Thanks.

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- Nov 23rd 2006, 12:13 PM #1chancerGuest

- Nov 23rd 2006, 12:46 PM #2
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 PM #3chancerGuest

- Nov 23rd 2006, 03:59 PM #4

- Nov 23rd 2006, 04:13 PM #5