Question 1

Assume that 15% of students at a university wear contact lenses.We randomly pick 200 students.What is the mean of the proportion of students in this group who may wear contact lenses? 
A) $\mu = 5.48 \%$
B) $\mu = 2.52 \%$
C) $\mu = 7.5 \%$
D) $\mu = 15 \%$
E) $\mu = 30 \%$

Accepted Answer

The mean of the proportion of students who wear contact lenses is equal to the population proportion, which is 15%.

Question 2

A survey found that 79% of a random sample of 1024 American 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 American adults who approve of cloning endangered animals.
A)2.49%
B)21.4%
C)2.09%
D)5.43%
E)4.56%

Accepted Answer

The margin of error (ME) can be calculated using the formula for a proportion: ME = Z * sqrt[(p(1-p))/n], where Z is the Z-score for the desired confidence level (1.645 for 90%), p is the sample proportion (0.79), and n is the sample size (1024). Calculating this gives approximately 2.09%.

Question 3

In a large class, the professor has each person toss a coin several times and calculate the proportion of his or her tosses that were heads.The students then report their results, and the professor plots a histogram of these several proportions.If the students toss the coin 100 times each, about 68% should have proportions between what two numbers?
Use the 68-95-99.7 Rule to provide the appropriate response.
A)0.34 and 0.67
B)0.45 and 0.55
C)0.16 and 0.84
D)0.495 and 0.505
E)0.05 and 0.1

Accepted Answer

The 68-95-99.7 Rule, or the empirical rule, states that for a normal distribution, about 68% of the data falls within one standard deviation (σ) of the mean (μ). For a fair coin toss, μ = 0.5 (since the probability of heads is 0.5). The standard deviation for a binomial distribution is calculated as √[p(1-p)/n], where p is the probability of success (0.5 for heads), and n is the number of trials (100 tosses). Thus, σ = √[0.5(1-0.5)/100] = √(0.25/100) = √(0.0025) = 0.05. Therefore, about 68% of the proportions should fall within one standard deviation of the mean, which is 0.5 ± 0.05, or between 0.45 and 0.55.

Question 4

A survey of shoppers is planned to see what percentage use credit cards.Prior surveys Suggest 54% 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)16,471
B)8589
C)20,745
D)6718.3
E)9543

Accepted Answer

To find the sample size needed for a proportion with a given margin of error and confidence level, use the formula: n = (Z^2 * p * (1-p)) / E^2. For 95% confidence, Z is 1.96. Here, p = 0.54 and E = 0.01. Calculating gives approximately 9543.

Question 5

Based on past experience, a bank believes that 7% of the people who receive loans will not make payments on time.The bank has recently approved 300 loans.What is the probability that over 8% of these clients will not make timely payments?
A)0.752
B)0.504
C)0.248
D)0.68
E)0.496

Accepted Answer

This is a binomial probability problem that can be approximated using the normal distribution because the sample size is large (n = 300). The mean (μ) of the distribution of the proportion of clients not making timely payments is np = 300 * 0.07 = 21. The standard deviation (σ) is sqrt(np(1-p)) = sqrt(300*0.07*0.93) ≈ 4.34. To find the probability of more than 8% (which is 24 out of 300) not making timely payments, we convert 24 to a z-score: z = (24 - 21) / 4.34 ≈ 0.69. Looking up 0.69 on the z-table gives a left-tail probability, so we subtract this from 1 to find the upper tail probability, which is approximately 0.248.

Question 6

When 346 college students are randomly selected and surveyed, it is found that 121 own a car. Construct a 99% confidence interval for the percentage of all college students who own a car.
A)(12.1%, 57.8%)
B)(29.9%, 40.0%)
C)(29.0%, 40.9%)
D)(28.4%, 41.6%)
E)(30.8%, 39.2%)

Accepted Answer

The sample proportion is 121/346. Using the formula for a confidence interval for a proportion and a 99% confidence level, the interval is calculated to be approximately (28.4%, 41.6%).

Question 7

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

Accepted Answer

The necessary sample size will increase by a factor of 9 because the standard error of the mean is inversely proportional to the square root of the sample size. To cut the standard error to 1/3 of its original value, the square of 3 (3^2 = 9) must be applied to the sample size.

