According to a Harris Interactive survey of 2401 adults conducted in April 2009, 25% of adults do not drink alcohol. Construct a 97% confidence interval for the corresponding population proportion.
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Q: In a survey of 3005 adults aged 57 through 85 years, it was found that 81.7% of them used at least…
A: Given : n = 3005 p̂=0.817
Q: In a random sample of 100 customers, the items that customers purchased were tracked. It was found…
A: Given,sample size(n)=100sample proportion(p^)=0.171-p^=1-0.17=0.83α=1-0.90=0.1α2=0.05Z0.05=1.645
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A: It is given that, x is 311 and n is 400.
Q: knowing
A: Given: Total number of drivers = 383 Drivers who buckle up = 297 Proportion of drivers, p =…
Q: The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5. What is the…
A: Given data is10,8,12,15,13,11,6,5sample size(n)=8sample mean(x¯)=10standard…
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A: Solution: Given information: n= 2558 Sample size of adults x = 1413 adults started paying bills…
Q: A university dean is interested in determining the proportion of students who receive some sort of…
A: It is given that Sample size n = 250 Number of students receiving scholarship, X = 70 The critical…
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A: Given X=1433 n=3254
Q: In a survey, 150 people were asked to identify their main source of news information; 93 indicated…
A: Given: The sample number of people that were asked to identify their main source of news information…
Q: In a survey of 3478 adults, 1462 say they have started paying bills online in the last year.…
A: From the provided information,Confidence level = 99%
Q: a survey of 3088 adults aged 57 through 85 years, it was found that 87.1% of them used at least…
A: The formula for computing confidence interval is, n is the sample size and P is the proportion.
Q: In a survey of 2703 adults, 1438 say they have started paying bills online in the last year.…
A: We have given that, Number of favorable cases(X) = 1438, Sample size(n) = 2703 Sample proportion…
Q: Out of 400 people sampled, 32 had kids. Based on this, construct a 95% confidence interval for the…
A: Solution: Let X be the number of people had kids and n be the sample number of people. From the…
Q: A medical director needs to know the proportion of patients who have a single bed in their room. A…
A: It is given that sample size (n) is 200 and p-hat value is 0.59.
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Q: Insurance companies are interested in knowing the population percent of drivers who always buckle up…
A: Solution
Q: A survey of 400 students at Red Rock College found that 180 planned to take summer classes. Find a…
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Q: In a survey of 1005 adults 526 stated construct a 95% confidence interval
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Q: In a survey of 1400 people, 75% indicated that they listened to their favorite radio station because…
A: In statistical inference there are two branches-1)hypothesis testing and 2)Estimation In estimation…
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Q: In a survey of 2282 adults, 749 say they believe in UFOs. Construct a 90% confidence interval for…
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Q: In a survey of 2002 adults in a recent year, 701 made a New Year's resolution to eat healthier.…
A: Given:
Q: Out of 200 people sampled, 42 had kids. Based on this, construct a 95% confidence interval for the…
A: For 95% confidence intervalα=1-CI=1-0.95=0.05So, zα2=z0.052=z0.025=1.96
Q: Of 127 randomly selected adults, 31 were found to have high blood pressure. Construct a 95%…
A: Given that, Of 127 randomly selected adults, 31 were found to have high blood pressure. X = 31 N =…
Q: In a survey of 3115 adults aged 57 through 85 years, it was found that 86.1% of them used at…
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Q: a survey of 1083 people, 585 answered "yes" to a particular question. Construct a 90% confidence…
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Q: In a survey of 2552 adults, 1423 say they have started paying bills online in the last year.…
A: Given that, In a survey of 2552 adults, 1423 say they have started paying bills online in the last…
Q: What proportion of the sample reported that they had been cyberbullied?
A: Givensample size(n)=14795No.of respondents say yes they had been cyberbullied(x1) = 2219No.of…
Q: A Pew Internet poll asked 4272 U.S. adults about their use of smart watches and fitness trackers. A…
A: Given,n=4272X=897
Q: fa family now li wo years from now? A suburban area? A rural area? uppose that at present, 40% of…
A: .
Q: n a randomly selected group of 750 automobile deaths, 240 were alcohol related. Construct a 95%…
A: Given that, x = 240, n = 750 The value of sample proportion is,
Q: 2. Construct a 95% confidence interval to estimate the proportion of Wingate female students that…
A: total number of female students =181 number of students who are left handed=15
Q: andom sample of 1200 high school freshmen was taken and it was found that 559 of them were reading…
A: The population proportion (p) will be proportion of high school freshman who reads at or above their…
Q: A survey of 800 women shoppers found that 15% of them shop on impulse. What is the 95% confidence…
A: n = 800p^ = 0.1595% CI for p = ?
Q: Construct a 99% confidence interval for the population proportion. Interpret the results.
A:
Q: In a sports poll, 142 of 500 randomly selected Americans indicated that they consider themselves to…
A: Given that, In a sports poll, 142 of 500 randomly selected Americans indicated that they consider…
Q: In a survey of 2693 adults, 1495 say they have started paying bills online in the last year.…
A: Given that, In a survey of 2693 adults, 1495 say they have started paying bills online in the last…
Q: In a survey of 3332 adults aged 57 through 85 years, it was found that 80.7% of them used at least…
A:
Q: In a random sample of 575 American goods, about 46.3% of them were transported by rail in 2017.…
A: Given, Sample size = 575 Sample proportion = 0.463 Critical value: Using z-table, the z-critical…
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- A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 99% confidence interval to estimate the true proportion of students on financial aid.Out of 300 people sampled, 279 had kids. Based on this, construct a 90% confidence interval for the true population proportion of people with kids.In a survey of 1,626,773 U.S. adults, 49,311 personally identified as lesbian, gay, bisexual, or transgender. Construct a 95% confidence interval for the population portion of U.S. adults who personally identify as gay, lesbian,bisexual or transgender.
- Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 391 drivers and find that 316 claim to always buckle up. Construct a 95% confidence interval for the population proportion that claim to always buckle up.In a survey of 3071 adults, 1484 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.In a survey of 2646 adults, 1401 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
- In a survey of 2847 adults, 1408 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.If 64% of a sample of 550 people leaving a shopping mall claims to have spent over $25 create and interpret a 95% confidence interval estimate for the proportion of shopping mall customers who spend over $25In a survey of 3342 adults, 1489 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
- In a survey of 3426 adults, 1487 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.Of 300 randomly selected medical students, 30 said that they planned to work in a rural community. Find 90% confidence interval for the true proportion of all medical students who plan to work in a rural community. The answers below represent approximate values.Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car. They randomly survey 387 drivers and find that 311 claim to always buckle up. Construct a 93% confidence interval for the population proportion that claim to always buckle up.