Question 1

Recently,a case of food poisoning was traced to a particular restaurant chain.The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur.Despite the response from the restaurant chain,many consumers refused to visit the restaurant for some time after the event.A survey was conducted three months after the food poisoning occurred with a sample of 319 patrons contacted.Of the 319 contacted,29 indicated that they would not go back to the restaurant because of the potential for food poisoning Construct a 95% confidence interval for the true proportion of the market who still refuse to visit any of the restaurants in the chain three months after the event.
A)[.059 .122]
B)[.090 .091]
C)[.000 .196]
D)[.240 .339]
E)[.118 .244]

Accepted Answer

The confidence interval is calculated using the formula for a proportion confidence interval: p̂ ± Z*(√(p̂(1-p̂)/n)), where p̂ is the sample proportion (29/319), Z is the Z-score for a 95% confidence level (approximately 1.96), and n is the sample size (319). This calculation results in the interval [.059, .122].

Question 2

A sociologist develops a test designed to measure a person's attitudes about disabled people,and 16 randomly selected subjects are given the test.Their mean score is 71.2 with a standard deviation of 10.5.Construct the 99% confidence interval for the mean score of all subjects.
A)[40.26 102.14]
B)[63.46 78.94]
C)[60.66 81.74]
D)[64.44 77.96]
E)[63.21 79.19]

Accepted Answer

The answer of A sociologist develops a test designed to...

Question 3

A manufacturer of dodge balls uses a machine to inflate its new balls to a pressure of 13.5 pounds (  = 0.1).When the machine is properly calibrated,the mean inflation pressure s 13.5 pounds,but uncontrollable factors can cause the pressure of individual dodge balls to vary.For quality control purposes,the manufacturer wishes to estimate the mean inflation pressure to within 0.025 pounds of its true value with a 99% confidence.What sample size should be used?
A)677
B)107
C)35
D)27
E)11

Accepted Answer

The answer of A manufacturer of dodge balls uses a...

Question 4

In the construction of a confidence interval,as the confidence level required in estimating the mean increases,the width of the confidence interval ______________.
A)Stays the same
B)Decreases
C)Increases

Accepted Answer

As the confidence level increases, the width of the confidence interval also increases. This is because a higher confidence level implies a larger range of values that could contain the true population mean. As a result, the margin of error (which is the distance between the sample mean and the upper/lower bounds of the interval) increases, leading to a wider confidence interval.

Question 5

The Ohio Department of Agriculture tested 203 fuel samples across the state in 1999 for accuracy of the reported octane level.For premium grade,14 out of 105 samples failed (they didn't meet ASTM specification and the FTC Octane posting rule).Find a 99% confidence interval for the true population proportion of premium grade fuel-quality failures.
A)[.045 .221]
B)[.068 .198]
C)[.023 .115]
D)[.047 .219]
E)[.100 .276]

Accepted Answer

The answer of The Ohio Department of Agriculture tested 203...

Question 6

At a sobriety checkpoint the sheriff's department screened 676 drivers and 6 were arrested for DWI.Find the 92% confidence interval for the true proportion of drivers who were DWI that evening.
A)[0.000 0.027]
B)[0.003 0.015]
C)[0.902 0.938]
D)[0.000 0.076]
E)[0.006 0.030]

Accepted Answer

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

p̂ ± z*√(p̂(1-p̂)/n)

where 
p̂ = 6/676 = 0.0089 (proportion of drivers arrested for DWI)
z = 1.75 (from the standard normal table for a 92% confidence interval)
n = 676

Substituting the values, we get:

0.0089 ± 1.75*√(0.0089*(1-0.0089)/676)

Simplifying, we get:

[0.0027, 0.0151]

Therefore, the 92% confidence interval for the true proportion of drivers who were DWI that evening is [0.003, 0.015]. The best choice is B.

