Confidence Interval

Violagirl

Junior Member
Joined
Mar 9, 2008
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87
A home shopping channel wants to know how much the average customer is spending on "Diammonique" jewlery this year. A random sample of 21 customers shopping during the diamonique show in 2010 found that the average amount spent was $84.24, and the standard deviation was 33.14.

A) Find the 98% confidence interval for the average amount spent on Diamonique jewlery in 2010.

I do know the correct formula in figuring out, however I am uncertain in determining the z-score for the confidence interval of 98%. From my notes, my prof told us that a 95% CI would have a z-score of 1.96 and a 99% CI would have a z-score of 2.575. How would one determine this for a CI of 98%?

B) How many people would they need to sample this year to get a 98% confidence interval where the margin of error is +/- $10.00.

I'm not sure what the first step would be in determining this problem. Any help would be greatly appreciated!
 
What? You can't look it up in a table?

I show:
0.95 ==> 1.960
0.96 ==> 2.054
0.97 ==> 2.170
0.98 ==> 2.326
0.99 ==> 2.576

Without a finite population, always \(\displaystyle n = \frac{z^{2}\cdot s^{2}}{d^{2}}\). Round UP.
 
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