Construct a 95% confidence interval ???

#1
Hi:) ,

I'm sort of confused with this statistics problem:


As part of a recent study conducted in France investigating the effectiveness of the drug mifepristone for terminating early pregnancy, 488 women were administered mifepristone followed 48 hours later by a single dose of a second drug, misoprostol. In 473 of these women, the pregnancy was terminated and the conceptus completely expelled.

a. Estimate the proportion of successfully terminated early pregnancies among women using the described treatment regimen.
473/488= .97 or 97% is this correct ???


b. Construct a 95% confidence interval for the true population proportion p.



We haven't covered this sort of problem in our class yet and I was wondering if someone could me along.

Thank You
 

JohnM

TS Contributor
#4
You need to scroll down to where it says "Confidence Interval for the Proportion.":)

Part a is correct.

In part b, the confidence interval is computed as:

p +/- [ z * sp ]

where p = sample proportion = 0.97
q = 1-p = 0.03
sp = standard error of the proportion = sqrt(pq/n) = sqrt(.97*.03/488)
z = two-tailed z score for the desired level of confidence --> for 95% confidence, z = 1.96
 
#5
Okay I think I got it. It should look something like this

Population proportion(p) = .97
x = # of successes = 473
^p= x/n= 473/488
^q= 1- ^q= 1-.97 = .03
σ = √ ((^p*^q)/n)) = √ ((.97 *.03)/ 488) = .01
zα/2 = 1.96

Margin of Error = zα/2 * √ ((^p*^q)/n)) = .01 * 1.96 = .0196

Confidence Interval for population p = ^p – E < ^p < ^p + E, so it should be
-1.9404 < .97 < 1.9796

I think this is correct???
 

JohnM

TS Contributor
#6
don't round off the margin of error - it should be 0.0151

then confidence interval = .97 +/- .0151

= (.9549, .9851)
 
#7
Okay I got the around the same as yours

Confidence Interval for population p = ^p – E < ^p < ^p + E=
.9547 < .97 < .9853

but I understand it now :yup:

Thank You :)