A survey asks, "If the husband in a family wants children, but the wife decides that she does not want any children, is it all right for the wife to refuse to have children?" Of 743 subjects, 569 said yes. Find a 99% confidence interval for the population proportion who would say yes using a TI 83 calculator.
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briefcaseTeacher (K-12)
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A survey question posed to 743 people gets 569 positive answers. We are asked to find a 99% confidence interval of the true population proportion.
(1) In a TI-83/84 calculator:
Hit Stat -> Tests -> A (1-PropZInt)
x=569 (number of "successes")
n=743 (sample size)
C-Level: 99 (given in problem)
<Enter> (Calculate)
Output:
1-PropZInt (type of test)
(.7258,.80583) (interval you seek)
`hat(p)=.7658142665` (p-hat is the sample proportion `569/743` )
n=743 (sample size)
The interval you want is .7258<p<.80583
(You will want to consult your text or instructor's notes for rounding rules.)
(2) To hand calculate, recognize that the interval is of the form:
`hat(p)-z_(alpha/2)sqrt((pq)/n)<p<hat(p)+z_(alpha/2)sqrt((pq)/n)`
where `hat(p)=569/743` is the sample proportion, `z_(alpha/2)` is the z-score associated with the given confidence level `alpha` , q is 1-p, n the sample size:
Since `alpha=.01, alpha/2=.005` and from a table (or use invnorm in a calculator) we get `z_(alpha/2)~~2.58`
Thus `569/743-2.58sqrt(((.7658)(.2342))/743)<p<569/743 + 2.58sqrt(((.7658)(.2342))/743)`
This gives the interval above.
Note that all this assumes that the population is approximately normal, the sample was truly random and representative of the population, the population was defined properly, the survey was conducted to minimize bias, etc.