enter image description here

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!!


1 Answer 1


(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 $1.96\cdot\dfrac{\sigma}{\sqrt{n^\prime}} = \dfrac{1.96}{2}\cdot\dfrac{\sigma}{\sqrt{n}}$.

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .