Thread: 99% confidence interval ????????????

Hi there, im doing report on mobile phone use and i want to use some clever statistics in it. Heres the basic info:

It was claimed that 65% of people wanted a regulation to ban mobile phone usage on public transport.


94 people, as far as possible selected randomly from the population, were asked if they supported this proposed rule. 42 were in agreement.


1) I want to make a 99% confidence interval for the proportion of people wanting to change the law.

2) Also i need to be sure whether my confidence intrerval supports my claim.

Any help would be great!

Originally Posted by bobchiba

Hi there, im doing report on mobile phone use and i want to use some clever statistics in it. Heres the basic info:

It was claimed that 65% of people wanted a regulation to ban mobile phone usage on public transport.


94 people, as far as possible selected randomly from the population, were asked if they supported this proposed rule. 42 were in agreement.


1) I want to make a 99% confidence interval for the proportion of people wanting to change the law.

2) Also i need to be sure whether my confidence intrerval supports my claim.

Any help would be great!

Lets assume that the normal approximation to the binomial is valid here, and we may make
large sample assumptions.

We have:

Pest = 42/94 ~= 0.447

Sigma_{Pest} = sqrt(94 (42/94) (52/94))/94 ~= 0.051

A 99% confidence interval for P (assumes sqrt(94 P (1-P))/94 ~= Sigma_{Pest}) is:

[Pest-2.575*Sigma_{Pest},Pest-2.575*Sigma_{Pest}] ~= [0.316, 0.578]

As 0.65 (which is 65%) does not lie in this interval, if your sample is a fair sample from
the population you are interested in, the hypothesis that P is 0.65 is not supported by
your data.

RonL