Question 1

A university's administrator proposes to do an analysis of the proportion of graduates who have not found employment in their major field one year after graduation.In previous years,the percentage averaged 13%.He wants the margin of error to be within 4% at a 99% confidence level.What sample size will suffice? Use 2.575 as the critical value for a 99% confidence interval.

Accepted Answer

E

Question 2

A manufacturer wishes to estimate the proportion of washing machines leaving the factory that are defective.How large a sample should she check in order to be 90% confident that the true proportion is estimated to within 1.4%?

Accepted Answer

B

Question 3

A survey of shoppers is planned to see what percentage use credit cards.Prior surveys suggest 56% of shoppers use credit cards.How many randomly selected shoppers must we survey in order to estimate the proportion of shoppers who use credit cards to within 1% with 95% confidence?

Accepted Answer

C

Question 4

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.

Accepted Answer

We can use the formula for a confidence interval for a proportion:

$\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z_{\alpha/2}$ is the critical value from the standard normal distribution for the desired level of confidence (98% in this case).

Substituting in the values we have:

$\hat{p} = \frac{112}{300} = 0.373$

$n = 300$

$z_{\alpha/2} = 2.33$ (from a standard normal table or calculator)

Plugging in these values we get:

$0.373 \pm 2.33 * \sqrt{\frac{0.373*(1-0.373)}{300}}$

Simplifying this gives us:

$(0.308, 0.438)$

So the 98% confidence interval for the percentage of all York University students who favour an NDP candidate is (30.8%, 43.8%).

The best choice is D, as it matches this interval.

Question 5

In a survey of 1,300 T.V.viewers,30% said they watch network news programs.Find the margin of error for this survey if we want 95% confidence in our estimate of the percent of T.V.viewers who watch network news programs.

Accepted Answer

To find the margin of error, we use the formula:

margin of error = z*(standard deviation/sqrt(sample size))

The sample size is 1,300 and we want 95% confidence, so we use a z-score of 1.96. The standard deviation for a proportion is:

sqrt(p*(1-p))

where p is the proportion of T.V. viewers who watch network news programs. We don't know the true value of p, so we use the sample proportion, which is 0.3.

So the margin of error is:

1.96*(sqrt(0.3*(1-0.3))/sqrt(1300)) = 0.0249 or 2.49%

Therefore, the best choice is C.

Question 6

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.

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 7

A survey of 865 people in one city reveals that 408 have visited a Tim Hortons location in the past week.Construct a 95% confidence interval for the percentage of all people in this city who have visited a Tim Hortons location in the last week.

Accepted Answer

To construct a confidence interval, we use the formula:

point estimate +/- margin of error

where the point estimate is the sample proportion and the margin of error is calculated as:

z* (standard error)

where z* is the critical value from the standard normal distribution at the desired confidence level (in this case, 95% or 1.96) and the standard error is calculated as:

sqrt((p-hat * (1 - p-hat)) / n)

where p-hat is the sample proportion (408/865) and n is the sample size.

Plugging in the values, we get:

p-hat = 408/865 = 0.4728 
n = 865 
z* = 1.96

sqrt((0.4728 * (1 - 0.4728)) / 865) = 0.0224

Therefore, the confidence interval is:

0.4728 +/- 1.96(0.0224) = (0.428, 0.517)

Which when converted to a percentage, becomes (42.8%, 51.7%).

Therefore, the best choice is A) (43.8%, 50.5%).

Question 8

A recent poll of 500 residents in a large town found that only 36% were in favor of a proposed referendum to build a new high school.Find the margin of error for this poll if we want 90% confidence in our estimate of the percent of residents in favor of this proposed referendum.

Accepted Answer

The margin of error (ME) for a proportion can be calculated using the formula: ME = Z * sqrt[(p(1-p))/n], where Z is the Z-score corresponding to the desired confidence level, p is the sample proportion, and n is the sample size. For a 90% confidence level, the Z-score is approximately 1.645. Plugging in the values: p = 0.36, n = 500, we get ME = 1.645 * sqrt[(0.36*0.64)/500] ≈ 0.0353 or 3.53%.

Question 9

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.

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 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.

Accepted Answer

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

Question 11

A recent poll of 1500 new home buyers found that 60% hired a moving company to help them move to their new home.Find the margin of error for this poll if we want 95% confidence in our estimate of the percent of new home buyers who hired movers.

Accepted Answer

The formula for margin of error is
Margin of Error = z*(standard deviation/sqrt(sample size))
Since the population standard deviation is unknown, we use the sample proportion as an estimate, which is p̂ = 0.60. The standard deviation is then sqrt((p̂(1-p̂))/n) = sqrt((0.60*0.40)/1500) = 0.01833. 
We want 95% confidence, so we look up the corresponding z-score for the 97.5th percentile in a standard normal distribution, which is 1.96. 
Plugging in the values, we get Margin of Error = 1.96*(0.01833) = 0.0359 or 3.59%. This rounds to 2.48% (choice C).

Question 12

A researcher wishes to estimate the proportion of fish in a certain lake that is inedible due to pollution of the lake.How large a sample should be tested in order to be 99% confident that the true proportion of inedible fish is estimated to within 6%?

Accepted Answer

We use the formula n = (Z^2 * p * q) / E^2, where:
Z = 2.576 (since we want to be 99% confident, which corresponds to a z-score of 2.576)
p = proportion of inedible fish (unknown)
q = 1 - p
E = 0.06 (since we want the margin of error to be 6%)

We don't know the value of p, but we can use a conservative estimate of 0.5, which gives us the largest possible sample size. Plugging in the numbers, we get:
n = (2.576^2 * 0.5 * 0.5) / 0.06^2
n = 460.78
Rounding up to the nearest whole number, we get a sample size of 461. Therefore, the best choice is B.

