Assign9 - . 2 ( 1 . 10 365 . 2 25 . 2 1 . 10 7 , 025 ....

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STA 3032 – Assignment 9 Problem 7.82: 63 347 . 62 10 48 645 . 1 10 48 2 2 05 . 0 = = = = z n Problem 7.84: 3 . 157745 7 . 135638 15 14380 ) 977 . 2 ( 146692 15 14380 ) 977 . 2 ( 146692 977 . 2 14 , 005 . 0 < < + < < - = μ t Problem 7.88: heats. two in the different not is content copper mean The . reject not Do . 5 043 . 2 7 1 3 1 003139 . 0 251 . 0 33 . 0 003139 . 2 7 3 ) 057 . 0 )( 1 7 ( ) 053 . 0 )( 1 3 ( 057 . 0 251 . 0 053 . 0 33 . 0 . 4 306 . 2 if Reject . 3 0 : . 2 0 : . 1 0 2 2 2 2 2 1 1 8 , 025 . 0 2 7 3 , 025 . 0 0 2 1 1 2 1 0 H t s s x s x t t t H H H calc p calc + - = = - + - + - = = = = = = = - = - - + Problem 7.90: 7.5. mean that hypothesis reject to evidence enough Not . reject not Do . 5 49 . 1 20 / 2 . 1 5 . 7 9 . 7 . 4 729 . 1 if Reject . 3 5 . 7 : . 2 5 . 7 : . 1 0 19 , 05 . 0 0 1 0 - = = H t t t H H H calc calc
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Problem 7.94: 98 . 11 22 . 8 8 25 . 2 ) 365 . 2 ( 1 . 10 8 25 . 2 ) 365
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Unformatted text preview: . 2 ( 1 . 10 365 . 2 25 . 2 1 . 10 7 , 025 . &lt; &lt; + &lt; &lt;-= = = t s x Problem 7.100: The dot plot of the data shows a large gap in the middle of the data, precisely where one would expect most of the data observations if a normal or t distribution were appropriate. In this case, it is not prudent to construct a CI based on the t distribution....
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This note was uploaded on 08/22/2010 for the course STA 3032 taught by Professor Sapkota during the Spring '08 term at University of Central Florida.

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Assign9 - . 2 ( 1 . 10 365 . 2 25 . 2 1 . 10 7 , 025 ....

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