(a) We need to calculate the standard deviation for the sample, size n=50. This is a binary case, because we are only considering brown eyes or not-brown eyes. The proportion p of brown eyes in the sample=15/50=0.3. The mean is np=15.
The standard deviation of the sample is taken to be √(p(1-p)/n)=√(0.3×0.7/50)=0.065 approx.
A confidence interval of 95% in a normal distribution is 95% the area beneath the symmetrical bell-shaped curve. The total area under the two tails of the distribution is 5% so each tail covers 2.5%, and this is the significance level we use. If we use the t distribution table we need to look up the required Z value (this is the number of standard deviations from the mean, to the left and right) according to the degrees of freedom, taken to be n-1, that is, 49. From the table Z=2.01.
This is the critical number of standard deviations from the mean proportion, which gives us 2.01×0.065=0.13 approx. The confidence interval is therefore 0.3±0.13=[0.17,0.43].
(b) 22%=0.22 which lies within the 95% confidence interval, so the claim can't be rejected, and by default the statistics seem to support the claim of 22% of the population having brown eyes.
(c) We need to find n which gives us a CI of [0.26,0.28]. Take 1.96 to be the critical Z value for large n. Therefore 1.96s=0.01, where s=√(p(1-p)/n)=0.01/1.96=0.0051 where p=0.27. 0.1971/n=0.000026, n=7581 approx.