Question 7

An environmental group at a local college is conducting independent tests to determine the distance a particular make of automobile will travel while consuming only 1 gallon of gas.A sample of five cars is tested and a mean of 28.2 miles is obtained.How many cars should the environmental group test if they wish to estimate $\mu$,mean miles per 1 gallon,to within 0.5 miles and be 99% confident? Assume a population standard deviation of 2.7 miles.
A)25
B)124
C)194
D)618
E)619

Accepted Answer

To determine the sample size needed, use the formula for sample size in estimating a population mean: n = (Z*σ/E)^2, where Z is the Z-value for the desired confidence level (2.576 for 99%), σ is the population standard deviation (2.7), and E is the margin of error (0.5). Plugging in these values gives n ≈ 194.

Question 8

A federal bank examiner is interested in estimating the mean outstanding defaulted loans balance of all defaulted loans over the last three years.A random sample of 20 defaulted loans yielded a mean of $67,918 with a standard deviation of $16,552.40.Calculate a 90% confidence interval for the mean balance of defaulted loans over the past three years.
A)[66,487 69,349]
B)[39,299 96,537]
C)[57,329 78,507]
D)[61,829 74,007]
E)[61,519 74,317]

Accepted Answer

The 90% confidence interval is calculated using the sample mean, standard deviation, and the t-distribution for a sample size of 20. The interval [61,519 74,317] correctly reflects these calculations.

Question 9

On a standard IQ test,the standard deviation is 15.How many random IQ scores must be obtained if we want to find the true population mean (with an allowable error of 0.5)and we want 97% confidence in the results?
A)4,239
B)283
C)212
D)131
E)66

Accepted Answer

The sample size needed for a confidence interval is given by the formula:n = (Z^2 * s^2) / e^2where:n is the sample sizeZ is the z-score for the desired confidence levels is the standard deviation of the populatione is the allowable errorIn this case, we have:Z = 2.576 (for a 97% confidence level)s = 15e = 0.5Plugging these values into the formula, we get:n = (2.576^2 * 15^2) / 0.5^2n = 4,239Therefore, the correct answer is A.

Question 10

In a poll of 1,004 adults,93% indicated that restaurants and bars should refuse service to patrons who have had too much to drink.Construct the 90% confidence interval for the proportion of all adults who feel the same way.
A)[.078 .108]
B)[.914 .946]
C)[.920 .940]
D)[.917 .943]
E)[.910 .950]

Accepted Answer

To construct the confidence interval, we can use the formula:

CI = p ± z*sqrt((p(1-p))/n)

where p is the sample proportion (0.93), z is the z-score corresponding to the desired confidence level (1.645 for 90% confidence), and n is the sample size (1004).

Plugging in the values, we get:

CI = 0.93 ± 1.645*sqrt((0.93*0.07)/1004)
   = 0.93 ± 0.014
   = (0.917, 0.943)

Therefore, the best choice is D, which contains the correct confidence interval for the proportion of all adults who feel the same way.

Question 11

An experiment was conducted to determine the effectiveness of a method designed to remove oil wastes found in soil.Three contaminated soil samples were treated.After 95 days,the percentage of contamination removed from each soil sample was measured with a mean of 49.3% and a standard deviation of 1.5%.If we wished to narrow the boundary around $\mu$ for a 98% confidence interval to within 0.5%,how many soil samples should be in our experiment.
A)437
B)33
C)9
D)6
E)3

Accepted Answer

To narrow the confidence interval to within 0.5% with a 98% confidence level, we use the formula for the sample size: n = (Z*σ/E)^2, where Z is the Z-score for 98% confidence, σ is the standard deviation, and E is the margin of error (0.5%). Using Z ≈ 2.33 for 98% confidence, σ = 1.5, and E = 0.5, we calculate n ≈ 437.

Question 12

An experiment was conducted to determine the effectiveness of a method designed to remove oil wastes found in soil.Three contaminated soil samples were treated.After 95 days,the percentage of contamination removed from each soil sample was measured with a mean of 49.3% and a standard deviation of 1.5%.Estimate the mean percentage of contamination removed at a 98% confidence level.
A)[48.36 50.24]
B)[43.27 55.33]
C)[47.29 51.31]
D)[47.88 50.72]
E)[46.47 52.13]

Accepted Answer

The answer of An experiment was conducted to determine the...

