Solution to Tutorial 3 - Solution to TUTORIAL 3 1. (R code)...

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Solution to TUTORIAL 3 1. (R code) (a) [3.061384, 3.341033] on average, with 95% confdence, the mean Freshman GPA is between 3.061384 and 3.341033 when their ACT test scores are 28 (b) [1.959355, 4.443063], with 95% confdence her GPA will be between 1.959355 and 4.443063 (c) Yes 2. (R code) (a) [16.67037, 19.72963], with 99% confdence, on average there are 16.67 to 19.73 broken ampuls aFter 2 times oF transFer [22.77964, 29.62036], with 99% confdence, on average there are 22.78 to 29.62 broken ampuls aFter 4 times oF transFer (b) [14.45319, 21.94681], with 99% confdence, there are 14.45 to 21.94 broken ampuls aFter 2 shipments 3. the α level used by the analyst was greater than 0.033, IF the α level had been 0.01, he shod accept H 0 4. t-test can test H 0 : β 1 = c For any constant c , but ±-test can only test H 0 : β 1 =0 5. (R code) (a) Set up the ANOVA table Response: y source DF SS MS ±-value p-value regression(x) 1 3.588 3.588 9.2402 0.002917 Residuals 118 45.818 0.388 Total 119 49.406 1
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(b) conduct an F-test for H 0 : β 1 =0w ith α =0 . 01 Since p-value is smaller than α . 01, we reject H 0 , i.e. β 1 is signifcantly diFerent ±rom 0.
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This note was uploaded on 10/04/2010 for the course STAT ST3131 taught by Professor Xiayingcun during the Fall '09 term at National University of Singapore.

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Solution to Tutorial 3 - Solution to TUTORIAL 3 1. (R code)...

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