Help please

chancer · November 23rd 2006, 12:13 PM

Thanks.

**galactus** · November 23rd 2006, 12:46 PM

This stuff shouldn't be a struggle even for a mathophobe.

Assuming normal distribution.

The formula $E=z\frac{\sigma}{\sqrt{n}}$ 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.

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

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

Plug them into your formula:

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

$375-3.27; 375+3.27$

$371.73<{\mu}<378.27$

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?.

chancer · November 23rd 2006, 12:50 PM

I think you underestimate your abilities.

THANK YOU FOR YOUR HELP!!!!

**topsquark** · November 23rd 2006, 03:59 PM

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

chancer · November 23rd 2006, 04:13 PM

I apologize.

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