Question

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). 18-to 24-year-olds The survey included 194 young adults (ages 18 to 24 ) and 53 of them said that they had used online dating. If we use this sample to estimate the proportion of all young adults to use online dating, the standard error is 0.032 . Find a 99 % confidence interval for the proportion of all US adults ages 18 to 24 to use online dating.

Answer 1

Solution :

Given that,

n = 194

x=53

SE=0.032

Point estimate = sample proportion = = x / n = 53/194=0.273

1 -   = 1- 0.273 =0.727

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645} ( Using z table )

Margin of error = E Z/2 *SE

= 1.645 *0.032

E = 0.0526

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.273-0.0526< p < 0.273+0.0526

0.2204< p < 0.3256

The 90% confidence interval for the population proportion p is 0.2204, 0.3256