# the difference with T-distribution and normal

• Aug 5th 2007, 05:30 PM
mkcar
the difference with T-distribution and normal
Hi, I'm confused, I don't know if I got the t- distribution right, and I can't fiqure out what value to use for the normal distribution table. Could someone please help me.

Construct the indicated confidence interval for the population mean. Using (a) a t-distribution (b) if you had incorrectly used a normal distribution which interval would be wider?

C = 0.95, x ( with the line on top) = 13.4, s= 0.85, n = 8

This is what I got for the t-distribution interval
3.13.4 + 2.364, is this even close?
• Aug 5th 2007, 08:30 PM
CaptainBlack
Quote:

Originally Posted by mkcar
Hi, I'm confused, I don't know if I got the t- distribution right, and I can't fiqure out what value to use for the normal distribution table. Could someone please help me.

Construct the indicated confidence interval for the population mean. Using (a) a t-distribution (b) if you had incorrectly used a normal distribution which interval would be wider?

C = 0.95, x ( with the line on top) = 13.4, s= 0.85, n = 8

This is what I got for the t-distribution interval
3.13.4 + 2.364, is this even close?

I don't understand what you mean by your t-distribution interval, it should be like what follows:

Such a confidence interval looks like:

$
[\bar{x}-k~s/\sqrt{n},~ \bar{x}+k~s/\sqrt{n}]
$

where in one case $k$ is determined from the table of the t-distribution with $n-1$ degrees of freedom, and in the other case from the normal table. These values are $k\approx 2.36$ for the t-distribution and $k=1.96$ for the normal distribution.

The t-distribution confidence interval is always wider.

RonL
• Aug 6th 2007, 02:01 PM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
I don't understand what you mean by your t-distribution interval, it should be like what follows:

Such a confidence interval looks like:

$
[\bar{x}-k~s/\sqrt{n},~ \bar{x}+k~s/\sqrt{n}]
$

where in one case $k$ is determined from the table of the t-distribution with $n-1$ degrees of freedom, and in the other case from the normal table. These values are $k\approx 2.36$ for the t-distribution and $k=1.96$ for the normal distribution.

The t-distribution confidence interval is always wider.

RonL

Putting in the numerical valuse we have:

$
\left[13.4 -\frac{2.36 \times 0.85}{\sqrt{8}}, 13.4 +\frac{2.36 \times 0.85}{\sqrt{8}}\right] = [12.69, 14.11]
$

RonL