# Sampling: the % of homeowners who own at least 2 TVs

It is planned 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: 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 $$E=0.01, \quad {Z}_{\tfrac{\alpha }{2}}=1.98\quad P = 0.5 \quad n=?$$ Then $$\operatorname{A confidence interval}=1.98\times \sqrt{\frac{0.5\left(1-0.5\right)}{n}}=0.01$$

$$0.01=1.98\sqrt{\dfrac{{{0.5}^{2}}}{n}}$$ $$n=9801 Thus the sample size should be equal n=9801 Am I right ? Any help would be much appreciated • what type of error are we talking about? it looks like your \alpha=.05 hence you want to calculate type II error. but in general you can't get a constant value for type II error, instead you get a type II error function which depends on distance from \bar x given in null hypothesis. what you calculated is limits of confidence interval around \hat x and you already used n=9801 as the sample size. so these calculations and the given question are far from making sense. Oct 3, 2014 at 8:51 • @Macond what should i do ? – Educ Oct 3, 2014 at 8:54 • you should be more clear in what is meant by "estimation error" Oct 3, 2014 at 8:55 • @Macond please see my update – Educ Oct 3, 2014 at 9:04 • I made an answer out of my last comment, I think it solves what you are asking for. Oct 3, 2014 at 9:11 ## 1 Answer I guess you are asked to find the necessary sample size, such that range between upper and lower limits of 95% confidence interval is 0.01. If this is the case you should calculate n in the following expression:$$0.01=2\cdot 1.98\sqrt{\dfrac{{{0.5}^{2}}}{n}}$$this modification to your original calculation is because right hand side corresponds to half of the confidence interval. Hence multiplying right hand side by two, or dividing left hand side by two gets you there.$$CI=\hat X \pm E $$you want$$ 0.01=|LowerLimit-UpperLimit|=|\hat X-E-(\hat X+E)|=2\cdot E $$according to your given values Z_{\alpha /2}=1.98 and p=0.5$$E=1.98\sqrt{\dfrac{{{0.5}^{2}}}{n}}2\cdot E=2\cdot 1.98\sqrt{\dfrac{{{0.5}^{2}}}{n}}=0.01

• Do you have any reference books which have the same exercise with solutions
– Educ
Oct 3, 2014 at 9:24
• I can't recommend a particular book, but any introductory applied statistics book should include these material, as these are basics of hypothesis testing. Oct 3, 2014 at 9:27