# You've asked 55 parents if they have more than one child. It turns out that 77 in 100 parents have more than one child. Compute the 99% confidence int

#### mn_sunil

##### New Member
You've asked 55 parents if they have more than one child. It turns out that 77 in 100 parents have more than one child. Compute the 99% confidence interval.

I have the below
n=55
sample mean =0.77
z99=2.58

Neither of the standard deviations are available, how do I calculate the intervals?

Can some one help me with the solution please.

Last edited:

#### hlsmith

##### Less is more. Stay pure. Stay poor.
Is there a reason why you wouldn't treat this as a binary variable (>1 Y/N)?

#### Dason

The numbers don't really make sense. First it was 55 then next thing is talking about 77 out of 100. I thought maybe they were just rephrasing the percentage from the 55 but 42/55 = .763 and 43/55 = .781 and neither round to 77/100.

#### joeb33050

##### Member
First statement
You've asked 55 parents if they have more than one child.
and then, second statement:
It turns out that 77 in 100 parents have more than one child.
then we have two statements, not mutually exclusive, and the question is:

Compute the 99% confidence interval.

This is a binomial distribution festival. To an online calculator>

example

P(X>/+x) with x = 35 is .9918, the confidence interval between 35 and 55 successes out of 55 tries = 99.18%
joe b.