Stats 201A HW1 Solution1.(a)H0:μ1=μ2, H1:μ1=μ2.t= (.00025)/(.002005674/√12) =.4318 withdf= 11.p-value=.6742. So accept the null at 5% level. The difference between the populationmeans of measurements is not significant.(b) The Analysis of Variance Table isDfSum SqMean Sq F value Pr(>F)inspector 11 2.8125e-05 2.5568e-061.2712 0.3488caliper1 3.7500e-07 3.7500e-070.1864 0.6742Residuals 11 2.2125e-05 2.0114e-06For caliper, the F-value is .1864 and p-value is .6742, so accept the null at 5% level.Again, the difference between the population means of measurements is not significant.(c) A 95% confidence interval of the mean difference is.00025±(2.201)(0.002006/√12) = (-0.001024,0.001524)2.(a) The Analysis of Variance Table isDf Sum Sq Mean Sq F valuePr(>F)Catalyst3 85.67628.5599.9157 0.001436 **Residuals 12 34.5622.880The p-value=.0014, so we reject the null at 1% level. There is strong evidence that the
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