Confidence Interval for population mean

• Jun 5th 2010, 01:28 AM
Musab
Confidence Interval for population mean
Confidence Interval for population mean ( Small Sample)

Question Statement:
The distance traveled, in hundred of kilometers by automobile, was determined for 20 individuals returning from vacations. The results are:
12.5, ............, (all the 20 numbers are given).

I am asked to calculate the 90% confidence interval for the population mean.

The instructor calculated the mean and the standard deviation, using the 20 numbers given, and he said that these are the population mean and population standard deviation. Then he calculated the sample mean and sample SD from those values. Is he correct ?

I think the mean and the standard deviation of those 20 number are already the sample mean and sample SD.

This is the only issue I have, I know how to work out the rest of the problem.
• Jun 5th 2010, 02:35 AM
SpringFan25
[quote]
The instructor calculated the mean and the standard deviation, using the 20 numbers given, and he said that these are the population mean and population standard deviation. Then he calculated the sample mean and sample SD from those values. Is he correct ?
[/tex]

as long as you meant to say, he used the sampled mean and sd as estimates of the population mean and sd, then yes - thats a normal thing to do.

Quote:

I think the mean and the standard deviation of those 20 number are already the sample mean and sample SD.
The thing you have to realise is that when you take a random sample, you dont know in advance which data will be in the sample. As a result mean of the sample is also a random variable. The question asks you to find:

"the sd of the sample mean"
not
"the sd of the data in one particular sample"

Once you have an estimate of the population standard deviation, you can estimate the sd of your sample means by writing them as the sum of some random variables $x_1 ...x_n$

$\bar{X} = \frac{x_1 + x_2 + x_3 ....}{n}$
$Var(\bar{X}) = Var\left(\frac{x_1 + x_2 + x_3 ....}{n}\right)$
$Var(\bar{X}) = \frac{Var\left(x_1 + x_2 + x_3 ....\right)}{n^2}$

$Var(\bar{X}) = \left(\frac{Var(x_1) + Var(x_2) + Var(x_3) ....}{n^2}\right)$ (provided the sample values are independant)

$Var(\bar{X}) = \left(\frac {nVar(X)}{n^2}\right)$ (because all the x values come from the same distribution)
$Var(\bar{X}) = \left(\frac {Var(X)}{n}\right)$
$SD(\bar{X}) = \left(\frac {SD(X)}{\sqrt{n}}\right)$
• Jun 5th 2010, 02:52 AM
Musab
Thanks SpringFan25,

So I have to calculate the sd using the numbers given and then divide the answer by $\sqrt{n}$ to get the sample sd, correct ?
• Jun 5th 2010, 03:02 AM
SpringFan25
its easy to tie yourself in knots with the terminology here.

to find the standard deviation of the sample mean, take the population s.d and divide by $\sqrt{n}$

You can use the s.d of the sample to estimae the population s.d