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For this problem, carry at least four digits after the decimal in your calculations. Answers may...

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding. In a survey of 1000 large corporations, 252 said that, given a choice between a job candidate who smokes and an equally qualified nonsmoker, the nonsmoker would get the job.

(a) Let p represent the proportion of all corporations preferring a nonsmoking candidate. Find a point estimate for p. (Round your answer to four decimal places.)

(b) Find a 0.95 confidence interval for p. (Round your answers to three decimal places.) lower limit upper limit

(c) As a news writer, how would you report the survey results regarding the proportion of corporations that would hire the equally qualified nonsmoker?

1)Report p̂ along with the margin of error.

2)Report p̂.

3)Report the margin of error.

What is the margin of error based on a 95% confidence interval? (Round your answer to three decimal places.)

I want to know how to do this on a TI 84 Calculator if possible. Thank you!

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Answer #1

(a) Let p represent the proportion of all corporations preferring a nonsmoking candidate. Find a point estimate for p.

p^=x/n=point estimate of p=x/n=252/1000=0.252

ANSWER:0.252

Solutionb:

95% confidence interval for p is

p^-zcrit*sqrt(p^(1-p^)/n,p^+zcrit*sqrt(p^(1+p^)/n

0.252-1.96*sqrt(0.252*(1-0.252)/1000,0.252+1.96*sqrt(0.252*(1-0.252)/1000

=0.225,0.279

ANSWER:

95% confidence interval for p lies in between

0.252 ,0.279

Solutionc:

1)Report p̂ along with the margin of error.

0.252 \pm 0.0269

Alternatuively

TI*83

go to tests>1-PropZint>x as 252 n as 1000C-level=0.95 you will get:

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