Homework4_solution

Homework4_solution - Homework #4 Solution 8.1 (a) Matched...

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Unformatted text preview: Homework #4 Solution 8.1 (a) Matched pairs (b) Independent samples (c) Matched pairs (d) Independent samples 8.2 (a) Experimental (b) Observational (c) Experimental (d) Experimental 8.3 (a) Matched pairs (b) Matched pairs (c) Independent samples (d) Independent samples 8.4 (a) Observational (b) Observational (c) Observational (d) Experimental 8.5 (a) The clouds in each group are not matched to one another on some characteristic. (b) Seeded cloud rainfalls tend to be larger than unseeded cloud rainfalls. 8.7 (a) The two different types of eyes are or belong to the same person. (b) The pairs tend to lie pretty close to the 45 degree line through the origin. Eyes with glaucoma do not appear to have thicker corneas than unaffected eyes. 8.9 (a) The pooled SD is 803 . 1 2 21 21 . 2 ) 1 21 ( 5 . 4 ) 1 21 ( 2 ) 1 ( ) 1 ( 2 1 2 2 2 2 1 1 =- +- +- =- +- +- = n n s n s n s Then a 95% CI for 2 1 - is given by ] 625 . 2 , 376 . [ 21 2 803 . 1 021 . 2 5 . 6 8 1 1 2 1 2 , 2 2 1 = - = + -- + n n s t y x n n Since the CI does not contain 0, we reject H and conclude that there is a significant difference between the two filters. (b) Since 463 . 21 5 . 4 1 2 1 1 = = = n s and 309 . 21 . 2 2 2 2 2 = = = n s The degrees of freedom are ) 1 ( ) 1 ( ) ( 2 2 2 1 2 1 2 2 1- +- + = n n v ) 1 21 ( ) 309 . ( ) 1 21 ( ) 463 . ( ) 309 . 463 . ( 2 2 2- +- + = 35 845 . 34 = Then a 95% CI for 2 1 - is given by : ] 608 . 2 , 392 . [ 21 . 2 21 5 . 4 032 . 2 5 . 6 8 2 2 2 1 2 1 2 , = + - = + - n s n s t y x v The results are similar, indicating that the assumption of equal variances is reasonable. 8.11 (a) Using 0026 . 12 = x , 0143 . 12 = y , 0079 . 1 = s , 0132 . 2 = s , 10 1 = n , and 5 2 = n , the pooled SD is 0099 . 2 5 10 ) 0132 . ( ) 1 5 ( ) 0079 . ( ) 1 10 ( 2 ) 1 ( ) 1 ( 2 2 2 1 2 2 2 2 1 1 =- + - + - =- +- +- = n n s n s n s Then the test statistic is 1616 . 2 5 1 10 1 0099 . 0143 . 12 0026 . 12 1 1 2 1- = +- = +- = n n s y x t Since 16 . 2 2 , 2 5 10 =- + t t , we reject H and conclude that there are significant differences between the methods. (b) Since 2 2 1 2 1 1 ) 0025 . ( 10 ) 0079 . ( = = = n s and 2 2 2 2 2 2 ) 0059 . ( 5 ) 0132 . ( = = = n s The degrees of freedom are ) 1 ( ) 1 ( ) ( 2 2 2 1 2 1 2 2 1- +- + = n n v ) 1 5 ( ) 0059 . ( ) 1 10 ( ) 0025 . ( ) ) 0059 . ( ) 0025 . (( 4 4 2 2 2- +- + = 5 481 . 5 = The test statistic is 817 . 1 5 ) 0132 . ( 10 ) 0079 . ( 0143 . 12 0026 . 12...
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Homework4_solution - Homework #4 Solution 8.1 (a) Matched...

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