# 6 - STAT 350 Fall 2008 Homework#6 Solution through...

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STAT 350 – Fall 2008 Homework #6 – Solution through Lecture #10 Text Exercises Chapter 7, Problems #38, 46*, 48*, 52 * do not assume equal variance! 7.38 () 0.05/ 2,8 1 3.1 30.2 2.365 27.608,32.792 8 s xt n ±⇒ ± 7.46 22 11 2 2 12 0.66 20 0.741194487 0.66 0.39 20 20 sn c ss nn == = + + ( ) ( ) 2 2 21 2 2 df 1 20 1 20 1 30.83 20 1 0.741194487 20 1 1 0.741194487 nc n c −− = −+ df is rounded down to 30. thus t * = 2.042 * / 2 0.66 0.39 8.74 4.96 2.042 3.43,4.13 20 20 xxt α −± + ⇒ − ± + Homework #6 – Solution Page 1 of 4

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7.48 a. high middle 470 460 450 440 430 420 Data Boxplot of middle, high b. Total Variable Count Mean StDev middle 17 438.29 15.14 high 11 437.45 6.83 22 11 2 2 12 15.14 17 0.760734951 15.14 6.83 17 11 sn c ss nn == = + + () ( ) ( ) 2 2 21 2 2 df 1 17 1 11 1 23.869 11 1 0.760734951 17 1 1 0.760734951 nc n c −− = −+ = = df is rounded down to 23 thus t * = 2.069 * / 2 15.14 6.83 438.29 437.45 2.069 17 11 7.87,9.55 xxt α −± + ⇒ ± + ⇒− The fact that the confidence interval for the difference in the means includes zero suggests that we cannot say with confidence that there is a true difference in the means of these two populations.
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## This note was uploaded on 02/23/2009 for the course STAT 350 taught by Professor Staff during the Fall '08 term at Purdue.

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6 - STAT 350 Fall 2008 Homework#6 Solution through...

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