Question 8

Based on past experience, a bank believes that 6% of the people who receive loans will not make payments on time.The bank has recently approved 300 loans.What is the standard deviation of the proportion of clients in this group who may not make timely payments? 
A) $\sigma = 1.88 \%$
B) $\sigma = 1.41 \%$
C) $\sigma = 1.37 \%$
D) $\sigma = 4.11 \%$
E) $\sigma = 4.24 \%$

Accepted Answer

The standard deviation of the proportion of clients in this group who may not make timely payments is:$\sigma = \sqrt{p(1-p)/n}$where:p is the proportion of clients who do not make timely payments (0.06)n is the number of loans (300)Plugging in these values, we get:$\sigma = \sqrt{0.06(1-0.06)/300} = 0.0137$or 1.37%

Question 9

In a survey of 300 T.V.viewers, 40% 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.
A)4.16%
B)7.28%
C)6.37%
D)5.54%
E)8.36%

Accepted Answer

The answer of In a survey of 300 T.V.viewers, 40%...

Question 10

In a poll of 929 voters in a certain city, 68% said that they backed a bill which would limit Growth and development in their city.The margin of error in the poll was reported as 3 Percentage points (with a 95% degree of confidence).Which statement is correct?
A)The reported margin of error is consistent with the sample size.
B)The sample size is too large to achieve the stated margin of error.
C)For the given sample size, the margin of error should be smaller than stated
D)The sample size is too small to achieve the stated margin of error.

Accepted Answer

The margin of error in a poll is influenced by the sample size and the confidence level. A sample size of 929 voters can achieve a margin of error of approximately 3 percentage points at a 95% confidence level, making the reported margin of error consistent with the sample size.

Question 11

Assume that 15% of students at a university wear contact lenses.We randomly pick 200 students.What is the probability that more than 16% of this sample wear contact lenses?
A)0.309
B)0.397
C)0.654
D)0.346
E)0.692

Accepted Answer

The answer of Assume that 15% of students at a...

Question 12

In a poll of 268 voters in a certain city, 60% said that they backed a bill which would limit growth and development in their city.The margin of error in the poll was reported as 6 percentage points (with a 95% degree of confidence).Which statement is correct?
A)The stated margin of error could be achieved with a larger sample size.
B)The reported margin of error is consistent with the sample size.
C)The stated margin of error could be achieved with a smaller sample size.
D)The sample size is too small to achieve the stated margin of error.
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 268 voters in...

Question 13

In a large class, the professor has each person toss a coin several times and calculate the proportion of his or her tosses that were heads.The students then report their results, and the professor plots a histogram of these several proportions.If the students toss the coin 80 times each, about 95% should have proportions between what two numbers?
Use the 68-95-99.7 Rule to provide the appropriate response.
A)0.025 and 0.975
B)0.39 and 0.62
C)0.1118 and 0.1677
D)0.4875 and 0.5125
E)0.2375 and 0.7375

Accepted Answer

The 68-95-99.7 Rule, also known as the Empirical Rule, applies to normal distributions and states that about 95% of the data falls within two standard deviations of the mean. For a fair coin toss, the mean proportion of heads is 0.5. The standard deviation for the proportion of heads in 80 tosses can be calculated using the formula for the standard deviation of a binomial distribution, which is sqrt[p(1-p)/n], where p is the probability of success (0.5 for a fair coin), and n is the number of trials (80). This gives sqrt[0.5*0.5/80] = sqrt[0.25/80] = sqrt[0.003125] ≈ 0.0559. Therefore, two standard deviations is 2*0.0559 = 0.1118. Adding and subtracting this from the mean (0.5) gives us the range 0.5 ± 0.1118, which is approximately 0.3882 to 0.6118, closest to the range given in option B (0.39 to 0.62).

Question 14

The number of hours per week that high school seniors spend on homework is normally distributed, with a mean of 10 hours and a standard deviation of 3 hours.60 students are chosen at random.Let represent the mean number of hours spent on homework for this group.Find the probability that is between 9.8 and 10.4.
A)0.547
B)0.5161
C)0.080
D)0.1528
E)0.3043

Accepted Answer

The sampling distribution of the sample mean for a sample size of 60, given the population is normally distributed, will also be normally distributed with a mean ($\mu_{\bar{x}}$) equal to the population mean ($\mu$) and a standard deviation ($\sigma_{\bar{x}}$) equal to the population standard deviation ($\sigma$) divided by the square root of the sample size ($n$), i.e., $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$. Here, $\mu = 10$ hours, $\sigma = 3$ hours, and $n = 60$. Thus, $\sigma_{\bar{x}} = \frac{3}{\sqrt{60}}$.To find the probability that $\bar{x}$ is between 9.8 and 10.4, we convert these values to Z-scores using the formula $Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$.For 9.8: $Z = \frac{9.8 - 10}{3/\sqrt{60}}$For 10.4: $Z = \frac{10.4 - 10}{3/\sqrt{60}}$Calculating these Z-scores and then using the standard normal distribution table or a calculator, we find the probabilities corresponding to these Z-scores and subtract the smaller from the larger to find the probability that $\bar{x}$ is between 9.8 and 10.4. This calculation yields a probability of approximately 0.547, making choice A correct.

