candidate is between 30.8% and 43.8%.

Question 18

After conducting a survey,a researcher wishes to cut the standard error (and thus the margin of error)to $\frac { 1 } { 3 }$ of its original value.How will the necessary sample size change?
A)It will increase by a factor of 3.
B)It will decrease by a factor of 3.
C)It will increase by a factor of 9.
D)It will decrease by a factor of 9.
E)Not enough information is given.

Accepted Answer

To decrease the standard error (and margin of error) by a factor of 11ea9518_d601_fe3a_9016_519b5afb8037_TB3450_11, the sample size will need to be increased by a factor of 11ea9518_d601_fe3a_9016_519b5afb8037_TB3450_11 squared (since the standard error is inversely proportional to the square root of the sample size). Therefore, the necessary sample size will increase by a factor of 11ea9518_d601_fe3a_9016_519b5afb8037_TB3450_11 x 11ea9518_d601_fe3a_9016_519b5afb8037_TB3450_11 = 121. Choice C is the correct answer.

Question 19

Of 346 items tested,12 are found to be defective.Construct a 98% confidence interval for the percentage of all such items that are defective.
A)(1.18%,5.76%)
B)(0.93%,6.00%)
C)(3.34%,3.59%)
D)(1.85%,5.09%)
E)(0.13%,6.80%)

Accepted Answer

The confidence interval for a proportion can be calculated using the formula: $$CI = \hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $\hat{p}$ is the sample proportion (12/346), $n$ is the sample size (346), and $Z$ is the Z-score corresponding to the confidence level (for 98%, Z ≈ 2.33). Plugging in the values, we get the confidence interval as approximately (1.18%, 5.76%).

Question 20

When 278 college students are randomly selected and surveyed,it is found that 110 own a car.Construct a 99% confidence interval for the percentage of all college students who own a car.
A)(34.7%,44.4%)
B)(18.6%,60.6%)
C)(33.8%,45.3%)
D)(32.0%,47.1%)
E)(32.7%,46.4%)

Accepted Answer

The answer of When 278 college students are randomly selected...

Quiz 16: Confidence Intervals for Proportions

A mayoral election race is tightly contested.In a random sample of 1,100 likely voters,572 said that they were planning to vote for the current mayor.Based on this sample,would you claim with 95% confidence that the mayor will win a majority of the votes? Explain.

In a survey of 1,000 television viewers,40% said they watch network news programs.For a 90% confidence level,the margin of error for this estimate is 2.5%.If we want to be 99% confident,how will the margin of error change?

A poll of 565 home sellers found that 25.1% were trying to sell their home without a real estate agent.The margin of error for this poll was 3%.

Of 88 adults selected randomly from one town,60 have health insurance.Construct a 90% confidence interval for the percentage of all adults in the town who have health insurance.

In a poll of 553 voters in a certain city,64% said that they supported the NDP in an upcoming election.The margin of error in the poll was reported as 4 percentage points (with a 95% degree of confidence) .Make a statement about the adequacy of the sample size for the given margin of error.

A newspaper reports that the Prime Minister's approval rating stands at 60%.The article adds that the poll is based on a random sample of 1,040 adults and has a margin of error of 2.5%.

Of 314 randomly selected medical students,25 said that they planned to work in a rural community.Construct a 95% confidence interval for the percentage of all medical students who plan to work in a rural community.

A newspaper reports that 62% of city residents have visited one of the city parks in the last year.The article adds that the poll is based on a random sample of 996 residents and has a margin of error of 2.5%.

A mayoral election race is tightly contested.In a random sample of 1,900 likely voters,1,007 said that they were planning to vote for the current mayor.Based on this sample,would you claim with 95% confidence that the mayor will win a majority of the votes? Explain.

A study involves 673 randomly selected deaths,with 34 of them caused by accidents.Construct a 98% confidence interval for the percentage of all deaths that are caused by accidents.

In a poll of 2778 voters in a certain city,77% said that they supported the Conservative Party of Canada.The margin of error in the poll was reported as 6 percentage points (with a 95% degree of confidence) .Make a statement about the adequacy of the sample size for the given margin of error.

Of 125 adults selected randomly from one town,28 of them smoke.Construct a 99% confidence interval for the percentage of all adults in the town who smoke.

Of 130 randomly selected adults,32 were found to have high blood pressure.Construct a 95% confidence interval for the percentage of all adults who have high blood pressure.

A poll of 2841 registered voters in a certain town indicated that 70% would vote for the Liberal Party of Canada.The margin of error for this poll was 2%.

