Confidence Intervals (Proportions)

Question 1

139 students at a college were asked whether they had completed their required English 101 course, and 78 students said "yes". Find the best point estimate for the proportion of students at the college who have completed their required English 101 course. Round to four decimal places.

Question 2

A poll of 1105 U.S. adults found that 47% regularly used Facebook as a news source. 

Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 90% level of confidence. Round all answers to 2 decimal places.

Margin of Error (as a percentage):

Confidence Interval:___ to ____

Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 95% level of confidence. Round all answers to 2 decimal places.

Margin of Error (as a percentage):

Confidence Interval:___ to ____

Find the margin of error and confidence interval for the percentage of U.S. adults who regularly use Facebook as a news source, at the 99% level of confidence. Round all answers to 2 decimal places.

Margin of Error (as a percentage):

Confidence Interval:___ to ____

The more error we allow, the less precise our estimate. Therefore, as the confidence level increases, the precision of our estimate

stays roughly the same

increases

decreases

Question 3

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 259 with 12% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. 

Question 4

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 184 with 61 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Question 5

Out of 600 people sampled, 540 preferred Candidate A. Based on this, estimate what proportion of the entire voting population (p) prefers Candidate A. 

Use a 90% confidence level, and give your answers as decimals, to three places.

Question 6

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 308 with 56% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Question 7

Assume that a sample is used to estimate a population proportion p. Find the 99% confidence interval for a sample of size 111 with 23% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

Question 8 

Out of 400 people sampled, 60 preferred Candidate A. Based on this, estimate what proportion of the voting population (p) prefers Candidate A.

Use a 95% confidence level, and give your answers as decimals, to three places.

Question 9

Giving a test to a group of students, the grades and gender are summarized below

Let p represent the population proportion of all female students who received a grade of A on this test. Use a 80% confidence interval to estimate p to four decimal places if possible.

Question 10

The result of a history test was collected, and the grades and gender are summarized below 

Let p represent the proportion of all female students who would receive a grade of C on this test. Use a 85% confidence interval to estimate to three decimal places.

Enter your answer as a tri-linear inequality using decimals (not percents)

Question 11

Express the confidence interval 77 % < p < 83 % in the form of pˆ ± E.

Question 12

Express the confidence interval 49.8 % ± 5 % in the form of a trilinear inequality.

Question 13 

In a survery of 176 households, a Food Marketing Institute found that 141 households spend more than $125 a week on groceries. Please find the 99.9% confidence interval for the true proportion of the households that spend more than $125 a week on groceries.

a. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

b. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. 

c. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

Question 14

We wish to estimate what percent of adult residents in Ventura County like chocolate. Out of 100 adult residents sampled, 80 like chocolate. Based on this, construct a 95% confidence interval for the proportion (p) of adult residents who like chocolate in Ventura County.

a. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

b. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

c. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

Question 15 

If n=270 and X = 216, construct a 90% confidence interval.

a. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. 

b. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

c. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

Question 16 

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 167 with 21% successes.

a. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. 

b. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places. 

c. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

Question 17

You are interested in constructing a 99% confidence interval for the proportion of all caterpillars that eventually become butterflies. Of the 416 randomly selected caterpillars observed, 49 lived to become butterflies.

a.  With 99% confidence the proportion of all caterpillars that lived to become a butterfly is between ____ and ____  .

b.  If many groups of 416 randomly selected caterpillars were observed, then a different confidence interval would be produced from each group. About ____ percent of these confidence intervals will contain the true population proportion of caterpillars that become butterflies and about ____ percent will not contain the true population proportion.

Question 18

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 583 randomly selected Americans surveyed, 351 were in favor of the initiative. Round answers to 2 decimal places where possible.

a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between  ____and  ____.

b. If many groups of 583 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About  ____percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about  ____percent will not contain the true population proportion.

Question 19

A smart phone manufacturer is interested in constructing a 99% confidence interval for the proportion of smart phones that break before the warranty expires. 85 of the 1517 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible.

a.  With 99% confidence the proportion of all smart phones that break before the warranty expires is between  ____and ____ .

b.  If many groups of 1517 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About ____ percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about  ____percent will not contain the true population proportion.

Question 20

A newsgroup is interested in constructing a 99% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 507 randomly selected Americans surveyed, 392 were in favor of the initiative.

a.  With 99% confidence the proportion of all Americans who favor the new Green initiative is between____ and____ .

b.  If many groups of 507 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About ____percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about ____percent will not contain the true population proportion.