1. ## Probability Question

An industrial engineer intends to use the mean of a random sample of size to estimate the average mechanical aptitude of workers. On basis of experience , the engineer assumes for such data , what can he assert with probability about maximum size of his error?

I dont have any idea mite what to do with this question .....Hope some body could help me ......

An industrial engineer intends to use the mean of a random sample of size to estimate the average mechanical aptitude of workers. On basis of experience , the engineer assumes for such data , what can he assert with probability about maximum size of his error?

I dont have any idea mite what to do with this question .....Hope some body could help me ......
An assertion with 99% probable error corresponding to a statistic, $\displaystyle S$, is $\displaystyle S\pm 2.58 \sigma$.
The probable error is $\displaystyle 2.58\sigma$=$\displaystyle (2.58)(6.2) \approx 16$.

If $\displaystyle S$ is standard deviation, then the error is huge.

If $\displaystyle S$ is the arithmetic mean, then it' not unusual.

At any rate $\displaystyle 2.58\sigma$ is what you are looking for.

3. Hi,

max_error=1.30=2.58*6.2/(150^1/2)

Sincerely
Hametceq

4. Originally Posted by hametceq
Hi,

max_error=.000709=2.58*6.2/(150^2)

Sincerely
Hametceq
For the 99% confindence limit? Sounds like a suicide.

5. It depends how do u read a question:
see:
"An industrial engineer intends to use the mean of a random sample of size to estimate the average mechanical aptitude of workers. On basis of experience , the engineer assumes for such data , what can he assert with probability about maximum size of his error?"

In my undrstanding, maximum size of error in this question: is a maximum distance of the individual worker aptitude (mu) from the estimated average aptitude or it is
absalute value of (mu-Xbar), so we need to find this distance so that no more than 1% of chances of P(|mu-Xabr|>max_error)=0.01

Please, if u disagree with me, I will understand. I am just sharing my views and not asserting on my right. There is no need to use insulting comments. For future, I will ignore any messages that contains abuse or insult.

Merry Christmass

6. Originally Posted by hametceq
It depends how do u read a question:
see:
"An industrial engineer intends to use the mean of a random sample of size to estimate the average mechanical aptitude of workers. On basis of experience , the engineer assumes for such data , what can he assert with probability about maximum size of his error?"

In my undrstanding, maximum size of error in this question: is a maximum distance of the individual worker aptitude from the estimated average aptitude or it is
absalute value of (X-Xbar), so we need to find this distance so that no more than 1% of chances of P(|X-Xabr|>max_error)=0.01

Please, if u disagree with me, I will understand. I am just sharing my views and not asserting on my right. There is no need to use insulting comments. For future, I will ignore any messages that contains abuse or insult.

Merry Christmass
My friend, please do not be upset. My comment was not directed personally against you. My point was that if the engineer asserts with 99% confidence of his error, he ought to make larger room for errors to be that confindent. If he were to offer errors as small is 1%, he ought to assert with 1% confidence; otherwise he would commit suicide.

Friend, please do not misunderstand. I do see you as fellow mathematician, do accept my sincere apology for not expounding my thought to avoid misunderstanding. After all mathematicians need to exchange ideas. Keep talking to me, I will keep listening. No hard feelings.

Have a merry Christmas and a happy new year.

7. I know what do you mean and I see how do u understand by saying error.
How u see is: that random worker aptitude deviates from the true aptitude of workers by no more than 16 with 99% conf. (which is true)
Just how I see is: that estimated average aptitude of workers deviates from the true aptitude of workers by no more than 1.30=16/sqrt(150) with 99% conf. (which is true also). It becomes small become of the sample size is big (obviouslly and I think we should use the information about the sample size).

By the way I made a mistake instead typing 2.58*6.2/(150^1/2) , I typed 2.58*6.2/(150^2). I will try to fix it now. No hard feelings and thank you very much for you clarification.

Regards
hametceq

8. Just how I see is: that estimated average aptitude of workers deviates from the true aptitude of workers by no more than 1.30=16/sqrt(150) with 99% conf. (which is true also).
Yes, that's correct.

Now, $\displaystyle S\pm 2.58\sigma$ is not $\displaystyle \overline X\pm 2.58\sigma$ but $\displaystyle \mu_X\pm2.58\sigma_{\overline X}$

I appreciate the discussion. With our hand shakes, we can say, "Peace on earth. Good will towards all men." Of course, that's including all women.

Merry Christmas to you again.

9. ## How did u get this?

Originally Posted by novice
An assertion with 99% probable error corresponding to a statistic, $\displaystyle S$, is $\displaystyle S\pm 2.58 \sigma$.
The probable error is $\displaystyle 2.58\sigma$=$\displaystyle (2.58)(6.2) \approx 16$.

If $\displaystyle S$ is standard deviation, then the error is huge.

If $\displaystyle S$ is the arithmetic mean, then it' not unusual.

At any rate $\displaystyle 2.58\sigma$ is what you are looking for.

Could u please tell me how to got 2.58