Inferential Statistics problem about a poll and samples

Question

The wording of the problem is a bit odd, but I'll try to keep true to the translation:

A poll about shopping malls seems
to bring implications to the state of small shops in the downtown area.
1008 adults were included in the poll.
A result says that more than 8 out of 10 persons would not to move near a
 shopping mall in the downtown area.
Also, 56% of the people stated that the shopping malls in the surroundings
 (suburban) offer a wider variety than those downtown.
Find a 90% confidence interval for the probability that an adult prefers
 suburban malls.

First of all, this seems to be a random sample of a Bernoulli distribution, whether a person prefers suburban malls or not. We are asked to find the parameter $p$ of said distribution along with its 90% confidence interval.

I thought that by the Law of Large Numbers, the $\mu$ of the people who prefer suburban malls is $.56n$. However I can't get around either $\sigma$ or $s$. I feel that the only thing close to giving the $s$ is the more than 8 out of 10 statement but can't get it down to a specific $s$.

Assuming that we get the $s$ correctly, then I would use a $t\ \text{distribution}$ and then simply find the confidence interval for $\overline{X}$ with the form $P(t_{.05} < t < t_{.95}) = 0.9$. Yet, I don't know how many degrees of freedom I would need to use, other than $1007$. So I would approximate this as a normal distribution $N(0,1)$, so it would become $P(Z_{.05} < z < Z_{.95}) = 0.9$.

So my question for this reasoning is whether it is correct, and if so, then how how can I get the information for the $s$, if necessary.

Thanks!

Answer 1

Instead of going overcomplicated, we have a Bernoulli distribution with estimator for $\hat{p}$, which is just $\overline{X}$. Similarly, the variance may be estimated with $\hat{p}\ (1-\hat{p})$. We know $\overline{X}=0.57$. Since our population is very large, using a normal distribution is suitable as an approximation to the distribution of $\overline{X}$. Therefore

$$ P(z_{(.5)} < Z < z_{(.95)}) = 0.9,\ Z \sim N(0,1) \\ P\left( z_{(.5)} < \frac{\overline{X}-p}{\frac{\overline{X}\ (1-\overline{X})}{\sqrt{n}}} < z_{(.95)}\right) = 0.9 \\ P\left( \overline{X} - z_{(.5)}\frac{\overline{X}\ (1-\overline{X})}{\sqrt{n}} < p < \overline{X} + z_{(.5)}\frac{\overline{X}\ (1-\overline{X})}{\sqrt{n}}\right) = 0.9 $$

Finally giving the interval $(0.534463, 0.585537)$.