What am I doing wrong in finding the confidence interval? 
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?
 A: Here are my results for calculating the probability of success, p value and confidence intervals. Besides the slight rounding differences, it looks likes 70% was the p value, which is the percent by which the results can be explained by the null hypothesis.
data:  60 and 125
number of successes = 60, number of trials = 125, p-value = 0.7207
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
0.3898361 0.5711333
sample estimates:
probability of success 
              0.48 

A: From what I can see, there are no errors in your method. It is possible that you made rounding error in your calculations. Doing this in R
> prop.test(60, 125, correct = FALSE)

    1-sample proportions test without continuity correction

data:  60 out of 125, null probability 0.5
X-squared = 0.2, df = 1, p-value = 0.6547
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
 0.3943277 0.5668649
sample estimates:
   p 
0.48 

Thus, the confidence interval is $(.394, .567)$. Since $.70$ is not in this interval, at $\alpha = .05$, you can conclude that you will reject the null hypothesis that $H_0: p = .70$ against the alternative that $H_0: p\ne .70$.
