Answered: I'm a survey of 2589 adults 1500 say…

I'm a survey of 2589 adults 1500 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proport

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter10: Statistics

Section10.3: Measures Of Spread

Problem 1GP

See similar textbooks

Related questions

Q: 47. Out of 400 people sampled, 372 had kids. Based on this, construct a 99% confidence interval for…

Q: Suppose we know that a confidence interval is (-.735, .125). Find the sample mean and the error…

A: Ans- Given, a confidence interval is (-.735, .125). Find the sample mean and the error bound.

Q: A study found that for a sample of cars traveling on Interstate 70, the average speed was 70 mph…

A: We have to find margin of error.

Q: A random survey of 70 male sophomores showed 16 preferred off campus living and out of 63 female…

Q: We randomly sample 300 people in a specific town. 96 of them like cats. Based on this, construct a…

A: The provided information are: Sample size=n=300 Number of favroable cases==x=96 to Construct the 95%…

Q: confidence interval for the proportion of unemployed adults who have been unemployed for more than…

A: Here given In a survey of 532 unemployed adults, 225 said that they had been unemployed for more…

Q: I want to estimate the average time it takes a tomato plant to bear fruit. I randomly sample 16…

A: Given,sample size(n)=16mean(x¯)=36standard deviation(σ)=17α=1-0.85=0.15α2=0.075Z0.075=1.4395…

Q: The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age…

A: Confidence interval for Proportions : The following is the formula for confidence interval of the…

Q: Let's say you survey a random sample of 70 people. You ask the people in your survey if they prefer…

A: given data n = 70 p = 0.53 95% ci for p margin of error (E) = ?

Q: In a study of 120 randomly selected families, 50 said they streamed their television content. Find…

Q: In a survey of 793 adults, 460 say they have started paying bills online in the last year. Construct…

A: Answer: From the given data, Sample size(n) = 793 X = 460 98% confidence interval,

Q: The Pew Research Center conducted a survey of 45535 adults in 40 countries to learn about major…

A: From the provided information, Sample size (n) = 45535 Number of respondents who are very concerned…

Q: In a survey of 589 adults, 233 say they have started paying bills online in the last year. Construct…

Q: A survey of 45,535 adults showed that 20,901 were very concerned abut climate change. What is the…

A: It is given asFavourable cases, x = 20901Sample size, n = 45535Confidence level = 99%

Q: Suppose we take a SRS of high school seniors and get a 95% confidence interval that 250 to 278…

Q: The first day of baseball comes in late March, ending in October with the World Series. Does fan…

A: given that,sample one, n1 =1001, p1= x1/n1=0.45sample two, n2 =1001, p2= x2/n2=0.51

Q: Joe’s Pizza Joint found that 64.2% of the 328 pizzas sold last week included pepperoni. Find a98%…

A: Obtain the 95% confidence interval for the proportion of Joe’s customers that order pepperoni. The…

Q: Construct a confidence interval for GDP in Sweden from 1990-2021.

A: Gross Domestic Product of the data from 1990-2021 1990 240 1991 241 1992 243 1993 210…

Q: The First National Bank of Wilson has 620 checking account customers. A recent sample of 70 of these…

A: Given that,N = 620n = 70x = 37Confidence interval(c) = 0.95The formula of the margin of error…

Q: In a survey of 3235 adults 1499 say they have started paying bills online in the last year. A 99%…

A: Solution: Let X be the number of adults say they have started paying bills online in the last year…

Q: As the sample standard deviation increases when all other parts of the confidence interval stay the…

A: We want tell the sample standard deviation increases when all other parts of the confidence interval…

Q: A sample of 400 homeowners includes 124 that expect their home to increase in value in the next…

A: Margin of error: ME=Zα/2 * √p̂(1-p̂)/n

Q: Out of 100 people sampled, 44 have lied to a family member in the past month. Based on this,…

Q: A survey found that 611 out of 663 college graduates cannot correctly spell the word aluminum. Find…

A: Given that A survey found that 611 out of 663 college graduates cannot correctly spell the word…

Q: confidence interval for a population mean is (56,78), with a margin of error of 11. What is the…

A: From provided information, Confidence interval for population mean is (56, 78), with margin of error…

Q: What are confidence intervals?

Q: A recent study by the McGraw-Hill company showed that 61% (1621) of college students use their…

A: Given, Proportion of college students use their smartphones for study = 0.61Total number of students…

Q: In a random survey of 2221 North Americans 689 of them had blood type A+. Construct a 94% Confidence…

Q: The American Heart Association is about to conduct in anti smoking campaign and wants to know the…

A: Solution: Out of a sample of 1421 Americans over 35, 412 smoke. Sample size: n = 1421 The number of…

Q: In a random sample, 136 of - persons given a flu vaccine experienced some discomfort. Construct a %…

Q: Compute the probability of successfully completing the task for an engineer having 24 months of…

A: Given the data of monthly experience of the task is a success or not for 2 engineers as shown below…

Question

100%

I'm a survey of 2589 adults 1500 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proport

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1: Given information

VIEW

Step 2: Calculate 99% confidence interval

VIEW

Solution

VIEW

Step by step

Solved in 3 steps with 6 images

SEE SOLUTION Check out a sample Q&A here

Similar questions

What is an experiment?
a survey of 3080 adults, 1471 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 3409 adults, 1445 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion.
We randomly sample 600 people in a specific town. 342 of them like cats. Based on this, construct a 95% confidence interval for the true population proportion of people in this town who like cats. Give your answers as decimals, rounded to four places.
In a survey of 2896 adults, 1441 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 6000 women, 3431 say they change their nail polish once a week. Construct a 99% confidence interval for the population proportion of women who change their nail polish once a week.
Construct a confidence interval for GDP in Sweden from 1990-2021.
A researcher takes a sample of 600 people and finds that 128 of them are paid the minimum wage. Construct a 95% confidence interval for the proportion of Seattle residents who earn the minimum wage. Then write a sentence explaining the Cl you found in the context of this researcher's findings.
In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Find the proportion of ASD in Arizona with a confidence level of 99%. Show the work.
In a survey of 3196 adults, 1405 say they have started paying bills online in the last year. Construct a 99% confidence interval for the population proportion. Interpret the results.
Suppose we take a SRS of high school seniors and get a 95% confidence interval that 250 to 278 seniors plan to attend prom. What is the point estimator and margin of error?
In a survey of 793 adults, 460 say they have started paying bills online in the last year. Construct a 98% confidence interval for the population proportion. (round to two decimals)