Question

How do you calculate a 95% confidence interval without the mean?

In 2007, the Pew Research Center assessed public opinion of the challenges of motherhood. Over a 4-week period, they surveyed 2020 Americans. They found 60% of respondents felt that it was more difficult to be a mother today than it was 20 or 30 years ago.

How do you calculate the 95% confidence interval to estimate the percentage of Americans who believe that it is more difficult to be a mother today than it was 20 or 30 years ago?

Answer 1

`[.58,.62]`

Explanation:

you can do confidence intervals using proportions. You can use normal distribution if the amount of items is greater than 10 for both outcomes. Variance is estimated as `p(1-p)`

`CI = z * se` where z is the z statistic you get from the z table at the 95% and `se` is the standard error . Since this is a two tail test it should be 1.96 thus
`CI = z *sqrt((p(1-p))/n)`

`CI = 1.96 * sqrt((.6(.4))/2020)`

`~~ .6 +- .021` or `[.58,.62]`