Question

In a survey of 607 males ages​ 18-64, 393 say they have gone to the dentist...

In a survey of

607

males ages​ 18-64,

393

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 p is

​(______,_______). ​(Round to three decimal places as​ needed.)

Interpret your results of both confidence intervals.

A. With the given​ confidence, it can be said that the population proportion of males ages​ 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval.

B. With the given​ confidence, it can be said that the population proportion of males ages​ 18-64 who say they have gone to the dentist in the past year is not between the endpoints of the given confidence interval.

C.With the given​ confidence, it can be said that the sample proportion of males ages​ 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval.

D. Which interval is​ wider?

(a) The​ 90% confidence interval

(b) The​ 95% confidence interval

Homework Answers

Answer #1

Solution:

The point estimate is 393/628=0.626

90% confidence interval has a z-value of 1.645 (5% in each tail) so now use the confidence interval formula:

CI = (mean proportion) +/- z* (sample standard deviation) where z* is 1.645 for 90% confidence:

(0.594,0.658) is the 90% CI

95% confidence interval has a z-value of 1.96 so now use the confidence interval formula:

CI = (mean proportion) +/- z* (sample standard deviation)


(0.588,0.664) is the 95% CI


The correct choice is B. CI are statements about the population. We know what the sample was; we are 90% or 95% confident that the true proportion, which we don't know and never will know, is in the interval constructed. It is not a probability.

d)
The 95% interval is wider, and one needs a wider interval if one wants to be more confident. One way to remember that is one needs a 100% interval if one wants to be 100% confident.

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
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 633 males ages​ 18-64, 393 say they have gone to the dentist...
in a survey of 633 males ages​ 18-64, 393 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.
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.)
Explain how to solve this please and thanks Construct 90% and 95% confidence intervals for the...
Explain how to solve this please and thanks Construct 90% and 95% confidence intervals for the population proportion. In a survey of 674 US males ages 18 to 64; 396 say they have gone to the dentist in the past year
In a survey of 3123 ​adults, 1492 say they have started paying bills online in the...
In a survey of 3123 ​adults, 1492 say they have started paying bills online in the last year. 1. Construct a​ 99% confidence interval for the population proportion. ______- 2. Interpret the results. (multiple choice) A. The endpoints of the given confidence interval show that adults pay bills online​ 99% of the time. B.With​ 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is between...
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...
In a survey of 2674 adults, 1470 say they have started paying bills online in the...
In a survey of 2674 adults, 1470 say they have started paying bills online in the last year. Construct a​ 99% confidence interval for the population proportion. Interpret the results. 1) The​ 90% confidence interval for the population proportion p is (_,_) (Round to three decimal places as needed.) Interpret your results. Choose the correct answer below. A. With​ 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in...
In a survey of 3398 adults, 1487 say they have started paying bills online in the...
In a survey of 3398 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. A 99% confidence interval for the population proportion is (___,___). (Round to three decimal places as needed.) Interpret your results. Choose the correct answer below. A. With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last...
In a survey of 3125 adults,1415 say they have started paying bills online in the last...
In a survey of 3125 adults,1415 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.) Interpret your results. Choose the correct answer below. A. With 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the last year is...
In a survey of 3, 487 ​adults, 1,424 say they have started paying bills online in...
In a survey of 3, 487 ​adults, 1,424 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.) Interpret your results. Choose the correct answer below. A. With​ 99% confidence, it can be said that the population proportion of adults who say they have started paying bills online in the...