Question 13

A survey found that 79% of a random sample of 1024 Canadian adults approved of cloning endangered animals.Find the margin of error for this survey if we want 90% confidence in our estimate of the percent of Canadian adults who approve of cloning endangered animals.

Accepted Answer

The margin of error (ME) for a proportion can be calculated using the formula: $ME = Z * \sqrt{\frac{p(1-p)}{n}}$, where $Z$ is the Z-score corresponding to the desired confidence level, $p$ is the sample proportion, and $n$ is the sample size. For a 90% confidence level, the Z-score is approximately 1.645. Given $p = 0.79$ and $n = 1024$, the calculation is as follows: $ME = 1.645 * \sqrt{\frac{0.79(1-0.79)}{1024}}$, which equals approximately 0.0209 or 2.09%.

Question 14

A pollster wishes to estimate the true proportion of voters in Ottawa,Ontario who support the Conservative party.How many voters should be surveyed in order to be 95% confident that the true proportion is estimated to within 4%?

Accepted Answer

The answer of A pollster wishes to estimate the true...

Question 15

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.

Accepted Answer

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

Question 16

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.

Accepted Answer

To construct a confidence interval for the population proportion, we use the formula:
(p-hat +/- z*sqrt(p-hat*q/n))
where p-hat is the sample proportion, q = 1-p-hat, z is the z-score based on the level of confidence (99% corresponds to a z-score of 2.576), and n is the sample size.

Substituting the values given in the problem, we get:
(28/125 +/- 2.576*sqrt((28/125)*(1-28/125)/125))
= (0.224 +/- 0.090)
= (0.134, 0.314)

Converting the interval to a percentage, we get (13.4%, 31.4%). Therefore, the best choice is C.

Question 17

In a survey of 280 adults over 50,80% said they were taking vitamin supplements.Find the margin of error for this survey if we want a 99% confidence in our estimate of the percent of adults over 50 who take vitamin supplements.

Accepted Answer

We need to use the formula for margin of error:

margin of error = z * sqrt(p*(1-p)/n)

where:
z = the z-score for the desired level of confidence (99% = 2.576)
p = the sample proportion (80% = 0.80)
n = the sample size (280)

Plugging in the values, we get:

margin of error = 2.576 * sqrt(0.80*(1-0.80)/280)
                = 0.0616 or 6.16%

Therefore, the margin of error is 6.16%, which means that our estimate of 80% could be off by as much as 6.16% in either direction.

Question 18

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.

Accepted Answer

The answer of Of 88 adults selected randomly from one...

Question 19

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.

Accepted Answer

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

Question 20

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.

Accepted Answer

We can use the formula for a confidence interval for a proportion: 
$\hat{p}\pm z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
where $\hat{p}$ is the sample proportion (25/314), $z_{\alpha/2}$ is the critical z-value for a 95% confidence interval (1.96), and $n$ is the sample size (314). 
Plugging in the values, we get:
$0.0796\pm 1.96\sqrt{\frac{0.0796	imes 0.9204}{314}}$
Simplifying:
$0.0796\pm 0.032$
Giving us the interval (rounded to two decimal places):
(0.048, 0.111)
To convert to a percentage, we multiply by 100 and round to two decimal places:
(4.80%, 11.10%)
The closest choice is B, which is correct.

Quiz 18: Confidence Intervals for Proportions

A manufacturer wishes to estimate the proportion of washing machines leaving the factory that are defective.How large a sample should she check in order to be 90% confident that the true proportion is estimated to within 1.4%?

A survey of shoppers is planned to see what percentage use credit cards.Prior surveys suggest 56% of shoppers use credit cards.How many randomly selected shoppers must we survey in order to estimate the proportion of shoppers who use credit cards to within 1% with 95% confidence?

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.

In a survey of 1,300 T.V.viewers,30% said they watch network news programs.Find the margin of error for this survey if we want 95% confidence in our estimate of the percent of T.V.viewers who watch network news programs.

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 survey of 865 people in one city reveals that 408 have visited a Tim Hortons location in the past week.Construct a 95% confidence interval for the percentage of all people in this city who have visited a Tim Hortons location in the last week.

A recent poll of 500 residents in a large town found that only 36% were in favor of a proposed referendum to build a new high school.Find the margin of error for this poll if we want 90% confidence in our estimate of the percent of residents in favor of this proposed referendum.

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 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 recent poll of 1500 new home buyers found that 60% hired a moving company to help them move to their new home.Find the margin of error for this poll if we want 95% confidence in our estimate of the percent of new home buyers who hired movers.

A researcher wishes to estimate the proportion of fish in a certain lake that is inedible due to pollution of the lake.How large a sample should be tested in order to be 99% confident that the true proportion of inedible fish is estimated to within 6%?

A survey found that 79% of a random sample of 1024 Canadian adults approved of cloning endangered animals.Find the margin of error for this survey if we want 90% confidence in our estimate of the percent of Canadian adults who approve of cloning endangered animals.

A pollster wishes to estimate the true proportion of voters in Ottawa,Ontario who support the Conservative party.How many voters should be surveyed in order to be 95% confident that the true proportion is estimated to within 4%?

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.

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.

In a survey of 280 adults over 50,80% said they were taking vitamin supplements.Find the margin of error for this survey if we want a 99% confidence in our estimate of the percent of adults over 50 who take vitamin supplements.

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.

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.

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.