Question 1

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

To find the margin of error (E), we use the formula:
E = z*sqrt(p*(1-p)/n)
where 
z = the z-score for 95% confidence level which is 1.96 (from the standard normal distribution table)
p = the sample proportion = 0.6
n = the sample size = 1500

Substituting the values, we get:
E = 1.96*sqrt(0.6*0.4/1500) = 0.0248 = 2.48%

Therefore, the margin of error is 2.48%, which is closest to option C (2.48%).

Question 2

Data in 1980 showed that about 42% of one country's population had never smoked cigarettes.In 2004,a national health survey in this country interviewed a random sample of 3000 adults and found that 51% had never been smokers.Create a 95% confidence interval for the proportion in 2004 of this country's adults who had never been smokers.

Accepted Answer

The 95% confidence interval for a proportion is calculated using the formula: $p \pm z\sqrt{\frac{p(1-p)}{n}}$, where $p$ is the sample proportion, $z$ is the z-score for the desired confidence level (approximately 1.96 for 95%), and $n$ is the sample size. Plugging in the values: $0.51 \pm 1.96\sqrt{\frac{0.51(1-0.51)}{3000}}$ gives an interval of approximately $0.492$ to $0.528$ or 49.2% to 52.8%.

Question 3

Five years ago,a company found that 8% of its employees commuted to work by car.A test for the current proportion of commuting employees was done on a random sample of 1000 employees,and found car commuting to be 12%.Create a 95% confidence interval for the proportion of employee commute by car.

Accepted Answer

To calculate a 95% confidence interval for the proportion, we use the formula: $\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p} = 0.12$, $Z = 1.96$ for 95% confidence, and $n = 1000$. Plugging these values in, we get $0.12 \pm 1.96 \sqrt{\frac{0.12(0.88)}{1000}}$, which simplifies to $0.12 \pm 0.02001$, or an interval from $9.99\%$ to $14.01\%$.

Question 4

A new manager,hired at a large warehouse,was told to reduce the 26% employee sick leave.The manager introduced a new incentive program for employees with perfect attendance.The manager decides to test the new program on a random sample size of 60 and found sick leave at 23%.Create a 95% confidence interval for the proportion of employee sick leave.

Accepted Answer

The answer of A new manager,hired at a large warehouse,was...

Question 5

The real estate industry claims that it is the best and most effective system to market residential real estate.A survey of randomly selected home sellers in Ontario found that a 99% confidence interval for the proportion of homes that are sold by a real estate agent is 70% to 80%.Explain what "99% confidence" means in this context.

Accepted Answer

"99% confidence" means that 99% of all random samples taken from the population of home sellers in Ontario will produce a confidence interval that contains the true proportion of homes sold by a real estate agent. In other words, if we were to take many random samples of home sellers in Ontario and construct a confidence interval for each sample using the same method, about 99% of these intervals would contain the true proportion of homes sold by a real estate agent. The 99% confidence interval in this survey ranges from 70% to 80%, which means that we are very confident (at the 99% level) that the true proportion of homes sold by a real estate agent in Ontario is between 70% and 80%.

Question 6

The real estate industry claims that it is the best and most effective system to market residential real estate.A survey of randomly selected home sellers in Ontario found that a 99% confidence interval for the proportion of homes that are sold by a real estate agent is 66% to 76%.Interpret the interval in this context.

Accepted Answer

This choice correctly interprets the 99% confidence interval as applying to all homes in Ontario, and not just to the sample from which it was calculated. It also acknowledges that the interval is based on a sample and not necessarily exact for the entire population.

Question 7

A weight loss centre provided a loss for 72% of its participants.The centre's leader decides to test a new weight loss strategy on a random sample size of 140 and found weight loss in 78% of the participants.Create a 95% confidence interval for the proportion of participant weight loss.

Accepted Answer

To create a 95% confidence interval for the proportion of participant weight loss, we can use the formula:

(sample proportion) +/- (critical value) x (standard error)

where the sample proportion is 0.78 (from the information given in the question), the critical value for a 95% confidence level is 1.96 (from a standard normal distribution table), and the standard error can be calculated as:

sqrt[(sample proportion x (1 - sample proportion)) / sample size]

= sqrt[(0.78 x 0.22) / 140]

= 0.043

Plugging in the values, we get:

0.78 +/- 1.96 x 0.043

= 0.78 +/- 0.084

Therefore, the 95% confidence interval for the proportion of participant weight loss is between 0.78 - 0.084 = 0.696 (or 69.6%) and 0.78 + 0.084 = 0.864 (or 86.4%).

Taking into account the given answer choices, the closest interval to this is answer choice E, where we are 95% confident that the proportion of participant weight loss is between 71.14% and 84.86%.

Question 8

In April of 2011,the unemployment rate in Canada was reported to be 7.6%.Suppose that Statistics Canada claims that the current unemployment rate is lower than that of April 2011.Suppose a test was conducted on current unemployment and a random sample size of 1000 and found unemployment at 3.8%.Create a 95% confidence interval for the proportion of unemployment.

Accepted Answer

To calculate a 95% confidence interval for the proportion, we use the formula for the confidence interval of a proportion: $p \pm z\sqrt{\frac{p(1-p)}{n}}$, where $p$ is the sample proportion (0.038), $n$ is the sample size (1000), and $z$ is the z-score corresponding to a 95% confidence level (approximately 1.96). Plugging in the values: $0.038 \pm 1.96\sqrt{\frac{0.038(1-0.038)}{1000}}$ gives an interval of approximately 0.0261 to 0.0499, or 2.61% to 4.99%.

Question 9

A company hopes to improve its engines,setting a goal of no more than 3% of customers using their warranty on defective engine parts.A random survey of 1200 customers found only 20 with complaints.Create a 95% confidence interval for the true level of warranty users among all customers.

