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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?

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    $\begingroup$ It might help if you provided the "right" answer. $\endgroup$ – Em. Mar 28 '16 at 9:21
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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.

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