Of 290 employees selected randomly from one company,12.41% of them commute by carpooling.Construct a 90% confidence interval for the percentage of all employees of the company who carpool.

A newspaper reports that 58% of farmers are planning to plant genetically modified crops this year.The article adds that the poll is based on a random sample of 798 farmers and has a margin of error of 4.5%.

A survey of 300 students at York University revealed that 112 favour an NDP candidate for M.P.Construct a 98% confidence interval for the percentage of all York University students who favour an NDP candidate.

After conducting a survey,a researcher wishes to cut the standard error (and thus the margin of error) to $\frac { 1 } { 3 }$ of its original value.How will the necessary sample size change?

Of 346 items tested,12 are found to be defective.Construct a 98% confidence interval for the percentage of all such items that are defective.

When 278 college students are randomly selected and surveyed,it is found that 110 own a car.Construct a 99% confidence interval for the percentage of all college students who own a car.

Quiz 16: Confidence Intervals for Proportions

A mayoral election race is tightly contested.In a random sample of 1,100 likely voters,572 said that they were planning to vote for the current mayor.Based on this sample,would you claim with 95% confidence that the mayor will win a majority of the votes? Explain.

In a survey of 1,000 television viewers,40% said they watch network news programs.For a 90% confidence level,the margin of error for this estimate is 2.5%.If we want to be 99% confident,how will the margin of error change?

A poll of 565 home sellers found that 25.1% were trying to sell their home without a real estate agent.The margin of error for this poll was 3%.

Of 88 adults selected randomly from one town,60 have health insurance.Construct a 90% confidence interval for the percentage of all adults in the town who have health insurance.

In a poll of 553 voters in a certain city,64% said that they supported the NDP in an upcoming election.The margin of error in the poll was reported as 4 percentage points (with a 95% degree of confidence) .Make a statement about the adequacy of the sample size for the given margin of error.

A newspaper reports that the Prime Minister's approval rating stands at 60%.The article adds that the poll is based on a random sample of 1,040 adults and has a margin of error of 2.5%.

Of 314 randomly selected medical students,25 said that they planned to work in a rural community.Construct a 95% confidence interval for the percentage of all medical students who plan to work in a rural community.

A newspaper reports that 62% of city residents have visited one of the city parks in the last year.The article adds that the poll is based on a random sample of 996 residents and has a margin of error of 2.5%.

A mayoral election race is tightly contested.In a random sample of 1,900 likely voters,1,007 said that they were planning to vote for the current mayor.Based on this sample,would you claim with 95% confidence that the mayor will win a majority of the votes? Explain.

A study involves 673 randomly selected deaths,with 34 of them caused by accidents.Construct a 98% confidence interval for the percentage of all deaths that are caused by accidents.

In a poll of 2778 voters in a certain city,77% said that they supported the Conservative Party of Canada.The margin of error in the poll was reported as 6 percentage points (with a 95% degree of confidence) .Make a statement about the adequacy of the sample size for the given margin of error.

Of 125 adults selected randomly from one town,28 of them smoke.Construct a 99% confidence interval for the percentage of all adults in the town who smoke.

Of 130 randomly selected adults,32 were found to have high blood pressure.Construct a 95% confidence interval for the percentage of all adults who have high blood pressure.

A poll of 2841 registered voters in a certain town indicated that 70% would vote for the Liberal Party of Canada.The margin of error for this poll was 2%.

Of 290 employees selected randomly from one company,12.41% of them commute by carpooling.Construct a 90% confidence interval for the percentage of all employees of the company who carpool.

A newspaper reports that 58% of farmers are planning to plant genetically modified crops this year.The article adds that the poll is based on a random sample of 798 farmers and has a margin of error of 4.5%.

A survey of 300 students at York University revealed that 112 favour an NDP candidate for M.P.Construct a 98% confidence interval for the percentage of all York University students who favour an NDP candidate.

After conducting a survey,a researcher wishes to cut the standard error (and thus the margin of error) to 13\frac { 1 } { 3 }31​ of its original value.How will the necessary sample size change?

Of 346 items tested,12 are found to be defective.Construct a 98% confidence interval for the percentage of all such items that are defective.

When 278 college students are randomly selected and surveyed,it is found that 110 own a car.Construct a 99% confidence interval for the percentage of all college students who own a car.

After conducting a survey,a researcher wishes to cut the standard error (and thus the margin of error) to $\frac { 1 } { 3 }$ of its original value.How will the necessary sample size change?