Question 13

In 1995,13,000 WWW users were surveyed and asked about their willingness to pay fees for access to websites.Of these,2,938 were definitely not willing to pay such fees.How large a sample is necessary to estimate the proportion of interest to within 2% in a 95% confidence interval?
A)18
B)307
C)1717
D)2000
E)2965

Accepted Answer

The sample size required can be determined using the formula:

n = (Z^2 * p * q) / E^2

Where:

Z = the Z-score corresponding to the desired level of confidence (1.96 for 95% confidence)

p = the estimated proportion of interest (unknown)

q = 1 - p

E = the desired margin of error (0.02)

We can estimate p by subtracting the proportion not willing to pay (2,938/13,000 = 0.225) from 1:

p = 1 - 0.225 = 0.775

Substituting these values into the formula:

n = (1.96^2 * 0.775 * 0.225) / 0.02^2 = 1716.79

Rounding up to the nearest integer, we get a sample size of 1717.

Question 14

In a survey of 308 adults,17% listed jewelry among the romantic gifts men want to receive.Construct the 98% confidence interval for the proportion of all adults who include jewelry among their responses.
A)[.025 .085]
B)[.169 .171]
C)[.120 .220]
D)[.140 .200]
E)[.050 .170]

Accepted Answer

To calculate the confidence interval, we can use the formula:

CI = p̂ ± Z*SE

Where:

p̂ = sample proportion = 0.17
Z = Z-value for a 98% confidence interval = 2.33
SE = standard error = √(p̂(1-p̂)/n), where n = sample size = 308

Plugging in the values, we get:

SE = √(0.17(1-0.17)/308) = 0.025

CI = 0.17 ± 2.33*0.025 = [0.120, 0.220]

Therefore, the best choice is C) [.120 .220].

Question 15

The Ohio Department of Agriculture tested 203 fuel samples across the state in 1999 for accuracy of the reported octane level.For premium grade,14 out of 105 samples failed (they didn't meet ASTM specification and the FTC Octane posting rule).How many samples would be needed to create a 99% confidence interval that is within 0.02 of the true proportion of premium grade fuel-quality failures?
A)4148
B)2838
C)1913
D)744
E)54

Accepted Answer

We want to find the sample size required to have a 99% confidence interval that is within 0.02 of the true proportion of premium grade fuel-quality failures.

Let p be the true proportion of premium grade fuel-quality failures. Then we want to find n such that:

margin of error = z*(sqrt(p*(1-p)/n)) <= 0.02

where z is the z-score corresponding to a 99% confidence level, which is approximately 2.576.

We don't know the value of p, so we can use the worst-case scenario where p = 0.5, which gives the largest possible margin of error:

2.576*(sqrt(0.5*(1-0.5)/n)) <= 0.02

Simplifying and solving for n, we get:

n >= 1912.74

Since we need a whole number of samples, the smallest sample size that satisfies this inequality is 1913. Therefore, the best choice is C) 1913.

Question 16

To investigate the rate at which employees with cancer are fired or laid off,a telephone survey was taken of 100 cancer survivors who worked while undergoing treatment.Seven (7)were either fired or laid off due to their illness.Construct a 90% confidence interval for the true percentage of all cancer patients who are fired or laid off due to their illness.
A)[0.0000 0.2034]
B)[0.0371 0.1029]
C)[0.0039 0.1361]
D)[0.0278 0.1122]
E)[0.0078 0.1400]

Accepted Answer

The sample proportion is 7/100 = 0.07. Using the formula for a confidence interval for a proportion, the 90% confidence interval is calculated as 0.07 ± 1.645 * sqrt((0.07*(1-0.07))/100), which results in the interval [0.0278, 0.1122].

Question 17

Recently,a case of food poisoning was traced to a particular restaurant chain.The source was identified and corrective actions were taken to make sure that the food poisoning would not reoccur.Despite the response from the restaurant chain,many consumers refused to visit the restaurant for some time after the event.A survey was conducted three months after the food poisoning occurred with a sample of 319 patrons contacted.Of the 319 contacted,29 indicated that they would not go back to the restaurant because of the potential for food poisoning.What sample size would be needed in order to be 99% confident that the sample proportion is within .02 of $\rho$,the true proportion of customers who refuse to go back to the restaurant?
A)14
B)38
C)129
D)1,371
E)1,777

Accepted Answer

The answer of Recently,a case of food poisoning was traced...

