ch05 - CHAPTER 5 Section 5-2 5-1. a) 1) The parameter of...

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Unformatted text preview: CHAPTER 5 Section 5-2 5-1. a) 1) The parameter of interest is the difference in fill volume, 1 2- 2) H : 1 2- = or 1 2 = 3) H 1 : 1 2- or 1 2 4) = 0.05 5) The test statistic is z x x n n 1 2 1 2 1 2 2 2 =-- + ( ) 6) Reject H if z < - z /2 = - 1.96 or z > z /2 = 1.96 7) x 1 = 16.015 x 2 = 16.005 = 0 1 = 0.02 2 = 0.025 n 1 = 10 n 2 = 10 z 2 2 16 015 16 005 0 02 10 0 025 10 0 99 =-- + = ( . . ) ( . ) ( . ) . 8) since -1.96 < 0.99 < 1.96, do not reject the null hypothesis and conclude there is no evidence that the two machine fill volumes differ at = 0.05. b) P-value = 2 1 0 99 2 1 08389 0 3222 ( ( . )) ( . ) .- =- = c) Power = 1- , where =-- + ---- + z n n z n n / / 2 1 2 1 2 2 2 2 1 2 1 2 2 2 = +-- - +- 10 ) 025 . ( 10 ) 02 . ( 04 . 96 . 1 10 ) 025 . ( 10 ) 02 . ( 04 . 96 . 1 2 2 2 2 = ( 29 ( 29 ( 29 ( 29 91 . 5 99 . 1 95 . 3 96 . 1 95 . 3 96 . 1- -- =-- -- = 0.0233 - 0 = 0.0233 Power = 1 - 0.0233 = 0.977 d) ( 29 ( 29 x x z n n x x z n n 1 2 2 1 2 1 2 2 2 1 2 1 2 2 1 2 1 2 2 2-- + - - + + / / ( 29 ( 29 16 015 16 005 196 0 02 10 0 025 10 16 015 16 005 196 0 02 10 0 025 10 2 2 1 2 2 2 . . . ( . ) ( . ) . . . ( . ) ( . )-- + - - + + - - 0 0098 0 0298 1 2 . . With 95% confidence, we believe the true difference in the mean fill volumes is between - 0.0098 and 0.0298. Since 0 is contained in this interval, we can conclude there is no significant difference between the means. e) Assume the sample sizes are to be equal, use = 0.05, = 0.01, and = 0.04 1 ( 29 ( 29 ( 29 ( 29 , 08 . 2 ) 04 . ( ) 025 . ( ) 02 . ( 33 . 2 96 . 1 z z n 2 2 2 2 2 2 2 2 1 2 2 / = + + = + + 2245 n = 11.79, use n 1 = n 2 = 12 5-2. 1) The parameter of interest is the difference in breaking strengths, 1 2- and = 10 2) H : 1 2 10- = or 1 2 = 3) H 1 : 1 2 10- or 1 2 4) = 0.05 5) The test statistic is z x x n n 1 2 1 2 1 2 2 2 =-- + ( ) 6) Reject H if z > z = 1.645 7) = 1 x 162.7 x 2 = 155.4 = 10 1 = 1.0 2 = 1.0 n 1 = 10 n 2 = 12 31 . 6 12 ) . 1 ( 10 ) . 1 ( 10 ) 4 . 155 7 . 162 ( z 2 2- = +-- = 8) Since 6.31 < 1.645 do not reject the null hypothesis and conclude there is insufficient evidence to support the use of plastic 1 at = 0.05. 5-3. a) 1) The parameter of interest is the difference in mean burning rate, 1 2- 2) H : 1 2- = or 1 2 = 3) H 1 : 1 2- or 1 2 4) = 0.05 5) The test statistic is z x x n n 1 2 1 2 1 2 2 2 =-- + ( ) 6) Reject H if z < - z /2 = - 1.96 or z > z /2 = 1.96 7) x 1 = 18.02 x 2 = 24.37 = 0 1 = 3 2 = 3 n 1 = 20 n...
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ch05 - CHAPTER 5 Section 5-2 5-1. a) 1) The parameter of...

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