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Category > Math Posted 23 Aug 2017 My Price 7.00

95% confidence level, 0(=0.05 and degrees of freedom = (n-1) =199, t= 1.972 Fherefore,

?or the 95% confidence level, 0(=0.05 and degrees of freedom = (n-1) =199, t= 1.972 Fherefore, p = 144.095 i 1.972(149.566/\/200) =144.095 i 20.856 Fhe confidence interval is [123.239, 164.951]. (Schi & Tao, 2008). 
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how do I get a 99% confidence level? What formula is used? Please show how its done. 
This question was created from QNT-561--Signature Assignment Wk 6 https://www.coursehero.com/file/22251289/QNT-561-Signature-Assignment-Wk-6/?focusQaId=10757643

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Martinakom
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Status NEW Posted 23 Aug 2017 03:08 PM My Price 7.00

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