# 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?

May 6, 2017

$\left[.58 , .62\right]$

#### 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 $p \left(1 - p\right)$

$C I = z \cdot s e$ where z is the z statistic you get from the z table at the 95% and $s e$ is the standard error . Since this is a two tail test it should be 1.96 thus
$C I = z \cdot \sqrt{\frac{p \left(1 - p\right)}{n}}$

$C I = 1.96 \cdot \sqrt{\frac{.6 \left(.4\right)}{2020}}$

$\approx .6 \pm .021$ or $\left[.58 , .62\right]$