Question

what is the 90% confidence interval for a population of 482, means of 164 standard deviation of 33

Answer 1

Task. What is the 90% confidence interval for a population of 482, means of 164 standard deviation of 33.

Solution. Thus we measure some numerical characteristic $X$ of these peoples and this characteristic has mean $\mu=164$ and standard deviation $\sigma=33$. We take a sample of $n=482$ people. Then the sample mean has normal distribution $N(\mu,\sigma/\sqrt{n})$.

Therefore the 90% interval has the form

$(\mu-\delta,\ \mu+\delta),$

where $\delta$ is choosen so that

$P(|X-\mu|<\delta)=0.9.$

Consider another random variable

$Z=\frac{X-\mu}{\sigma/\sqrt{n}}.$

Then $Z$ has standard normal distribution $N(0,1)$ and the values of its distribution function

$F(t)=P(Z\leq t)$

can usually be found in any book in Probability theory.

Notice that

$X-\mu=Z\sigma/\sqrt{n},$

and so

$P(|X-\mu|<\delta)=P(|Z\sigma/\sqrt{n}|\leq\delta)=P(|Z|\leq\delta\sqrt{n}/\sigma)=0.9$

Denote

$a=-\delta\sqrt{n}/\sigma.$

Since the normal distribution is symmetrical,

$P(|Z|<a)=1-2F(a),$

whence

$P(|X-\mu|<\delta)=1-2F(a)=0.9,$

whence

$F(a)=(1-0.9)/2=0.05.$

Then from tables of values of $F$ we obtain that

$a=1.645.$

Thus

$a=\delta\sqrt{n}/\sigma=1.645,$

whence

$\delta=1.645\cdot\frac{\sigma}{\sqrt{n}}=\frac{1.645*33}{\sqrt{482}}=2.4726.$

Therefore 90% confidence interval is the following:

$(164-2.4726,164+2.4726)$

$(161.5274,166.4726).$