In a random sample of 81 items taken from a large consignment some were found to be defective. If the standard error of the proportion of defective items in the sample is 1/18 find 95% confidence limits of the percentage of defective items in the consignment.
My try: I tried to calculate the value of p using the given S.E which was approximately equal to 40/81. And then at 95% confidence, the limits I obtained are 38.5% and 60.3%. Is this the correct answer?
$p$ is not "about" $40/81$ but it is exactly $0.5$
$$\sqrt{\frac{p(1-p)}{81}}=\frac{1}{18}$$
That is equivalent to
$$(p-0.5)^2=0$$
Giving exactly
$$p=0.5$$
Using the approx Confidence Interval you get
$$p=(50\pm 10.89)\%$$