Confidence Interval question no standard divination given to find the standard error

Question

A survey of 800 Irish households finds that 54% of households have oil-fired central heating. Infer a 95% confidence interval for the percentage of the total population of Irish households that have oil-fired central heating.

This is how I have worked on this problem using the formula to find the standard error

Notation

$N = 800$

$P = 54$

$95\% = 1.96$

Workings out

(1) Find the standard error

$$\frac{\sqrt{54(1-.54)}}{800} *100= 0.062$$

(2) Multiply the confidence interval by the standard error

$54\% \pm 1.96 * 0.062$

(3) Answer

$54\% \pm 0.015$

I would like to know if this is the correct anwser or am I even going the right way about it.

The problems contains the standard deviation I can do but I am struggling with this

Answer 1

You are almost right. The approximated limits of the confidence interval are

$\Large{ \hat p\pm z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}},$

where $\alpha$ is the significance level of $1-0.95=0.05$

$z_{1-\frac{\alpha}{2}}=z_{1-\frac{0.05}{2}}=z_{0.975}=1.96 \ \checkmark$

$\frac{\sqrt{54(1-.54)}}{800}\cdot 100$

It is not necessary to multiply it by $100$. But the 800 at the denominator has to be unter the square root as well.

$\sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}=\sqrt{\frac{ 0.54\cdot (1- 0.54)}{800}}\approx 0.0176$

Thus the limits of the confidence interval are

$0.54\pm 1.96\cdot 0.0176=0.540\pm 0.035$

Answer 2

Recall the formula for the confidence interval of a proportion $\left(\hat{p}-Z_{1-\frac{\alpha}{2}}\sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}},\hat{p}+Z_{1-\frac{\alpha}{2}}\sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}\right)$, as you can see $\hat{p}=0.54$, $n=800$ and $Z=1.96$ for a confidence of $95\%$. replacing on the formula we have the confidence interval for this problem: $\left(0.54-1.96\sqrt{\frac{0.54.(1-0.54)}{800}},0.54+1.96\sqrt{\frac{0.54.(1-0.54)}{800}}\right)=(0.505,0.574)$.

So we can conclude that we a confidence of $95\%$, $p\in(0.505,0.574)=(50.5\%,57.4\%)$