Question 15

Assume that 10% of students at a university wear contact lenses.We randomly pick 200 students.What is the standard deviation of the proportion of students in this group who may wear contact lenses? 
A) $\sigma = 4.5 \%$
B) $\sigma = 2.24 \%$
C) $\sigma = 2.12 \%$
D) $\sigma = 4.47 \%$
E) $\sigma = 4.24 \%$

Accepted Answer

The standard deviation of the sample proportion is calculated using the formula $\sigma = \sqrt{\frac{p(1-p)}{n}}$, where $p = 0.10$ and $n = 200$. This gives $\sigma = \sqrt{\frac{0.10 \times 0.90}{200}} = 0.0212$ or 2.12%.

Question 16

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 15%.He wants the margin of error to be within 5% at a 99% confidence level.What sample size will suffice?
A)17
B)138.3
C)407
D)196
E)  339

Accepted Answer

The answer of A university's administrator proposes to do an...

Question 17

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)(0.13%, 6.80%)
B)(3.34%, 3.59%)
C)(0.93%, 6.00%)
D)(1.85%, 5.09%)
E)(1.18%, 5.76%)

Accepted Answer

The sample proportion of defective items is 12/346. Using the formula for a confidence interval for a proportion and a Z-score for 98% confidence, the interval is calculated as (1.18%, 5.76%).

Question 18

Based on past experience, a bank believes that 4% of the people who receive loans will not make payments on time.The bank has recently approved 300 loans.What is the mean of the proportion of clients in this group who may not make timely payments? 
A) $\mu = 4 \%$
B) $\mu = 3.46 \%$
C) $\mu = 1.1 \%$
D) $\mu = 2 \%$
E) $\mu = 12 \%$

Accepted Answer

The mean of the proportion of clients who may not make timely payments is the expected rate of not making payments on time, which is 4%.

Question 19

The college daily reported: " 300 students living in university housing were polled. 180 said that they were satisfied with their living conditions. Based on this survey we conclude that $60 \%$ of students living in dormitories are satisfied. The margin of error of the study is $\pm 6$ percentage points (with a $95 \%$ degree of confidence). Which statement is correct?
A) A larger sample should be used to achieve the stated margin of error.
B) The margin of error is consistent with the sample size.
C) The sample size is too small to achieve the stated margin of error.
D) The stated margin of error could have been 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 The college daily reported: " 300 students...

Question 20

The number of hours per week that high school seniors spend on computers is normally distributed, with a mean of 4 hours and a standard deviation of 2 hours.60 students are chosen at random.Let represent the mean number of hours spent on the computer for this group.Find the probability that is between 4.2 and 4.4.
A)0.041
B)0.159
C)0.317
D)0.720
E)0.775

Accepted Answer

The answer of The number of hours per week that...

Quiz 13: Sampling Distribution Models and Confidence Intervals for Proportions

Assume that 15% of students at a university wear contact lenses.We randomly pick 200 students.What is the mean of the proportion of students in this group who may wear contact lenses?

A survey found that 79% of a random sample of 1024 American 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 American adults who approve of cloning endangered animals.

A survey of shoppers is planned to see what percentage use credit cards.Prior surveys Suggest 54% 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?

Based on past experience, a bank believes that 7% of the people who receive loans will not make payments on time.The bank has recently approved 300 loans.What is the probability that over 8% of these clients will not make timely payments?

When 346 college students are randomly selected and surveyed, it is found that 121 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 marginof error) to of its original value.How will the necessary sample size change?

Based on past experience, a bank believes that 6% of the people who receive loans will not make payments on time.The bank has recently approved 300 loans.What is the standard deviation of the proportion of clients in this group who may not make timely payments?

In a survey of 300 T.V.viewers, 40% 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.

In a poll of 929 voters in a certain city, 68% said that they backed a bill which would limit Growth and development in their city.The margin of error in the poll was reported as 3 Percentage points (with a 95% degree of confidence) .Which statement is correct?

Assume that 15% of students at a university wear contact lenses.We randomly pick 200 students.What is the probability that more than 16% of this sample wear contact lenses?

In a poll of 268 voters in a certain city, 60% said that they backed a bill which would limit growth and development in their city.The margin of error in the poll was reported as 6 percentage points (with a 95% degree of confidence) .Which statement is correct?

Assume that 10% of students at a university wear contact lenses.We randomly pick 200 students.What is the standard deviation of the proportion of students in this group who may wear contact lenses?

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.

Based on past experience, a bank believes that 4% of the people who receive loans will not make payments on time.The bank has recently approved 300 loans.What is the mean of the proportion of clients in this group who may not make timely payments?