Thanks.

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- Nov 23rd 2006, 12:13 PMchancerHelp please
Thanks.

- Nov 23rd 2006, 12:46 PMgalactus
This stuff shouldn't be a struggle even for a mathophobe.

Assuming normal distribution.

The formula $\displaystyle E=z\frac{\sigma}{\sqrt{n}}$ gives you the margin of error.

The z score that corresponds to a 95% CI is 1.96.

n=144, the number in your sample.

$\displaystyle \overline{x}$=375, the sample mean.

$\displaystyle {\sigma}$=20, the standard deviation.

Plug them into your formula:

$\displaystyle 1.96\frac{20}{12}=3.2\overline{66}$

$\displaystyle 375-3.27; 375+3.27$

$\displaystyle 371.73<{\mu}<378.27$

You can say with 95% confidence that the mean price of all TV's(from the store population) will be between $371.73 and $378.26.

Try it with the 99% CI. See what happens to the interval. Does it widen?. Narrow?.

The z-score that corresponds to 99% is 2.575. Do you have a book with a table to look these up in?. - Nov 23rd 2006, 12:50 PMchancer
I think you underestimate your abilities.

THANK YOU FOR YOUR HELP!!!! - Nov 23rd 2006, 03:59 PMtopsquark
Chancer, you didn't happen to delete the problem you posted, did you?

We'd prefer it if you left the problem there. Someone else might also be able to benefit from it. :)

-Dan - Nov 23rd 2006, 04:13 PMchancer
I apologize.