Accepted Answer

The sample proportion is 20/1200 = 0.0167. Using the formula for a confidence interval for a proportion, the 95% confidence interval is approximately 0.9% to 2.4%, which is below the 3% goal.

Question 10

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 11

The real estate industry claims that it is the best and most effective system to market residential real estate.A survey of randomly selected home sellers in Ontario found that a 90% confidence interval for the proportion of homes that are sold by a real estate agent is 75% ± 6%.Explain why each of these conclusions is not correct.
a)Between 69% and 81% of all homes in Ontario are sold by a real estate agent.
b)90% of all random samples of home sellers in Ontario will show that 75% of homes are sold by a real estate agent.
c)90% of all random samples of home sellers in Ontario will show that 69% to 81% of homes are sold by a real estate agent.
d)We are 90% sure that between 69% and 81% of homes in this sample were sold by a real estate agent.

Accepted Answer

The answer of The real estate industry claims that it...

Question 12

A university wants to increase its retention rate of 4% for graduating students from the previous year.After implementing several new programs during the last two years,in 2014 the university reevaluated its retention rate using a random sample of 352 students and found the retention rate at 5%.Create a 95% confidence interval for the proportion of students (in 2014)who had been retained.

Accepted Answer

The 95% confidence interval for the proportion of students who had been retained is calculated using the formula (p̂ ± z*√((p̂(1-p̂))/n)), where p̂ is the sample proportion, z* is the z-value for the 95% confidence level (1.96), and n is the sample size. Substituting the given values, we get (0.05 ± 1.96*√((0.05(0.95))/352)), which simplifies to (0.0272, 0.0728). Therefore, we can say with 95% confidence that the proportion of students who had been retained in 2014 is between 2.72% and 7.28%. The answer is D.

Question 13

What is the difference between the standard deviation and the standard error in the sampling distribution of a sample proportion?

Accepted Answer

The answer of What is the difference between the standard...

Question 14

In a survey of 1300 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 need to use the formula:

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

where:
- z is the z-score for the confidence level (95% confidence level has a z-score of 1.96)
- p is the estimated proportion (30% or 0.3)
- n is the sample size (1300)

So, plugging in the values, we get:

margin of error = 1.96*sqrt(0.3*(1-0.3)/1300) ≈ 0.025 or 2.49%

Therefore, the best choice is C) 2.49%.

Question 15

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

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

Margin of Error = z * (sqrt(p*q/n))

where p is the proportion of adults over 50 who take vitamin supplements (0.8), q is the complementary proportion (1 - 0.8 = 0.2), n is the sample size (280), and z is the z-score that corresponds to the desired level of confidence (99%).

Using a z-score calculator, we find that the z-score for a 99% confidence level is approximately 2.576.

Substituting the values into the formula, we get:

Margin of Error = 2.576 * (sqrt(0.8*0.2/280)) = 0.0616 or 6.16%

Therefore, the best choice is C, with a margin of error of 6.16%.

Question 16

According to the 2004 Canadian Community Health Survey,23.1% of Canadians aged 18 or older were considered obese; that is,they had a body mass index (BMI)of 30 or higher.Suppose a current survey in 2014 of 1000 random Canadians found that 245 were obese.Create a 95% confidence interval for the proportion of Canadians that are currently obese.

Accepted Answer

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

CI = p ± z*sqrt((p*(1-p))/n)

where p is the sample proportion, z is the z-score corresponding to the desired level of confidence (1.96 for 95% confidence), and n is the sample size.

Plugging in the values given:

p = 245/1000 = 0.245

z = 1.96

n = 1000

CI = 0.245 ± 1.96*sqrt((0.245*(1-0.245))/1000)

CI = 0.245 ± 0.025

CI = (0.22, 0.27)

Therefore, we can say, with 95% confidence, that the true proportion of obese Canadians is between 21.8% and 27.2%. The best choice is D.

Question 17

A city claims that 75% of all low income housing is 1500 sq.ft.The tenants believe the proportion of housing this size is smaller than the claim,and hire an independent firm to test a random sample size of 120 and found 1500 sq.ft.housing at 66%.Create a 95% confidence interval for the proportion of housing 1500 sq.ft.

Accepted Answer

To calculate a 95% confidence interval for the proportion, we use the formula: $$CI = \hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$ where $\hat{p} = 0.66$, $n = 120$, and $Z_{\alpha/2} = 1.96$ for a 95% confidence level. Plugging these values into the formula gives us the interval $(0.5752, 0.7448)$ or $57.52\%, 74.48\%$.

Question 18

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 find the confidence interval, we use the formula:
CI = p̂ ± Z*√(p̂(1-p̂)/n)
where p̂ is the sample proportion (408/865), Z is the Z-score for a 95% confidence level (1.96), and n is the sample size (865).
Plugging in these values, we get:
CI = 0.471 ± 1.96*√[(0.471)(0.529)/865]
CI ≈ (0.438, 0.505)
So the best choice is A, which represents this interval.

Question 19

The management of a water park has concerns about the chlorine level of 0.4% mg/mL increasing to an unsafe level.The management tests the hypothesis that the park's chlorine proportions have remained the same.An independent agency ran a test on a random sample size of $10000$ and found the chlorine level at 0.5% mg/mL.Create a 95% confidence interval for the proportion of chlorine in the water.

Accepted Answer

The sample proportion is 0.5%. Using the formula for a confidence interval for a proportion, the interval is calculated as 0.5% ± 1.96 * sqrt((0.5% * (1 - 0.5%)) / 10000), which results in the interval 0.362% to 0.638%.

Question 20

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%.

Quiz 16: Confidence Intervals for Proportions