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OpenStudy (day_2010):

Quick question on how to find a Z probability when given a confidence of 90%. I keep getting something different than my professor.

OpenStudy (day_2010):

OpenStudy (day_2010):

OpenStudy (day_2010):

My professor gets z=1.645

OpenStudy (day_2010):

I get z=.1736

OpenStudy (day_2010):

Any Ideas? I'm just stuck really on this one problem.

OpenStudy (agent0smith):

90% confidence means 10% in the tails, 5% in the right tail and 5% in the left tail. So look for 0.95 (in the central section of numbers) and find the z score that matches it. You're reading the table wrong. Don't look for 0.05 in the top row of numbers. You're looking for 0.95 in the central numbers, and then you see which z score matches. It's much easier to show this in person...

OpenStudy (agent0smith):

http://cosstatistics.pbworks.com/f/1281154582/6368.png See here, a probability of 0.6368 corresponds to a z score of z= 0.35. So find 0.95 in the middle area, then get the z score for that.

OpenStudy (agent0smith):

What you did is find a probability of 0.1736, which means a z score of z=0.45

OpenStudy (day_2010):

Z=.8289?

OpenStudy (agent0smith):

No...

OpenStudy (day_2010):

Yes, my professor said to take half of the confidence 90% and that gives you Z=.45

OpenStudy (day_2010):

Then I looked that up and got.1736

OpenStudy (agent0smith):

No... you take half of the TAILS. Which is 10%. You don't take half of the 90%. And with 95% confidence, the tail is 5%, so half that is 2.5%, or 0.025 on the z chart

OpenStudy (day_2010):

So how did he come up with Z=1.645?

OpenStudy (day_2010):

Nevermind, I figured it out myself. Thanks for trying to help me.

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