# Thread: the difference with T-distribution and normal

1. ## 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?
Thank you in advance.

2. 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?
Thank you in advance.
I don't understand what you mean by your t-distribution interval, it should be like what follows:

Such a confidence interval looks like:

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

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

The t-distribution confidence interval is always wider.

RonL

3. 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:

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

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

The t-distribution confidence interval is always wider.

RonL
Putting in the numerical valuse we have:

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

RonL

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