Question 18

There is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area with a standard deviation of 18.52.Construct a 99% confidence interval for the true mean number of full-time employees at office furniture dealers.
A)[11.014 33.896]
B)[11.277 33.633]
C)[12.284 32.626]
D)[12.513 32.397]
E)[20.072 24.838]

Accepted Answer

The 99% confidence interval is calculated using the formula: mean ± (z-score * (standard deviation/√n)). For 99% confidence, the z-score is approximately 2.576. Plugging in the values: 22.455 ± (2.576 * (18.52/√22)) gives the interval [11.277, 33.633].

Question 19

A botanist measures the heights of 16 seedlings and obtains a mean and standard deviation of 72.5 cm and 4.5 cm,respectively.Find the 95% confidence interval for the mean height of seedlings in the population form which the sample was selected.
A)[62.91 82.09]
B)[70.30 74.70]
C)[70.10 74.90]
D)[70.54 74.46]
E)[71.90 73.10]

Accepted Answer

We will use the formula: 
CI= x̄ ± z(α/2) * (σ/√n)
where:
x̄ = sample mean = 72.5
σ = standard deviation = 4.5
n = sample size = 16
α = level of significance = 0.05 (for 95% confidence interval, α = 1-0.95 = 0.05)
z(α/2) = z-score corresponding to α/2=0.025 from the standard normal distribution table.
z(0.025) = 1.96 
Putting the values in the formula, we get
CI = 72.5 ± 1.96 * (4.5/√16)
CI = 72.5 ± 1.96 * 1.125
CI = 72.5 ± 2.204
CI = [70.296, 74.704]
Therefore, the best choice is C: [70.10 74.90], as it is closest to the calculated confidence interval.

Question 20

In 1995,13,000 WWW users were surveyed and asked about their willingness to pay fees for access to websites.Of these,2,938 were definitely not willing to pay such fees.Construct a 95% confidence interval for the proportion definitely unwilling to pay fees.
A)[0.286 0.302]
B)[0.219 0.233]
C)[0.220 0.232]
D)[0.212 0.241]
E)[0.214 0.245]

Accepted Answer

The answer of In 1995,13,000 WWW users were surveyed and...

Quiz 8: Confidence Intervals

A sociologist develops a test designed to measure a person's attitudes about disabled people,and 16 randomly selected subjects are given the test.Their mean score is 71.2 with a standard deviation of 10.5.Construct the 99% confidence interval for the mean score of all subjects.

In the construction of a confidence interval,as the confidence level required in estimating the mean increases,the width of the confidence interval ______________.

At a sobriety checkpoint the sheriff's department screened 676 drivers and 6 were arrested for DWI.Find the 92% confidence interval for the true proportion of drivers who were DWI that evening.

On a standard IQ test,the standard deviation is 15.How many random IQ scores must be obtained if we want to find the true population mean (with an allowable error of 0.5) and we want 97% confidence in the results?

In a poll of 1,004 adults,93% indicated that restaurants and bars should refuse service to patrons who have had too much to drink.Construct the 90% confidence interval for the proportion of all adults who feel the same way.

In 1995,13,000 WWW users were surveyed and asked about their willingness to pay fees for access to websites.Of these,2,938 were definitely not willing to pay such fees.How large a sample is necessary to estimate the proportion of interest to within 2% in a 95% confidence interval?

In a survey of 308 adults,17% listed jewelry among the romantic gifts men want to receive.Construct the 98% confidence interval for the proportion of all adults who include jewelry among their responses.

There is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area with a standard deviation of 18.52.Construct a 99% confidence interval for the true mean number of full-time employees at office furniture dealers.

A botanist measures the heights of 16 seedlings and obtains a mean and standard deviation of 72.5 cm and 4.5 cm,respectively.Find the 95% confidence interval for the mean height of seedlings in the population form which the sample was selected.

In 1995,13,000 WWW users were surveyed and asked about their willingness to pay fees for access to websites.Of these,2,938 were definitely not willing to pay such fees.Construct a 95% confidence interval for the proportion definitely unwilling to pay fees.