Question:
In a certain bio-engineering experiment, a successful outcome was achieved 60 times out of 125 attempts.
Construct a 95% confidence interval for the probability, p, of success in a single trial.
My Attempt
We know that the confidence interval is:
$$(p-ks,p+ks)$$ Where s is the standard error.
I found the k value using Matlab code:
k=norminv(0.975)=1.9600
Also:
$$p=60/125=0.48$$
Standard Error:
$$s=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.48\cdot0.52}{125}}=0.04469$$
We get the Confidence Interval:
$$(0.392,0.568)$$
But my answer is wrong. Is there anything I'm missing out here?
In the next part:
The researchers expected a successful outcome 70% of the time. Is the data consistent with this hypothesis?
Obviously, 0.7 doesn't lie in the Confidence Interval which I calculated (which is wrong). So, how do we approach such a problem?
If $X = 60$ successes in $n = 125$ binomial trials, maybe your text is using the confidence interval that 'appends 2 successes and 2 failures' to the data before computing the CI. Such CIs have been shown to have more accurate coverage probabilities than the traditional ones. Then $p^+ = (60 + 2)/(125 + 4) = .4806$ and $n^+ = 129$ so that the CI is $( 0.394, 0.567).$
If that is not the 'correct answer' you're looking for, please do tell us what it is.