It plans to conduct a study on the percentage of homeowners who have at least two TVs.
What should be the sample size if we want to ensure that $95\%$ of estimation error is less than $0.01$?
The original French text version:
i have some old exercise in paper of Inferential Statistics, i'm trying to rewrite it with solution in latex to stock it in my googledrive but i don't have any idea because It's been a long time since I studied. Any help would be much appreciated
as Mr Paul
we want to calculate $$\operatorname{A confidence interval}=\overline{X}\pm E\quad \textrm{with}\quad E={{Z}_{\tfrac{\alpha }{2}}}\sqrt{\dfrac{P(1-P)}{n}}$$ or $${Z}_{\tfrac{\alpha }{2}}=1.98\quad P = 0.5 \quad n=?$$ Then $$2\times 1.98\times \sqrt{\frac{0.5\left(1-0.5\right)}{n}}=0.01$$
Thus the sample size should be equal $9801$
Am I right ?
A confidence interval for the values of a random variable (like the % of homeowners in a sample who own at least 2 TV's) is of the form
$$\overline{X}\pm E$$
where $\overline{X}$ is the sample mean and E is the error. The form of E depends on the type of variable $X$ you are looking at but in your case, for a large sample of values n, it is given by
$$E={{Z}_{\tfrac{\alpha }{2}}}\sqrt{\dfrac{P(1-P)}{n}}$$
Here $P$ is an estimate of the mean $\%$ of homeowners who own at least $2$ TV's (this interval is called the "confidence interval for proportions" in most stats books). This might be done by taking a sample of homeowners but if no sample is taken the best estimate you have is $50\%$ or $P = 0.5$. The number ${Z}_{\tfrac{\alpha }{2}}$ can be looked up in a table and has the value $1.98$ for $\frac{\alpha}{2}=0.05$ or $5\%$ significance.
You now want to solve
$$0.01=1.98\sqrt{\dfrac{{{0.5}^{2}}}{n}}$$
for $n$.