# 95% Confidence interval, and sample size

Sorry in advance, the question is quite long. At first I had no idea how to even start this question but as I was reading through textbook, I've found that the sample size necessary for the CI to have a width $$w$$ is $$n=\left(2z_{\alpha/2}\frac{\sigma}{w}\right)^2$$

and the half width of the 95% CI is called the bound on the error or estimation and is denoted as $$1.96\frac{\sigma}{\sqrt{n}}$$

and if its normally distributed, do I have to consider CLT(central limit thm) as well?? any help would be appreciated!!

(a) $$67.3\%$$.

Letting $$x$$ be the sample height of the plant after $$8$$ weeks (i.e. the plant that belongs to the researcher),
$$u$$ be the population mean of plant sizes after $$8$$ weeks,
$$\sigma$$ be the standard deviation of an observation,
$$n$$ being the sample size: $$\dfrac{x-u}{\dfrac{\sigma}{\sqrt{n}}}$$ ~$$N(0,1)$$ asymptotically (by CLT).\

The width is therefore $$1.96\cdot\dfrac{\sigma}{\sqrt{n}}$$. Halving the width: $$0.98\cdot\dfrac{\sigma}{\sqrt{n}}$$.

Since $$P(Z>|0.98|)=0.673$$, this is the answer (refer to std normal tables); just as $$P(Z>|1.96|)=0.95$$.

(b) $$n' = 4\cdot\dfrac{n}{\sigma^2}$$. Just solve for $$1.96\cdot\dfrac{\sigma}{\sqrt{n^\prime}} = \dfrac{1.96}{2}\cdot\dfrac{\sigma}{\sqrt{n}}$$.

• shouldn't b) be $n'=4n$ Commented Mar 9, 2016 at 17:10
• Yes it should,, Commented Mar 10, 2016 at 2:33