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# sol1 - Stats 201A HW1 Solution 1(a H0 1 = 2 H1 1 = 2 t...

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Stats 201A HW1 Solution 1. (a) H 0 : μ 1 = μ 2 , H 1 : μ 1 = μ 2 . t = ( . 00025) / ( . 002005674 / 12) = . 4318 with df = 11. p-value=.6742. So accept the null at 5% level. The di ff erence between the population means of measurements is not significant. (b) The Analysis of Variance Table is Df Sum Sq Mean Sq F value Pr(>F) inspector 11 2.8125e-05 2.5568e-06 1.2712 0.3488 caliper 1 3.7500e-07 3.7500e-07 0.1864 0.6742 Residuals 11 2.2125e-05 2.0114e-06 For caliper, the F-value is .1864 and p-value is .6742, so accept the null at 5% level. Again, the di ff erence between the population means of measurements is not significant. (c) A 95% confidence interval of the mean di ff erence is . 00025 ± (2 . 201)(0 . 002006 / 12) = ( - 0 . 001024 , 0 . 001524) 2. (a) The Analysis of Variance Table is Df Sum Sq Mean Sq F value Pr(>F) Catalyst 3 85.676 28.559 9.9157 0.001436 ** Residuals 12 34.562 2.880 The p-value=.0014, so we reject the null at 1% level. There is strong evidence that the
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