Question

Let p be the population proportion for the following condition. Find the point estimates for p...

Let p be the population proportion for the following condition. Find the point estimates for p and q.

A study of 4938 adults from country​ A, 3179 think mainstream media is more interested in making money than in telling the truth.

The point estimates are: _____(p) & _____ (q). ​(Round to three decimal places as​ needed).

Part 2:

In a survey of 2334 adults in a recent​ year, 1347 say they have made a New​ Year's resolution.

Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals.

​(Round to three decimal places as​ needed.)

Part 3:

In a survey of 3172 ​adults,1425 say they have started paying bills online in the last year.

Construct a​ 99% confidence interval for the population proportion. Interpret the results.

A​ 99% confidence interval for the population proportion is _______. ​(Round to three decimal places as​ needed.)

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Let p be the population proportion for the following condition. Find the point estimates for p...
Let p be the population proportion for the following condition. Find the point estimates for p and q. In a survey of 10271027 adults from country​ A, 242242 said that they were not confident that the food they eat in country A is safe. The point estimate for​ p, ModifyingAbove p with caretp​, is nothingm. ​(Round to three decimal places as​ needed.)
In a survey of 2572 adults in a recent​ year, 1308 say they have made a...
In a survey of 2572 adults in a recent​ year, 1308 say they have made a New​ Year's resolution. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. (round to three decimal places as needed) The​ 90% confidence interval for the population proportion p is? The 95% confidence interval for the population proportion p is?
In a survey of 609609 males ages​ 18-64, 397397 say they have gone to the dentist...
In a survey of 609609 males ages​ 18-64, 397397 say they have gone to the dentist in the past year. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. If​ convenient, use technology to construct the confidence intervals. The​ 90% confidence interval for the population proportion p is ​(nothing​,nothing​). ​(Round to three decimal places as​ needed.)
In a survey of 2800 adults, 1476 say they have started paying bills online in the...
In a survey of 2800 adults, 1476 say they have started paying bills online in the last year. Construct a​ 99% confidence interval for the population proportion. Interpret the results. A​ 99% confidence interval for the population proportion is ( ), ( ) ​(Round to three decimal places as​ needed.)
In a survey of 3352 ​adults, 1490 say they have started paying bills online in the...
In a survey of 3352 ​adults, 1490 say they have started paying bills online in the last year. Construct a​ 99% confidence interval for the population proportion. Interpret the results. A​ 99% confidence interval for the population proportion is (?,?). (Round to three decimal places as​ needed.)
1. In a survey of 3230 ​adults, 1418 say they have started paying bills online in...
1. In a survey of 3230 ​adults, 1418 say they have started paying bills online in the last year. Construct a​ 99% confidence interval for the population proportion. Interpret the results. A​ 99% confidence interval for the population proportion is (),() . ​(Round to three decimal places as​ needed.)
In a survey of 619 males ages 18-64, 395 say they have gone to the dentist...
In a survey of 619 males ages 18-64, 395 say they have gone to the dentist in the past year. Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. The 90% confidence interval for the population proportion p is ( __ _, ___ ). (Round to three decimal places as needed.) The 95% confidence interval for the population proportion...
In a survey of 2460 adults in a recent​ year, 1491 say they have made a...
In a survey of 2460 adults in a recent​ year, 1491 say they have made a New​ Year's resolution. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. The​ 90% confidence interval for the population proportion p is
In a survey of 2363 adults in a recent​ year, 1366 say they have made a...
In a survey of 2363 adults in a recent​ year, 1366 say they have made a New​ Year's resolution. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. A. The​ 90% confidence interval for the population proportion p is (__,__)
In a survey of 2437 adults in a recent​ year, 1302 say they have made a...
In a survey of 2437 adults in a recent​ year, 1302 say they have made a New​ Year's resolution. Construct​ 90% and​ 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. The​ 90% confidence interval for the population proportion p is left parenthesis nothing comma nothing right parenthesis,. ​(Round to three decimal places as​ needed.) The​ 95% confidence interval for the population proportion p is left parenthesis nothing comma nothing right...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT