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A survey conducted in July 2015 asked a random sample of American adults whether they had ever used online dating (either an online dating site or a dating app on their cell phone). The survey included 411 adults between the ages of 55 and 64, and 50 of them said that they had used online dating. If we use this sample to estimate the proportion of all American adults ages 55 to 64 to use online dating, the standard error is 0.016. Find a 99% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating? Round your answers to three decimal places.

User Dubby
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Answer:
(0.081,\ 0.163)

Explanation:

Let
\hat{p} be the sample proportion.

As per given , we have

n= 411


\hat{p}=(50)/(411)=0.121654501217\approx0.122

Standard error : se=0.016

Critical value for 99% confidence :
z_(\alpha/2)=2.576

Confidence interval for population proportion :-


\hat{p}\pm z_(\alpha/2) (se)\\\\=0.122\pm (2.576)(0.016)\\\\=0.122\pm0.041216\\\\=(0.122-0.041216,\ 0.122+0.041216)\\\\=(0.080784,\ 0.163216)\approx(0.081,\ 0.163)

Hence, the 99% confidence interval for the proportion of all US adults ages 55 to 64 to use online dating :
(0.081,\ 0.163)

User Ajnas Askar
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