Finding the size of sample required.

Question

Suppose that light bulbs made by a standard process have an average of $2000$ hours with a standard deviation of $250$ hours , and suppose that it is considered worthwhile to replace the process if the mean life can be increased by at least 10%.

An engineer wishes to test a proposed new process and he is willing to assume that the standard deviation of the distribution of lives is about the same as the standard process. How large a sample should he examine if he wishes the probability to be about 0.01 that he will fail to adopt the new process if in fact it produces bulbs with a mean life of $2250$ hours.

My shot at the problem :

Let $\bar{X}$ denotes the sample mean of the lives of the bulbs produced in the new process. Now for failure :

$\bar{X}< 2000 + (0.1)2000$ => $\bar{X}-2000 < (0.1)2000$

So , $P(\bar{X}-2000 < (0.1)2000) = 0.01$

=> $P((\bar{X}-2000)^2 < (200)^2) = 0.01$

So using chebyshev's inequality we have :

$1 - \dfrac{E[(\bar{X}-2000)^2]}{200^2} = 0.01$

( $E(\bar{X}) = 2250$ and $Var(\bar{X}) = \dfrac{250^2}{n}$ )

Is the interpretation okay ? ( It doesn't seem so according to me , as $n$ comes out to be negative)

Can anyone help ?

Answer 1

You should be employing the CLT, and you got a little mixed up with the first step I think.

Since we need a $10$ percent increase in mean, we'd require a new sample mean $\geq 2200$. Then taking our true mean of $2250$, we have that our rejection criteria is $\bar{X} - \mu < -50$.

So all we want is

$$ P\left(\bar{X} - \mu < -50\right) = 0.01. $$ Now divide the left hand side by $\frac{\sigma}{\sqrt{n}}$, and by the CLT we have that $$\frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \sim N(0,1).$$

You should be able to find $n$ from there, recalling that $\sigma$ is given.